Algebra - Korey And Math, Teaching Mathematics · Algebra, Math-ematical Models, and Problem...

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Korey Nishimoto

Math 103Notes

Korey NishimotoMath Department, Kapiolani Community College

January 29, 2019

ContentsMath 103

Contents

1 Algebra, Mathematical Models, and Problem Solving . . . . . . . . . . . . 61.1 Algebraic Expressions, Real Numbers, and Interval Notation . . . . . . . 6

1.1.1 Translate English Phrases Into Algebraic Expressions. . . . . . . 61.1.2 Use mathematical models. . . . . . . . . . . . . . . . . . . . . 71.1.3 Recognize the sets that make up the real numbers.. . . . . . . . 71.1.4 Use Interval Notation and the Inequality Symbols. . . . . . . . . 9

1.2 Operations with Real Numbers and Simplifying Algebraic Expressions. . 111.2.1 Operations using real numbers. . . . . . . . . . . . . . . . . . . 111.2.2 Algebraic properties. Commutative, Associative, and

Distributive Properties. Find Opposites and inverse. Absolutevalue. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.2.3 Use the Order of Operations. . . . . . . . . . . . . . . . . . . . 161.2.4 Simplify Algebraic Expressions vs Evaluate Algebraic

Expressions.. . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.3 Graphing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.3.1 Plotting Points and Graphing equations in the rectangularcoordinate system. . . . . . . . . . . . . . . . . . . . . . . . . 19

1.4 Scientific Notation, and conversion between fractions and decimals. . . . 221.4.1 Convert Between Fractions and Decimals. . . . . . . . . . . . . 221.4.2 Convert Between Scientific and Decimal Notation. . . . . . . . . 221.4.3 Perform computations with scientific notation. . . . . . . . . . . 24

1.5 More On Rectangular Coordinates . . . . . . . . . . . . . . . . . . . . 252 Functions and Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1 Introduction to Functions . . . . . . . . . . . . . . . . . . . . . . . . . 272.1.1 Find the domain and range of a relation. . . . . . . . . . . . . . 272.1.2 Determine whether a relation is a function. . . . . . . . . . . . 292.1.3 Evaluate a function. . . . . . . . . . . . . . . . . . . . . . . . 30

2.2 Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.1 Does a Graph Represent a Function . . . . . . . . . . . . . . . 32

2.3 The Algebra of Functions . . . . . . . . . . . . . . . . . . . . . . . . . 352.3.1 Arithmetic With Functions . . . . . . . . . . . . . . . . . . . . 352.3.2 Domains of Functions . . . . . . . . . . . . . . . . . . . . . . . 36

2.4 Linear Functions and Slope . . . . . . . . . . . . . . . . . . . . . . . . 392.4.1 Equation of a Line . . . . . . . . . . . . . . . . . . . . . . . . 402.4.2 Finding the Equation of a Line . . . . . . . . . . . . . . . . . . 402.4.3 Graphing a Line . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.5 The Point-Slope Form of the Equation of a Line . . . . . . . . . . . . . 462.5.1 Model Data With Linear Functions and Make Predictions. . . . 462.5.2 Parallel and Perpendicular Lines. . . . . . . . . . . . . . . . . . 47

3 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.1 Systems of Linear Equations in Two Variables . . . . . . . . . . . . . . 513.2 Problem Solving and Applications Using Systems of Equations. . . . . . 583.3 Systems of Linear Equations in Three Variables . . . . . . . . . . . . . 66

3.3.1 Cases of Systems of 3 Equations With 3 Variables . . . . . . . . 663.3.2 Solving Systems of 3 Equations With 3 Variables . . . . . . . . 67

4 Inequalities and Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . 734.1 Solving Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 73

4.1.1 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.1.2 Application of Linear Inequalities. . . . . . . . . . . . . . . . . 76

ContentsMath 103

4.2 Compound Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . 774.2.1 And Compound Inequalities . . . . . . . . . . . . . . . . . . . 774.2.2 Or Compound Inequalities . . . . . . . . . . . . . . . . . . . . 78

4.3 Equations and Inequalities Involving Absolute Value . . . . . . . . . . . 804.3.1 Equations With Absolute Value . . . . . . . . . . . . . . . . . 804.3.2 Inequalites With Absolute Value . . . . . . . . . . . . . . . . . 83

4.4 Linear Inequalities in Two Variables . . . . . . . . . . . . . . . . . . . 874.4.1 Systems of Linear Inequalities . . . . . . . . . . . . . . . . . . 88

5 Polynomials, Polynomial Functions, and Equations . . . . . . . . . . . . . . 915.1 Introduction to Polynomial Functions. . . . . . . . . . . . . . . . . . . 91

5.1.1 Adding, Subtracting, and Evaluating Polynomials . . . . . . . . 925.1.2 Tail Ends of Polynomials . . . . . . . . . . . . . . . . . . . . . 94

5.2 Multiplication of Polynomials . . . . . . . . . . . . . . . . . . . . . . . 965.2.1 General Multiplication . . . . . . . . . . . . . . . . . . . . . . 96

5.3 Greatest Common Factors and Factor by Grouping. . . . . . . . . . . . 1015.3.1 Greatest Common Factor (GCF) . . . . . . . . . . . . . . . . . 101

5.4 Factoring Trinomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 1055.4.1 Factor by Guess and Check . . . . . . . . . . . . . . . . . . . . 1055.4.2 Factor by Grouping . . . . . . . . . . . . . . . . . . . . . . . . 1075.4.3 Factor Using the Quadratic Equation. . . . . . . . . . . . . . . 1095.4.4 Factor by Substitution . . . . . . . . . . . . . . . . . . . . . . 113

5.5 Factoring Special Forms . . . . . . . . . . . . . . . . . . . . . . . . . . 1155.5.1 Difference of Squares . . . . . . . . . . . . . . . . . . . . . . . 1155.5.2 Perfect Square Trinomials . . . . . . . . . . . . . . . . . . . . 1175.5.3 Sum and Difference of Cubes . . . . . . . . . . . . . . . . . . . 118

5.6 A General Factoring Strategy . . . . . . . . . . . . . . . . . . . . . . . 1215.6.1 Random Factoring Problems . . . . . . . . . . . . . . . . . . . 121

6 Quadratic Equations and Functions . . . . . . . . . . . . . . . . . . . . . . . 1256.1 The Square Root Property and Completing the Square. . . . . . . . . . 125

6.1.1 Complete the Square . . . . . . . . . . . . . . . . . . . . . . . 1286.2 The Quadratic Equations and The Quadratic Formula . . . . . . . . . . 133

6.2.1 Quadratic Equation. . . . . . . . . . . . . . . . . . . . . . . . 1336.2.2 Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . 137

6.3 Quadratic Functions and Their Graphs . . . . . . . . . . . . . . . . . . 1456.4 Equations Quadratic in Form . . . . . . . . . . . . . . . . . . . . . . . 1516.5 Polynomial Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . 156

6.5.1 Polynomial Inequalities . . . . . . . . . . . . . . . . . . . . . . 1567 Rational Expressions, Functions, and Equations . . . . . . . . . . . . . . . 160

7.1 Rational Expressions and Functions: Multiplying and Dividing . . . . . 1607.1.1 Multiplying Rational Expressions. . . . . . . . . . . . . . . . . 1637.1.2 Dividing Rational Expressions . . . . . . . . . . . . . . . . . . 164

7.2 Adding and Subtracting Rational Expressions . . . . . . . . . . . . . . 1677.3 Complicated Rational Expressions . . . . . . . . . . . . . . . . . . . . 1727.4 Division of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 1757.5 Synthetic Division and the Remainder Theorem . . . . . . . . . . . . . 1807.6 Rational Equations and Inequalities. . . . . . . . . . . . . . . . . . . . 184

7.6.1 Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . 1847.6.2 Rational Inequalities . . . . . . . . . . . . . . . . . . . . . . . 189

7.7 Formulas and Applications of Rational Equations . . . . . . . . . . . . 1957.8 Modeling Using Variation . . . . . . . . . . . . . . . . . . . . . . . . . 198

ContentsMath 103

8 Radicals, Radical Functions, and Rational Exponents . . . . . . . . . . . . 2048.1 Radical Expressions and Functions . . . . . . . . . . . . . . . . . . . . 2048.2 Rational Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2078.3 Multiplying and Simplifying Radical Expressions . . . . . . . . . . . . . 2138.4 Adding, Subtracting, and Dividing Radical Expressions . . . . . . . . . 2178.5 Multiplying with More Than One Term and Rationalizing Denominators 2228.6 Radical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2298.7 Complex Numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

8.7.1 Homework 8.7 Complex Numbers . . . . . . . . . . . . . . . . 243A Appendix Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . 248B Appendix Some More Notation . . . . . . . . . . . . . . . . . . . . . . . . . 259

Algebra, Math-ematical Models,and Problem Solv-ing

Math 103 Algebraic Expressions, Real Numbers, and Interval Notation

1 Algebra, Mathematical Models, and ProblemSolving

This section will be a quick review of material that is a pre-requisite for this class. If youneed to review, please refer to this section and/or come see me or tutors for help. We willnot spend much time in this section.

1.1 Algebraic Expressions, Real Numbers, and Interval Notation

The most important thing in mathematics which is not conveyed very clearly, is the im-portance of reading and english skills. The main reason why mathematics students have adifficult time in math is their inability to read the problem. I remember when I was tak-ing math classes and the professor would use mathematical terms that they assumed werecommon knowledge, but I had no idea what they were saying. I would understand one wordthen get lost in the next few seconds. By the time I caught up, we were three problems in.I cannot stress the importance of reading mathematics.

1.1.1 Translate English Phrases Into Algebraic Expressions.

The operations are the symbols that you have been using consistently for all your math life.The operations are +, −, ÷ (ab ), × (· or parenthesis). Each of these operations have a leftand a right hand side. The goal when translating english phrases into algebraic expressionsis to determine the right and left hand side, and then put in on the corresponding side ofthe operation used.Definition 1.1.1: The following english phrases can be translated as such,

1. A sum of a and b is the end result of adding a and b.

a+ b = sum

2. A product of a and b is the end result of multiplying a and b.

a · b = product

3. A difference of a and b is the end result of subtracting a and b.

a− b = difference

4. A quotient of a and b is the end result of dividing a and b.a

b= quotient

Example: 1.1.1:

Translate the following english phrases into algebraic expressions.

1) The sum of 5 and 3.

5+3 is the sum.

2) The quotient of 17 and -4.

17−4 is the quotient.

3) The difference of a number n and- 3

2 .

n− (− 32 ) is the difference.

4) The product of x and the differenceof 3 and y.

x(3− y) is the difference.



5) Fifteen more than the quotient of anumber and 5.

15 + n5 is fifteen more than the

quotient.

6) Two less than the difference of anumber n and the product of 3

2 andm.

(n− 32 ·m)− 2 is two less than the

difference.

Most books have you perform exercises like example 1.1.1. However, it is very importantto be able to read mathematical symbols into english. You need to be able to do it bothways. For example, how do you read x

2 − 4? Four less than the quotient of a number x and2. Many people cannot do this and is the true essence of reading mathematics.

1.1.2 Use mathematical models.

1.1.3 Recognize the sets that make up the real numbers.

Before we can discuss these sets, we need to understand the mathematical symbol ∈.Definition 1.1.2: The mathematical symbol ∈ reads as inside of (in). So 5 ∈ {1, 2, 3, 4, 5}says that 5 can be found inside of {1,2,3,4,5}. Similarly to 6=, a line through ∈ reads as notin, and is written 6∈. This says that we cannot find the element, for example; 5 6∈ {1, 2, 3, 4}.

The real numbers are the numbers that you have been using your entire life. Thesenumbers however can be broken up into different parts. they areNote: Set notation is a

list of elements or “things”in a set (collection). Setbuilder notation is readbest read through an ex-ample. A = {x | x ∈Z and 5 ≤ x ≤ 10}. Thisreads A is equal to the setof x such that x is an in-teger and is between or in-cluding 5 and 10.

Definition 1.1.3: The mathematical sets of numbers written in set notation and set buildernotation are,

1. The set of integers=Z = {. . . ,−3,−2,−1, 0, 1, 2, 3, . . .} (Set notation)The integers are ± the whole numbers.

2. The set of natural numbers=N = {1, 2, 3, 4, 5, . . .}(Set notation)These are also called the counting numbers. The numbers we use to count.

3. The set of whole numbers=W = {0, 1, 2, 3, 4, 5, . . .}(Set notation)This is the natural numbers plus the element 0.

4. The set of rational numbers=Q = {ab | a, b ∈ Z and b 6= 0} (Set builder notation)This is the set of all numbers that can be written as a fraction.

5. The set of irrational numbers=Q′ = {x | x 6∈ Q} (Set builder notation)This is the set of all numbers that cannot be written as a fraction (π,

√2, and e.

6. The set of real numbers=R = {x | x is on a number line}. (Set builder notation)All numbers that we can find on a number line. This Q and Q′ combined. That is allnumbers that can be written as a fraction and all numbers that cannot be written asa fraction.

A visual representation of these sets can be draw in something called a Venn diagram.It helps visualize the size and containment of each of the sets. You will notice that each setis contained in another set. For example; everything in the natural numbers can be foundin the whole numbers, and every one of these sets is inside of the real numbers.



Real numbersReal numbers

Rational numbersRational numbers

IntegersIntegers

Whole numbersWhole numbers

NaturalNaturalnumbersnumbers

IrrationalsIrrationals

Recognizing the sets that we use every time we do algebra is very important, especiallywhen discussing equations, and solutions to those equations.

Sets can be combined using two symbols ∩ and ∪. The symbol ∩ represents the word“and” while ∪ represents the word “or”. “And” in english translates as both must happen.For example; lets go to a dinner and a movie. This means that we will go to a dinner aswell as a movie. The word “or” translates as one, the other, or both. So if we use the sameexample but instead use the word “or”; lets go to dinner or a movie. This will translateas we can go to just dinner, just a movie, or both a movie and dinner. This fundamentaltranslation of the words will allow you to use them mathematically.

Example: 1.1.2:

1. {a, b, c, d} ∪ {1, 2, 3, 4}∪ represents an “or” statement. This means we want the elements in one, theother, or both. This comes out to {a, b, c, d, 1, 2, 3, 4}.

2. {2, 4, 8, 10} ∩ {1, 2, 3, 4}∩ represents an “and” statement. This means we want the elements in both.This ends up being the elements {1, 4}. These are the only elements that theyboth have in common.

3. {a, b, c, d} ∩ {ab, ac, ad}∩ represents an “and” statement. This means we want the elements in both.This ends up having nothing since there are no elements that are in both. Eventhough a can be seen in both, the element in the first set is a while the elementin the second is ab. They are not the same. This is true for all the letters. Thisset can be written as the empty set (a set that is empty) ∅.

4. {a, b, c, d} ∪ {ab, ac, ad}∪ represents an “or” statement. This means we want the elements in one, theother, or both. This comes out to {a, b, c, d, ab, ac, ad}.



1.1.4 Use Interval Notation and the Inequality Symbols.

Now that we understand the real numbers and the sets that make up the real line, we candiscuss interval notation. Interval notation is a way of writing an interval (a space betweentwo things; a gap) on the line.

Interval Notation:

Let’s start of with a question. How would we write

−5 −4 −3 −2 −1 0 1 2 3 4 5

••

in words as well as symbolically? In words we can see that we have all real numbers betweenand including -2 and 4. This is written as [−2, 4].Definition 1.1.4: Some useful information about interval notation.

1. Interval notation can be written using two symbols, (), [], (], or [) . The parenthesistell us the number is not included in the interval, while the brackets tells us that thenumbers are included in the interval.

2. Graphically this is shown using filled in dots and circles. Dots at the end of the intervalare shaded in are included and not shaded dots or circles are not included.

3. In interval notation ∞ and −∞ always use parenthesis.

4. If there are two or more intervals that need to be joined together, we seperate the twointervals with ∪. This is called the union and is read as or (in one side, the other side,or both). Ex: [1, 3] ∪ [8, 10) All real numbers in between and including 1 and 3 or thereal numbers between 8 and 10, including 8 but not 10.

Example: 1.1.3:

1)

−5−4−3−2−1 0 1 2 3 4 5

•

(−∞, 4]

2)

−5−4−3−2−1 0 1 2 3 4 5

◦ ◦

(−3, 0)

3)

−5−4−3−2−1 0 1 2 3 4 5

◦

(1,∞)

4)

−6−5−4−3−2−1 0 1 2 3 4 5 6

•• ◦ ◦

[−6,−3) ∪ (3, 5]

Note: Interval notation isalways read left to right. Inequality Symbol:

Notice in example 1.1.3 we have infinities, interval without the endpoints, and unions ofintervals. Using these problems we will learn how to write intervals using inequalities. First,what is an inequality?Definition 1.1.5: An inequality is like an equation but instead uses the symbols ≤, <, ≥, or>. These read as, in order, less than or equal to (smaller or the same as), less than (strictlysmaller), greater than or equal to (larger or the same as), and greater than (strictly larger).We use ≥ and ≤ when the end points are in the interval.

Most books or classes teach inequalities by saying, the crocodile or Pac man is hungryso the mouth is eating the larger number. If this makes sense to you then use it. I think of



it differently. Very simply, the arrow points to the smaller. So 3 < 4, is correct because itpoints to the smaller.

We will use 1, 2, 3, and 4 from example 1.1.3, to write the intervals using inequalities.Example: 1.1.4:

1) (−∞, 4]Notice that we want all numbersthat are smaller than or equal to 4.(Again going back to readingsymbols instead of translatingenglish into symbols.) Since thisnumber varies we will call it avariable n. So n is smaller than orequal to 4, and where does thearrow point? It points to thesmaller number, in this case n.

n ≤ 4 or 4 ≥ n.

Notice that the translation of “ Wewant all numbers that 4 is largerthan or equal to is the same as theone previously stated. Either caseworks but having the variable onthe left is standard.

2) (−3, 0)We want all numbers in between -3and 0, but not including thosenumbers. Again a variable isneeded, lets use m. Notice this alsomeans that the number we wantmust be larger than -3 but smallerthan 0. So

−3 < m < 0 or 0 > m > −3.

Both are correct but we like to keepnumbers in order, so again the leftis preferred. We may also translatethis as

−3 < m and m < 0

or0 > m and m > −3.

3) (1,∞)We want a number to be strictlylarger than 1. Let that variable bey.

y > 1 or 1 < y.

4) [−5,−3) ∪ (3, 5]In this case we have to options forthe number that we want. It can belarger or equal to -5 but smallerthan -3, or bigger than 3 butsmaller than or equal to 5. Let ourvariable be p.

−5 ≤ p < −3 or 3 < p ≤ 5.

In example 1.1.4 parts 2 and 4 we used the words “and” as well as the word “or”. Youmust be very careful. These words do not mean the same thing and will give you differentintervals. The word “and” means that the number is in both intervals at the same time.The word “or” allows the number to be in both at the same time, but also allows for thenumber to be in one interval or the other interval only.


Math 103 Operations with Real Numbers and Simplifying Algebraic Expressions

1.2 Operations with Real Numbers and Simplifying Algebraic Expressions

The most basic part of any mathematics class is operation sense and simplification. Asstated in the first section, your ability to read is the most important part of math peopleare missing. The next part is basic arithmetic skills.

1.2.1 Operations using real numbers.

Addition and What it Means

Imagine you have a 5 or 6 year old baby sister who is asking you for help with their mathhomework. They have a problem which says add 5 and 2 together. What is the total? Yoursister gets the number 7 as their answer but asks why she gets 7. Could you explain thereason why the sum of 5 and 2 is seven? Many responses are along the lines of well 5+2=7. We cannot explain why something is true by repeating the question. One way to explainwhat addition is, is a count. Addition is a physical count of object. In the example abovewe would have �� + �� = a total of 7 squares. This idea may not seem importantnow, but is quite useful when discussing fractions, units, and variables.

Subtraction and What it Means

Subtraction can be thought of as“take away”, decrease, remove, or reduce. All of thesewords will translate into subtraction. Lets refer back to the little sister example where yourlittle sister is now asking about a problem about subtraction. Their problem is 5-2. Againyour sister knows that the answer is 3 but does not understand why. We can explain thisby writing the squares out again. In english this can be read as, there are 5 squares and wewant to take away two of them so how many are left. ��−�� = 6 � 6 �� = ��=a total of 3 squares left.

Multiplication and What it Means

For this section lets use some examples and see what multiplication is. Lets consideradding the following expression. 2 + 2 + 2 + 2. The answer is 8. What about the expression4+4? This is also 8.

1. What is the base number used in 2+2+2+2?2

2. How many times did we add 2 in 2+2+2+2?We added 2, 4 times.

3. Is there a way to multiply the answers from 1 and 2 to get the desired answer?2 · 4 = 8

4. What is the base number used in 4+4?4

5. How many times did we add 4 in 4+4?We added 4,2 times.

6. Is there a way to multiply the answers from 4 and 4 to get the desired answer?4 · 2 = 8

7. What would you say multiplication is?Multiplication is repeated addition.

We may also see this using an area model. To find the area of a rectangle you multiplythe length by the width (A = l · w). Lets consider the example 4 · 2. The length will be oflength 4 and the width will be of length 2. We can represent this as

Notice that the total number of squares is 8. This method also shows the repeated ad-dition method. If we look at the columns of the right rectangle, we have 4 columns of 2



→

4 4

22

Figure 1: First break the length and width up into 4 and 2 parts. Then count the numberof squares.

squares, or add two four times (2+2+2+2=8). If we look at the rows we then have two rowsof 4 squares, giving us addition of 4 two times (4+4=8).

Division and What it Means

Division is a way to group items. The quotient is the number of groups of a certainsize. In a fraction 8

4 , the top number (numerator) 8 is the number of dots, while the bottomnumber (denominator) 4 is the size of the group. This says we have 8 dots grouped by sizefour. This is why we get 2 as our answer. There are 2 groups of 4 in 8 dots. This even workswhen the numerator is smaller than the denominator. The fraction 2

4 says that in two dotsthere is only 1

2 if a group of four. This will also describe division by zero and zero in thenumerator.

Theorem 1.2.1:

1. If you have a numerator of zero and the denominator is not zero, then thequotient is zero.Notice that if the numerator is zero, then we are trying to group zero objectinto a group of a certain size. When we do this there are no groups. Hence0n = 0 as long as n 6= 0.

2. If the denominator is zero, the quotient is undefined.If the denominator is zero, we want to group all the items into a group of sizezero. This is impossible and does not make sense. This is why n

0 is undefined.

Fraction Arithmetic

The rules of fractions are one of the most important rules to know. Because people donot like fractions, they tend not to remember these rules and causes problems for them evenwhen they are not using fraction. We will use fraction rules all throught.Note: Make sure that you

repeat these rules to your-self over and over again.Also do not cut the state-ment short. This causesproblems. Saying flip andmultiply confuses peoplebecause “flip” is not veryprecise.

Theorem 1.2.2:

The rules of fraction arithmetic.

1. When adding fractions, check if the denominators are the same. If not, findthe LCD and make the denominators the same. Then add the top and keepthe bottom.

2. When subtracting fractions, check if the denominators are the same. If not,find the LCD and make the denominators the same. Then subtract the topand keep the bottom.

3. When multiplying fraction, multiply straight across.

4. When dividing fractions, flip the bottom fraction, then take that number andmultiply straight across to the top fraction.



Example: 1.2.1:

Evaluate the following.

1) 34 + 5

2

34 + 5

2 = 34 + 10

4

= 138

2) 35 − 2

35 − 2 = 3

5 −105

= −75

3) 173 ·

324

173 ·

324 = 51

72

or = 17 · 33 · 24

= 1724

4)23− 6

5

23− 6

5= 2

3 · −56

= − 53 · 3

= −59

Note: Don’t forget thatpairs of negatives in multi-plication and division can-cel each other out to makea positive number.

−2 · (−3) = 6

−3−5 = 3

5

Exponential Expressions

Exponential expressions were created so that you wouldn’t have to write out terms.Similarly to how multiplication is repeated addition and is a short hand to writing multipleaddition operations, exponents is short hand for writing multiple multiplications. If we have2 · 2 · 2 · 2 · 2, we are multiplying 2 five times. This gives us 25. In general the formula isan = (base)number of times base is repeated. Since exponents are repeated multiplication, we cancreate rules which will help us in calculations.

Theorem 1.2.3:

Exponent rules:1. Product rule: The product rule of exponents can be easily shown through an

example. All we need to do is write it out. a2 · a3 = a · a · a · a · a = a5. If wemultiply exponents with the same base, then we keep the base and raise it tothe power of the sum of the former exponents.

an · am = an+m.

2. Quotient rule: The quotient rule of exponents is also easily shown through anexample. a4

a2 = a·a·a·aa·a = a ·a = a2. The quotient rule of exponents say, if we are

dividing two exponents with the same base, then keep the base and subtractthe bottom exponent to the top exponent.

an

am= an−m.

3. Zero-exponent rule: This rule can be showed using the quotient rule. Noticethat a zero exponent is always a difference of two exponents that are the same.That is a0 = a2−2 = a2

a2 = a·aa·a = 1. We used 2-2 for the sake of ease, but you

may use any number you would like. This says that

a0 = 1.

Any number besides 0 raised to the zero power is one. The expression 00 is



indeterminate and you will learn about this in calculus.

4. Negative exponent rule: This similarly can be shown through the quotient rule.

1a3 = 1

a · a · a= a · aa · a · a · a · a

= a2

a5 = a2−5 = a−3.

The negative exponent rule says that you may write the exponential expressionin the denominator if you remove the negative attached to the exponent.

a−n = 1an.

I think of this as erase the negative and write 1The rest .

5. Fraction and exponent rule: A fraction with an exponent has the property thatthe exponent distributes to the numerator and the denominator. This can beseen through the definition of exponents. (ab )3 = a

b ·ab ·

ab = a·a·a

b·b·b = a3

b3 .(ab

)n= an

bn

This also works with any multiplication in the numerator and denominator.(a · bc · d

)n= an · bn

cn · dn

6. Exponents to exponents rule: An exponent raised to another exponent hasthe property of multiplication. We can see this through the product rule.(a2)3 = a2 · a2 · a2 = a2+2+2 = a6.

(an)m = an·m

Similarly as in 5, we can do this with multiple multiplications.

((a · b)n)m = (an · bn)m = an·m · bn·m

Example: 1.2.2:

Simplify the exponential expressions. Write without negative exponents.

1) 3 · 35

Product with the same base, so addpowers.

31+5 = 36

2) a5

a3

Quotient with the same base, sosubtract the bottom power from thetop power.

a5−3 = a2

3) n−7 · n2

Product with the same base, so addpowers.

n−7+2 = n−5

= 1n5

4) y12

y2

Quotient with the same base, sosubtract the bottom power from thetop power.

y12−2 = y−

32

= 1y

32



5) m2

m− 32·m3

First subtract the − 32 power from

the 2 power, then add the power ofthree.

m2−(− 32 ) ·m3 = m

42 + 3

2 ·m3

= m72 ·m3

= m72 +3

= m72 + 6

2

= m132

6) x34 ·x

54

x2

First add the 34 power and the 5

4power, then subtract the 2 power.

x34 + 5

4

x2 = x2

x2

= x2−2

= x0

= 1

7)(( 3

2)2)2

Exponent raised to an exponent, sowe multiply the powers. Then wedistribute the power.(

32

)2·2=(

32

)4

= 34

24

= 8116

8)(

3·103

5·102

)2

Quotient so subtract the 2 from thethree, then distribute the 2.(

35 · 103−2

)2=(

35 · 101

)2

=(

35

)2(·101)2

= 32

52 · 102

= 925 · 102

= 925 · 100

= 9 · 4

= 36

1.2.2 Algebraic properties. Commutative, Associative, and DistributiveProperties. Find Opposites and inverse. Absolute value.

The use of the algebraic properties is to simplify expression or make them easier than theoriginal problem being solved. Math is an art, in the sense that it takes a creative mind tosee a problem and manipulate it so that it is easy.

Theorem 1.2.4:

The commutative and associative properties only hold for addition and multiplica-tion. They do not work for division and subtraction.

1. Commutative property: a + b = b + a and a · b = b · a. This says that we canchange the order of addition and multiplication. Ex: 5 · 15 · 2 = 5 · 2 · 15 =10 · 15 = 150.

2. Associative property: (a + b) + c = a + (b + c) and (a · b) · c = a · (b · c).This says that we can change the placement of the parenthesis of addition and



multiplication. Ex: (3 + 38) + 2 = 3 + (38 + 2) = 3 + 40 = 43.

3. Distributive property: a(b+ c) = a · b+ a · c. The distributive property says, ifthere is a multiplication of an object to a sum of objects, then the multiplicationmust be done to each of the objects in the sum. Ex: (3 + x)(x + y) = (3 +x)x+ (3 + x)y.

I do not believe that these are properties that need specifically designed problems forpractice. We will see them continuously though out the book.

Opposites and Inverse

Definition 1.2.1: We will define opposite and inverse at the same time.

1. Opposite: The opposite of a number n is the negative version −n. The opposite of2 is −2 and the opposite of −3 is −(−3) = 3.

2. Inverse: The inverse of a number n is the number multiplied to n so that the productis 1. n · 1

n = 1. The opposite of 2 is 12 , and the opposite of − 3

4 is 1− 3

4= 1 · − 4

3 = − 43 ,

since − 34 · −

43 = 1.

Similarly I do not believe that we need to have more example problems than this. Theopposite number will be used in the next topic of absolute values and the inverse was alreadydiscussed using negative exponents.

Absolute Value

Note: As a rule of thumb,if you think you canjust erase or delete thingsin mathematics, you arewrong. If you cannot jus-tify why it disappeared,then it should not havedisappeared.

Most people have learned the absolute value by saying, oh its the positive of the number,just get rid of the negative sign. This is a really bad way of teaching the absolute value. Itneither describes the graph nor the true essence of the piece wise function.Definition 1.2.2: The absolute value |x| is given by the piecewise function,

f(x) = |x| ={x x ≥ 0−x x < 0

This says if the value in the absolute value is negative, then multiply it by a negative, ortake the opposite value. If the value in the absolute value is positive, then keep the samevalue. The absolute value is discussing the distance away from 0. The numbers -2 and 2 areboth two steps away from 0, and therefore | − 2| = −(−2) = 2 = |2|.

1.2.3 Use the Order of Operations.

Have you ever heard or Parenthesis Exponents Multiplication Division Addition Subtractionor PEMDAS? This is the acronym used to assist in memorizing the order of operations. Iprefer to write it as (PE)(MD)(AS). Each parenthesis represents a group starting from leftto right. Most people do order of operations wrong when they have a problem like 2÷3 ·4. Ifyou read PEMDAS from left to right, we would do multiplication before division. Howeverthis is not the case. We actually do division, then multiplication. This can be rectified bychanging the way you think/write PEMDAS to (PE)(MD)(AS).Definition 1.2.3: To follow (PE)(MD)(AS), we must

1. Look at the expression and determine if there are PE, MD, or AS. PE is the highestlevel and should be done first, then MD, and lastly AS. So the highest level found first

2. If there is more than one of the highest level found, then you must do it left to right.

3. Repeat steps 1 and 2.



Example: 1.2.3:

1) 3(4 + 3 · 4)− 2Parenthesis, multiply 3 and 4, add4, multiply by 4, and subtract 2.

3(4 + 3 · 4)− 2 = 3(4 + 12)− 2

= 3(16)− 2

= 48− 2

= 46

2) (4÷ 3× 2)2 ÷ 4Parenthesis, (there are two in thesame level D and M, so left toright) divide by 3, multiply by 2,exponent of 2, then divide by 4.

(4÷ 3× 2)2 ÷ 4 = (43 × 2)2 ÷ 4

= (83)2 ÷ 4

= 649 ÷ 4

= 169

1.2.4 Simplify Algebraic Expressions vs Evaluate Algebraic Expressions.

There are two things commonly done with algebraic expressions. We either simplify orevaluate these expressions. We must first define what an algebraic expression is.Definition 1.2.4: An algebraic expression is an expression constructed with variables,algebraic numbers, an mathematical operations.

Since algebraic expressions have variables, we have two options. We can either simplifythem, or evaluate them if we know the value of the variables. Simplification is very importantin mathematics because simply it makes things simpler. That is why it is called simplification(to make simple). This can be used throughout, even during the evaluation of algebraicexpressions process. You cannot evaluate a problem which specifically asks for simplification.For the sake of ease, simplify always, evaluate only if you need to.

When simplifying you may only combine by subtraction and addition if the variablesare exactly the same. This is called like terms. Think of variables as units. We have 5exes or 5x, we add 3 exes to it to get a total of 8 exes or 8x. This is because addition andsubtraction are forms of counting. We cannot add 5x and 3x2. This is like saying trying tocount 5 cats and 3 flowers. You won’t be able to say 8 cats or 8 flowers. The only thing youcan say is there are 5 cats and 3 flowers. Some examples of things to simplify are

• Reduce fractions of all types.

• Combine like term.

• Use exponent rules.

• Factor or distribute.

When you evaluate an expression, people make many mistakes. Most of these mistakescan be fixed using step one.

Instructions: 1.2.1:In order to evaluate an expression,

1. Put parenthesis around every variable.

2. Copy what you want to substitute in to each parenthesis.



Example: 1.2.4:

Determine if the problem is a simplification or an evaluation problem. Then simplifyor evaluate accordingly.

1) 4a− 8b− 16

This is a simplification problem andthere is nothing that can becombined or made simpler. This issimplified.

2) m2+2mm

This is simplification problem wherewe combine the m2 and 2m2, thenm from the top and bottom usingexponent rules.

m2 + 2m2

m= 3m2

m

= 3m2−1

= 3m

3) x = 3, y = 4,

4x− 8y − 16

This is an evaluation problem sincewe know what x and y are.

4( )− 8( )− 16 = 4(3)− 8(4)− 16

= 12− 32− 16

= −36

4)34 (3ω+2θ)

( 32 )2(3ω+2θ)

This is a simplification problem.34 (3ω + 2θ)( 3

2)2 (3ω + 2θ)

= 34 ·

22

32

= 13

5) g = (x+ 1)2

g−1x

This is an evaluation andsimplification problem combined.We will first evaluate knowing g,then simplify.

( )− 1x

= ((x+ 1)2)− 1x

= (x+ 1)(x+ 1)− 1x

= x2 + 2x+ 1− 1x

= x(x+ 2)x

= x+ 2

6) ω = 32, θ = 1534 (3ω+2θ)

( 32 )2(3ω+2θ)

This is a evaluation problem sincewe know ω and θ. However, we cansimplify this equation beforeevaluating and get the end result of4. Now that we have 1

3 there isnothing to plug in. If you did plugin, it would work, however it is notthe easiest way to do this problem.


Math 103 Graphing Equations

1.3 Graphing Equations

The graphing of equations gives us a visual representation of what is happening. This isnot a way to give exact answers. Since we cannot draw perfectly, graphs will always giveus estimations. Even if we use a computer, if you visually try to approximate the input oroutput of a equation, you will not be able to say that it is correct. If we do this it is anapproximation. In order to get exact answers, you must evaluate the equation.

1.3.1 Plotting Points and Graphing equations in the rectangular coordi-nate system.

Plotting Points:

Points in the rectangular coordinate system are written as (input, output). It is normallytaught as (x, y). This however is not a very useful way of teaching rectangular coordinates.The input is not always x and the output is not always y. It is much more efficient toremember something that is always true.

The graphs are representing these inputs and outputs. The horizontal axis representsinput (left or right), and the vertical axis represents the output (height). So the (input, )tells you how far left or right to go, and the ( , output) tells you how far up or down togo.Note: Make sure that you

do not connect the dots.You can only connect thedots if you know that thepoints in between are alsopart of the graph.

Example: 1.3.1:

Plot the coordinate points on the graph.

1. (1,3), (3,4), (5,7), (−1,−4), (−5, 7), (0,0)

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

••

•

•

•

•input

output

2.

Input Output−3 5−5 −52 43 75 −4

Remember that we have (input, output) in our rectangular coordinates. Wecan rewrite the table to a problem similar to part 1. (−3, 5), (−5,−5), (2, 4),(3, 7) and (5,−4)



−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

•

•

•

•

•

input

output

Graphing Equations:

The difference between points and equations is the values that we have for the input andoutput. Equations use points to create a graph of the equation. We will demonstrate thisthrough plotting different variety of points to get a clear picture of a graph.

Example: 1.3.2:

Graph a = b2. Let’s try plotting 2 points, then 9 points to see what happens. Letstry plugging in 1 and -1 for b. We get (1,1) and (-1,1). Plotting these points we get

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

•• x

y

If we connected the dots we would get the blue line.Lets try plugging in 9 different points. I will organize the data in a table.

Input=b Output=a or b2

0 01 1-1 12 4-2 43 9-3 94 16-4 16



−5 −4 −3 −2 −1 1 2 3 4 5

−20

−15

−10

−5

5

10

15

20

• ••••

••

••

x

y

Notice that the graph is becoming more accurate. Since we know that we can plugin any input into b2, we may connect the dots to get the graph

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

x

y


Math 103 Scientific Notation, and conversion between fractions and decimals.

1.4 Scientific Notation, and conversion between fractions and decimals.

Scientific notation is a nice way of writing small and large numbers. For example; insteadof writing 120,000,000 we could write 1.2 × 108. Not only is it useful for writing numbersnicely, it is also useful because we have exponent rules which will allow us to simplify thesenicely. Another very useful form is fractions, which we have already covered. We need tobe able to convert from decimals to fractions, and decimals to scientific notation.

1.4.1 Convert Between Fractions and Decimals.

The easier of the two discussed above is converting between fractions and decimals. Toconvert a fraction into a decimal is very easy. All you need to do is long division. Therewon’t be any more discussion on this topic. There are many references online if you areunable to do long division. The other way requires a little more.

Theorem 1.4.1:

To convert a decimal into a fraction you must

1. Start with a decimal.

2. Move the decimal as far right as possible.

3. Write the numeral without the decimal over one 1 and as many zeros as youmoved the decimal.

If you can read the decimal, then you can also write out the fraction form directly.

Example: 1.4.1:

Covert each decimal into its fraction form.

1) 32.1234

1) 32.12342) 321234.3) 321234

10,000 .

If you can read the number, then itsays thirty two and one thousandtwo hundred thirty four tenthousandths. This is the same asthe fraction above.

2) .000152

1) .0001522) 000152.3) 000152

1,000,000 = 1521,000,000

3) 234

1) 2342) 234.3) 234

1

4) 12.00042

1) 12.000422) 1200042.3) 1200042

100,000

1.4.2 Convert Between Scientific and Decimal Notation.

I think about conversion between scientific notation and decimal notation. I think the waythat it is taught is to memorize which direction the decimal moves, either left or right.I don’t like doing this because it changes depending if you are converting from scientificnotation to decimal notation, or from decimal notation to scientific notation. Again if thisis the way that you know how to do it, there is nothing wrong with it and you should



continue to perform it the way that works. However if you are continuously forgetting andcan’t seem to get it, then don’t continue to do it the same way. Try something different.

I always start with scientific notation and use that to either determine the sign of thepower, or where to move the decimal.

Theorem 1.4.2:

Convert from a decimal to scientific notation To convert from a decimal toscientific notation, always start in scientific notation form. That is

1. Move the decimal after the first non-zero number from left to right.

2. Write that decimal multiplied by 10Number of times the decimal was moved.

3. Determine if the exponent is negative or positive by looking at the decimal in2, and see if it needs to get bigger or small to get back to the original.

(a) If it needs to get bigger, then the exponent is positive.(b) If it needs to be smaller, then the exponent is negative.

Convert from decimal notation to scientific notation: To convert from decimalnotation to scientific notation, you need to understand what a exponent does. Usinga) and b) we can determine this. Part a) should make sense because 10Positive power ismultiplication. This makes numbers bigger. The number 10Negative power = 1

10 Power

which is actually division. Division by whole numbers makes things smaller. This isintuitive.

1. Move the decimal as many times as the power so that the new number is smalleror bigger according to the sign of the exponent.

Example: 1.4.2:

Either convert into scientific notation or into decimal form.

1) 9.2000000001× 109

Positive powers of 10 make numbersbigger. So the answer is9200000000.1 > 9.2000000001

2) 123456789

1) 1.234567892) 1.23456789× 108

3) Starting from1.23456789→ 123456789 ismaking the number bigger. Sothe exponent is positive, andthe answer is 1.23456789× 108

3) .000000000000001

1) 12) 1× 1015

3) Staring from1→ .000000000000001, thenumber is getting smaller. Sothe exponent is negative, andthe solution is 1× 10−15

4) 5.123400001× 103

Positive power of 10 makes numberslarger, so the answer is5123.400001 > 5.123400001.



5) 2.0000001× 10−6

Negative powers of 10 makenumbers smaller, so the answer is.0000020000001 < 2.0000001

6) 0.001224

1) 1.2242) 1.224× 103

3) Starting from1.224→ 0.001224 the numberis getting smaller. So theexponent is negative, and thesolution is 1.224× 10−3.

7) 9898989.1

1) 9.89898912) 9.8989891× 106

3) Since 9.8989891→ 9898989.1is getting bigger, the exponentis positive. The answer is9.8989891× 106.

8) 3.231× 10−3

Negative powers of ten make thenumber smaller, so .003231 < 3.231is the solution.

1.4.3 Perform computations with scientific notation.

In order to perform computations with scientific notation we need to use the basic ideathat addition is a count, exponent rules, and the conversions between decimals and scientificnotation. These are best seen through examples.Note: I understand the

thought, I would just plugit into a calculator. Youneed to understand whatis happening to ensureyour intuition is accurate.The calculator or com-puter can give wrong num-bers because the personinputing the numbers isdoing it wrong. You needthe intuition to know thatthe number is not correct.You build this up by doingmathematics through penand paper.

Example: 1.4.3:

1) 64000000 · 0.0008

64000000 · 0.0008 = 6.4 · 107 · 8 · 10−4

= 6410 · 8 · 107 · 10−4

= 51210 · 107−4

= 51.2 · 103

= 5.12 · 101 · 103

= 5.12 · 104


Math 103 More On Rectangular Coordinates

2) 64000000÷ 0.0008

64000000÷ 0.0008 = 6.4 · 107

8 · 10−4

= 6.48 · 107−(−4)

=64108 · 1011

= 810 · 1011

= 8 · 1010

3) 2× 105 + 4× 103 − (−6× 105 + 2× 103 + 9× 104)

= 2× 105 + 4× 103 + 6× 105 − 2× 103 − 9× 104)

= 2× 105 + 6× 105 + 4× 103 − 2× 103 − 9× 104

= 8× 105 + 2× 103 − 9× 104

= 800, 000 + 2000− 90, 000

= 712, 000

= 7.12× 105

4) −2×103

5×10−2−9×10−2

−2× 103

5× 10−2 − 9× 10−2 = −2× 103

−4× 10−2

= −2−4 × 103−(−2)

= 12 × 105

= .5× 105

= 5× 10−1 × 105

= 5× 104

1.5 More On Rectangular Coordinates

I will state a few formulas which will not be proven. We will then use these with lines.Theorem 1.5.1:

The sum of the squares of the legs of a right triangle is equal to the square of thehypotenuse. That is

a2 + b2 = c2.


Math 103 More On Rectangular Coordinates

c

a = x2 − x1

b = y1 − y2

(x1, y1)

(x2, y2)(x1, y2)

Notice that we can substitute for a and b in theorem 1.5.1. This will give us the distanceformula.

Theorem 1.5.2:

The distance between two points (x1, y2) and (x2, y2) is given by

c2 = (x2 − x1)2 + (y1 − y2)2

c =√

(x1 − x2)2 + (y1 − y2)2

We may switch x1 and x2, because the square is the same.

Functions andLinear Functions

Math 103 Introduction to Functions

2 Functions and Linear Functions

Goals:1. Find the domain

and range of arelation.

2. Determine whethera relation is a func-tion.

3. Evaluate a function.

This is the beginning of material that is specific to this class. An introduction to functionsand lines is the most fundamental part of calculation based mathematics. You have been us-ing functions an lines for most of you mathematical careers. Even if it was not presented thisway, you learned these as early as 6th or 5th grade. We will discuss this with the intentionof having a deeper understanding of the material. This will help you retain the information.I believe that math should be taught through understanding instead of calculations. This isa very difficult thing to do and many people resist because this is not how they were taughtwhen they were young. However it will benefit you in the long run and you should do yourbest to learn math as a thought process instead of a calculation based system.

2.1 Introduction to Functions

We must first define what the domain and range are.

Definition 2.1.1: We will define both the domain and the range together.

1. The domain is the set of all possible inputs. This is written in either set notation orset builder notation discussed in chapter 1.

2. The range is the set of all possible outputs. This is written in either set notation orset builder notation discussed in chapter 1.

2.1.1 Find the domain and range of a relation.

Example: 2.1.1:

Going back to example 1.3.1, we have the points (1,3), (3,4), (5,7), (−1,−4), (−5, 7),(0,0).The domain is the set of all inputs, or the list of all parts in the first coordinate(input,output).

D = {1, 3, 5,−1,−5, 0}

The range is the set of all outputs, or the list of all parts in the second coordinate(input,output).

R = {3, 4, 7,−4, 7, 0}

We may also do this using a graph. The definition does not change. We still want a listof all the inputs for domain, and a list of all the outputs for the range. Don’t forget thatthe input is on the horizontal axis and the vertical axis is the output. This is restricted tothe values on each axis that is being used. For example, in example 2.1.1 the number 2 isphysically on the axis but is not in the domain because it is not being used by the points.

Example: 2.1.2:

Find the domain and the range.

1. The relation below is given as points.



−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

•

•

•

•

•

input

output

The domain is all possible inputs. These are all the values on the horizontalaxis that are being used in the graph.

D = {−5,−3, 2, 3, 5}

The range is all possible outputs. These are all the values on the vertical axisthat are being used in the graph.

R = {−5,−4, 4, 5, 7}

2. The relations is another type of relationship. We will see the difference in theanswer.

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−6−5−4−3−2−1

123456

•◦◦

x

y

Notice that the domain is all real numbers. This is written as (−∞,∞). Thisis because any input will allow us to get an output value. You can test this byputting your pencil on an output value, then move your pencil vertically. Eachof these will hit the graph. The range however does not have this property.Take your pencil on an output value. Notice that if we move it left and rightthere is a section which will not hit the graph. This section is [4, 5]. This isnot in the range. So the range is all numbers except [4, 5]. This is writtenas (−∞, 4) ∪ (5,∞). Notice that this interval starts from −∞ and ends at 4,but does not include it. It then starts again at 5 but does not include it, andcontinues up to ∞.

There is a difference between the two different ways of writing the domain and the range.



in the first example, we have a list that we can physically write, so we use {}. The secondexample has all the numbers in-between. Therefore, we need an interval.

2.1.2 Determine whether a relation is a function.

Definition 2.1.2: A function is relationship between its domain and range. It has theproperty that every input corresponds to exactly one output. It does not matter the otherway. A function may have one output corresponding to multiple inputs. You need to checkthat each input has one and only one output.

Example: 2.1.3:

Determine if the following are examples of functions. Justify your answer.

1. Let the left ellipse represent the domain, and the right ellipse represent therange.

•

•

•

•

•

•

•

•

Domain Range

We ask if the picture follows the definition of a function. Does every thing inthe domain go to one and only one thing in the range? No. The third pointgoes to the second and third point. It does not go to one and only one thingin the range. This does not represent a function.

2. Let the left ellipse represent the domain, and the right ellipse represent therange.

•

•

•

•

•

Domain Range

Does every thing in the domain go to one and only one thing in the range?Yes. Each point in the domain goes to exactly one point in the range, the onlypoint there. Hence this represents a function.

3. (1,3), (3,4), (5,7), (−1,−4), (−5, 7), (0,0).Does every thing in the domain go to one and only one thing in the range? Yes.There are no repeats in the domain, so there is no possibility that the thingsin the domain go to different things in the range.



2.1.3 Evaluate a function.

Functions will be written using function notation. We write functions as f(x), and is readas f of x. This is true using any letter. If we mean to multiply the variables, we will writeit as hx = h · x = (h)(x). This is different than h(x), h of x. It is important to be ableto interchange the definitions of input and output. This goes back to the idea of readingmathematics.Definition 2.1.3: The input and range have been used in multiple ways so far. We will makea list for sake of ease.

Input OutputIndependent variable Dependent variable

Horizontal axis Vertical axis(input, ) ( , output)f(input) output=f(x)

Domain, set of all inputs Range, set of all outputs

We will look at multiple problems which ask us to evaluate the function given differentways of expressing the input and out put.

Example: 2.1.4:

Evaluate the functions by plugging in for either the input or the output. Don’t forgetto put parenthesis around what you are plugging in for.

1) input=3, g(n) = 4n− 1

g(3) = 4(3)− 1

= 12− 1

= 11

2) output=2, f(x) = 3x+ 2

f(x) = 2

(2) = 3x+ 2

2− 2 = 3x

0 = 3x

x = 0

3) p(a) = 0, p(a) = 32a− 3

(0) = 32a− 3

3 = 32a

3 · 23 = a

a = 2

4) y = 2, h(y) = 3y3 + 2y + 2

h(2) = 3(2)3 + 2(2) + 2

= 3(8) + 4 + 2

= 24 + 6

= 30



5) ( , 12 ), ω(θ) = θ+1

2

ω(θ) = 12(

12

)= θ + 1

2

1 = θ + 1

θ = 0

6) (−1, ), h(k) = (k100 + 1)2

h(−1) = ((−1)100 + 1)2

= (1 + 1)2

= 22

= 4


Math 103 Graphs of Functions

2.2 Graphs of Functions

Goals:1. Graph functions by

plotting points.

2. Use the vertical linetest to identify func-tions.

3. Obtain informationabout a functionfrom its graph.

4. Identify the domainand range of a func-tion from its graph.

We have already learned how to graph functions in section 1.3. We will now learn howto determine if a graph represents a function or not, approximations using graphs, anddetermining domain and ranges using graphs.

2.2.1 Does a Graph Represent a Function

What is the definition of a function? A function has one and only one output for each input.Lets think about this in terms of a graph of a function. What would happen if a input hadto outputs? We would have for example (1,1) and (1,2). The input one goes to one as wellas two. This would result in a graph of

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

••

x

y

Graphs that are not functions have points which are in the same input coordinate.Theorem 2.2.1:

If you have a graph, you can determine if the graph represents a function by checkingif the graph passes the vertical line test. This says for any vertical line you can draw,the graph will only touch the vertical line once. Another way to think of this is to

1. Hold your pencil vertically.

2. Start from the left and slowly move your pencil to the right.

3. If the graph touches your pencil more than once at any given time, then thegraph does not represent a function. Like the example above.

4. If the graph only touches your pencil once at any given time, then the graphrepresents a function.

Example: 2.2.1:

Determine if the graph represents a function. If it is find its domain and range.



1)

−5 −4 −3 −2 −1 1 2 3 4 5

−5−4−3−2−1

12345

•

•

x

y

This is a function, because it passesthe vertical line test. The domain isall the input values. This startsfrom the left at −3 and ends at 4.So the domain is (−3, 4). Its rangeis from (0, 4).

2)

−5 −4 −3 −2 −1 1 2 3 4 5

−5−4−3−2−1

12345

x

y

This is not a function, because itfails the vertical line test.

3)

−5 −4 −3 −2 −1 1 2 3 4 5

−5−4−3−2−1

12345

• • x

y

This is a function, because it passesthe vertical line test. The domain isfrom (−2, 2) and the range is from(0, 2).

4)

−5 −4 −3 −2 −1 1 2 3 4 5

−5−4−3−2−1

12345

x

y

This is not a function, because itfails the vertical line test.

Lets use one and three from example 2.2.1, to obtain some information about a givensituation.



Example: 2.2.2:

1) Let the function given by the graphin example 2.2.1 part 1 represent thefollowing situation. The dragon Dro-gon from Game of Thrones, has startedbreathing fire over the region. The point(0,0) represents the day that he startedhis incineration of the land, where theinput is the number of days from theday he started, and the output repre-sents the number of acres burned.

a) What is the domain and range of the function, and what do they represent?The domain is [−3, 4] and it represents the days before and after the startof Drogon’s attack. The range is [0, 4]. This represents the number of acresburned.

b) What is the practical meaning of the point (0, 0)?This points represents 0 days since the attack started and 0 acres burned. Thestart of the attack.

c) What is the practical meaning of the point (4, 4)?This points represents 4 days since the attack started and 4 acres burned. Theend of the attack.

d) What is the practical meaning of the point (−2, 0)? This points represents 2days before the attack started and 0 acres burned.

2) Let the function given by the graph in example 2.2.1 part 3 represent the followingsituation. In the year 2015, a company was at its height of profit in millions. Thefunction represents the relationship between the company’s profit and time from2015.

a) What is the domain and range of the function, and what do they represent?The domain is [−2, 2], and it represents the times between and including 2013-2017. The range is [0, 2]. This represents the profits of the company in millions.

b) What is the practical meaning of the point (0, 2)?This points represents the year 2015, that had 2 million dollars in profit . Theyear the company was most profitable.

b) What is the practical meaning of the point (−2, 0)?This points represents the year 2013, that had 0 dollars in profit . The yearthe company was created.

b) What is the practical meaning of the point (2, 0)?This points represents the year 2017, that had 0 dollars in profit . The yearthe company went bankrupt.


Math 103 The Algebra of Functions

2.3 The Algebra of Functions

Goals:1. Use the algebra of

functions to com-bine functions.

2. Find the domain ofa function.

It is always interesting to see peoples reaction to the algebra of functions. People tend tochange their basic arithmetic rules when discussing this section. The rules of mathematicsdo not change. The rules of arithmetic and fractions do not change when doing things withvariables and functions. If you need a review, please go back to section 1.2.

2.3.1 Arithmetic With Functions

The operations that can be used with functions are the same as with number. There howeveris one more called the composition.Definition 2.3.1: These operations are written as

1. (f + g)(x) = f(x) + g(x)

2. (fg)(x) = f(x) · g(x)

3. (f − g)(x) = f(x)− g(x)

4.(fg

)(x) = f(x)

g(x) g(x) 6= 0.

5. (f ◦ g)(x) = f(g(x))

The domain of each is the set of values that are in the domain of both f and g.The same rules apply from chapter 1. We may combine like terms, the associative,

distributive, and commutative property hold, power rules, and when we plug in don’t forgetparenthesis. The most common reason why students in the STEM side of mathematics havea difficult time is because of their algebra skills. Many people do not understand materiallike trigonometry or calc1-4, because they forget how to add fractions or use the power rules.Understand those materials first and the rest of math will be much easier.

A basic example of this would be

1. (3f + f)(x) = 3f(x) + f(x) = 4f(x)Addition is a count. There are in total 4 functions f(x).

2. (3f − f)(x) = 3f(x)− f(x) = 2f(x)Similarly to addition. There are 2 functions f(x) left if we take one away from thethree.

3. (f2 · f3)(x) = (f(x))2 · (f(x))3 = (f(x))2+3 = (f(x))5.Multiplication with the same base add powers.

4.(f2

f3

)(x) = (f(x))2

(f(x))3 = (f(x))2−3 = (f(x))−1 = 1f(x)

Division with the same base subtracts powers, and the negative one power puts thefunction in the denominator.

Example: 2.3.1:

Find the sum, product, difference, quotient, and composition of each pair of functions.Always the first function then the second. Then evaluate each at the given value.



1. f(x) = 3x+ 3, g(x) = 3x, x = 3

a) Sum

(f + g)(x) = f(x) + g(x)= 3x+ 3 + 3x= 6x+ 3

(f + g)(3) = 6(3) + 3= 21

b) Product

(fg)(x) = f(x) · g(x)= (3x+ 3)(3x)= 9x2 + 9x

(fg)(3) = 9(3)2 + 9(3)= 9(9) + 27= 108

c) Difference

(f − g)(x) = f(x)− g(x)= 3x+ 3− (3x)= 3

(f − g)(3) = 3

d) Quotient(f

g

)(x) = f(x)

g(x)= 3x+ 3

3x= 3(x+ 3)

3x= x+ 3

x(f

g

)(3) = (3) + 3

(3)= 6

3= 2

e) Composition

(f ◦ g)(x) =f(g(x))= 3(3x) + 3= 9x+ 3

(f ◦ g)(3) = 9(3) + 3= 30

2. h(m) = m2+1, g(m) = m+1, m = 1

a) Sum

(h+ g)(m) = h(m) + g(m)= (m2 + 1) + (m+ 1)= m2 +m+ 2

(h+ g)(1) = (1)2 + (1) + 2= 4

b) Product

(hg)(m) = h(m) · g(m)= (m2 + 1)(m+ 1)= m3 +m2 +m+ 1

(hg)(1) = (1)3 + (1)2 + (1) + 1= 4

c) Difference

(h− g)(m) = h(m)− g(m)= (m2 + 1)− (m+ 1)= m2 −m

(h− g)(1) = (1)2 − (1)= 0

d) Quotient(h

g

)(m) = h(m)

g(m)

= m2 + 1m+ 1(

h

g

)(1) = (1)1 + 1

(1) + 1= 1

e) Composition

(h ◦ g)(m) = h(g(m))= (m+ 1)2 + 1= (m+ 1)(m+ 1) + 1= m2 + 2m+ 1 + 1= m2 + 2m+ 2

(h ◦ g)(1) = (1)2 + 2(1) + 2= 5

2.3.2 Domains of Functions

The domain of a function still follows the definition previously stated, all possible inputvalues. Since we know that the things we are considering are functions, we also know thatif we have a value in the domain, it must go to one and only one output value.



•

•input

•

•

•

•

•

•

f(input)

Domain Range

For functions, what are some restrictions that you have seen throughout your mathe-matics career? We have discussed one in the first section. We know that we cannot havedivision by zero, and we may not have a fractional exponent, whose denominator is even,and interior negative.

Theorem 2.3.1:

The domain of a function is (−∞,∞) minus the input values that give use divisionby zero and an even nth root of a negative number. That is, no input value thatwould give us ( )

0 or even√

negative number. To do this

1. Set the denominator of a function equal to zero and solve for all values. Thedomain cannot have these numbers in it.

2. Set the inside any even root to be less than 0 and solve. These values cannotbe in the domain as well.

Example: 2.3.2:

Find the domain of the following functions

1) (h ◦ g)(m) if h(m) = 3m2 and

g(m) = m2

m+1 .

(h ◦ g)(m) =3( m2

m+1 )2

= 32

m2

m+ 1

The only thing we have to worryabout is division by zero. Thishappens when m+ 1 = 0, m = −1.The domain is (−∞,∞) minus −1or (−∞,−1) ∪ (−1,∞)

2) (p · q)(n) if p(n) = 3n 12 and

q(n) = 1n .

(p · q)(n) = 3n 12 · 1

n

= 3n 12

n

= 3n 12−1

= 3n− 12

= 3 1n

12

= 3 1√n

We have two restrictions,√n = 0

and√n < 0. So n cannot be 0 and

cannot be less than zero. Thedomain is (−∞,∞) minus 0 andanything less than zero or (0,∞)



3) (a+ b)(c) if a(c) = 3c2 + 3 andq(n) = c3 + 2c+ 1.

(a+ b)(c) = 3c2 + 3 + c3 + 2c+ 1

= c3 + 3c2 + 2c+ 4

There is no division and nofractional exponents. Hence norestrictions on the domain or(−∞,∞).

4)(θω

)(r) if θ(r) = 3r + 3 and

ω(r) = 6r + 6.

(θ

ω

)(r) = 3r + 3

6r + 6

= 3(r + 1)6(r + 1)

= 12

The domain of a function isdependent on the originalstatement, not the simplifiedversion. The domain here has onerestriction, 6r + 6 = 0, r = −1. Sothe domain is (−∞,∞) minus −1or (−∞,−1) ∪ (−1,∞).


Math 103 Linear Functions and Slope

2.4 Linear Functions and Slope

Let’s first define what a line is.

Goals:1. Use intercepts to

graph a linear func-tion in standardform.

2. Compute a linesslope.

3. Find a lines slopeand y-intercept fromits equation.

4. Graph linear func-tions in slope-intercept form.

5. Graph horizontal orvertical lines.

6. Interpret slope asrate of change.

7. Find a functions av-erage rate of change.

8. Use slope and y-intercept to modeldata.

Definition 2.4.1: A line is a straight one-dimensional figure having no thickness and ex-tending infinitely in both directions. A line is sometimes called a straight line or, morearchaically, a right line (Casey 1893), to emphasize that it has no“wiggles” anywhere alongits length. While lines are intrinsically one-dimensional objects, they may be embedded inhigher dimensional spaces. In either case lines are functions whose variable has apower (degree) of 1. A line is uniquely determined by any two points (x1, y1) and (x2, y2)that lie on the line. That is there is one and only one line which goes through these twopoints.

−8 −6 −4 −2 2 4 6 8

−10

−5

5

10

(x1, y1)•

(x2, y2)•

x

y

Figure 2: Graph of a line through the points (2, 3) and (5, 6). Given by the equationy = x+ 1.

This section is labeled linear functions, because every linear equation is a function. Thisis not true the other way around. Not all functions are linear (lines). It is actually thecase that most functions are not lines. One of the most important and basic parts of linearfunctions is the slope.Definition 2.4.2: The slope of a line is given by the variable m = y2−y1

x2−x1= y1−y2

x1−x2= rise

run .Either one is fine. The slope tells you how fast the graph is going up or down. The rise tellsyou to go up (+) or down (−) and the run tells you to go left (−) or right (+). The slopecalculated in this way gives us the slope of a linear function that goes through the points(x1, y1) and (x2, y2).

The slope of a line represents the average rate of change of the linear function. You maythink of this as how fast the function is changing in it’s y values compared to its x values,or how fast is the function changing in its output, with respects to what we plug in (input).This is not restricted to linear equations only. You may find the average rate of changebetween any two points. This will give you the slope of the line connecting these two points.In terms of a function, we could write the average rate of change of f as

m = f(x1)− f(x2)x1 − x2

.

Example: 2.4.1:

Find the average rate of change of between the two points for the given situation.



1) You are given the points (3, 5) and(7, 1).The average rate of change or slopeof the line connecting (3, 5) and(7, 1) is given bym = 5−1

3−7 = 4−4 = −1.

2) Find the average rate of change onthe interval [0, 3] for the graph

−5 5

−10

−5

5

10

f(x) = x2 x

y

The average rate of change on theinterval [0, 3] is given byf(0)−f(3)

0−3 = (0)2−33

0−3 = −9−3 = 3. This

gives us the slope of the green linethat connects (0, 0) and (3, 9).

2.4.1 Equation of a Line

There are a few different ways the equation of a line to written. I will first write possibleformulas for a line and we will see that they are all the same.

1. Standard Form: Ax + By = C, where A,B, and C are real numbers and A and Bare not both equal to zero.

2. Slope Intercept Form: y = mx + b, where m is the slope of the line and (0, b) isthe vertical intercept. Notice that this equation is called the Slope Intercept Form,because we need the slope, m = y1−y2

x1−x2= y2−y1

x2−x1, and the vertical intercept (0, b) to

create the equation.

3. Point Slope Form: y−y1 = m(x−x1), where, m = y1−y2x1−x2

= y2−y1x2−x1

, is the slope and(x1, y1) is any point on the line. Notice that this equation is called the Point SlopeForm, because we need the slope, m, and a point (x1, y1) on the line to create theequation.

Proof. If the equation of the line is written in standard form Ax + By = C, then we cansolve for y to get

y = C −AxB

= −ABx+ C

B.

By letting −AB = m and CB = b we get the slope intercept form y = mx + b. If we can

show that the point slope form can also be written in slope intercept form, then we haveshown that they are all the same. If equation of the line is written in point slope formy − y1 = m(x− x1), then similarly solve for y we get

y = m(x− x1) + y1 = mx−mx1 + y1.

Don’t forget that x1 and y1 are numbers that form a point on a line, so by lettingmx1+y1 = bwe get the slope intercept form.

2.4.2 Finding the Equation of a Line



Equation of a Line:

The reason why I discussed this is because the way that you write a line is irrelavent. Iwill enforce writing the equation in slope intercept form. If it is not stated in the problemor by me as a correction in class, then you can write the equation of a line in any form youwould like. You will most likely use slope intercept form and point slope form, however,notice in each case the name tells you what you need.

Example: 2.4.2:

Write the equation of a line in slope intercept form using the given information.

1. We know the slope m = − 12 and b = 3.

Look at the requirements for slope intercept form. We need to find the slope andthe vertical intercept. Notice that this is give by slope=m = − 1

2 and verticalintercept (0, b) = (0, 3). Since we have both of these, we can automaticallywrite the equation of a line in slope intercept form

y = −12x+ 3.

2. We have the slope m = 34 , and a point on the line (2, 1).

Notice in this problem we know slope=m = 34 and a point on the line (2, 1).

These are the things needed in the point slope form. So we have

y − 1 = 34(x− 2)

y = 34x−

32 + 1

y = 34x−

12

3. We have two points on the line (4, 3) and (1, 2).Notice in this problem we are missing slope which is required in either form wewould use. We can solve for this using the formula above, m = y1−y2

x1−x2, using

(x1, y1) = (4, 3) and (x2, y2) = (1, 2).

m = 3− 24− 1 = 1

3

Now that we have the slope we can find the intercept using either point or wecan use the slope intercept form and write it in slope intercept form. We willdo both

(a) Finding b.

y = 13x+ b

2 = 131 + b Using point (1,2) and plugging in.

2− 13 = b

b = 53

=⇒ y = 13x+ 5

3



(b) Slope intercept form.

y − y2 = 13(x− x2)

y − 2 = 13(x− 1) Using point (1,2) and plugging in.

y = 13x−

13 + 2

y = 13x+ 5

3

These are the possible situations. We have everything we need as in example 2.4.2 part1 and 2, or we don’t have the information to find the equation of the line directly. So wewill then

1. Find the slope using m = y1−y2x1−x2

. If it is given, then part one is done.

2. Plug in a point using either the slope intercept form or point slope form.

(a) If you used the slope intercept form, then solve for b and plug back in so you canwrite the equation in slope intercept form.

(b) If you used the point slope form, then solve for y and simplify so the equation isin slope intercept form.

Reading the Equation of a Line:

Remember the equation of a line is any one or two variable equation whose variableshave exponent 1. This is true for x = c, y = c, and ax+ by = c. Which of the three differentequations of a line has the most useful information given? Slope intercept form. Notice thatin all the examples, we wrote the equations in slope intercept form. This is because we areable to look at the equation and determine the slope and the vertical intercept instantly.

Example: 2.4.3:

Determine the slope and the vertical intercept of the line.

1) y = 3x+ 2The slope m = 3 and the verticalintercept is (0, 2).

2) 32x = 1

4y + 1

14y = 3

2x− 1

y = 4(

32x− 1

)y = 6x− 4

The slope m = 6 and the verticalintercept is (0,−4).

3) yx = 2y = 2x, so the slope m = 2 and thevertical intercept is (0, 0).

4) x = 1This is a vertical line and has slopem = y1−y2

1−1 = y1−y20 = undefined.

This line is vertical at x = 1 anddoes not touch the vertical axis,hence no vertical intercept.



5) 25y + 2x

3 = 2

25y = −2x

3 + 2

y = 52

(−2x

3 + 2)

y = −53x+ 5

The slope m = − 53 with vertical

intercept (0, 5).

6) y = 4This is a horizontal line and hasslope m = 4−4

x1−x2= 0

x1−x2= 0. This

line is horizontal at y = 4 andtouches the vertical axis at (0, 4).

2.4.3 Graphing a Line

Remember definition 2.4.1 above tells us that any time you have two points, you can de-termine a unique line. Because of this, when you graph a line, the goal will always be tofind two points that are on the line. However you do this if fine. The book has you findthe vertical and horizontal intercept. This is not always the easiest way. Just remember thegoal is to find two points on the graph. You may do this by plugging in and solving for twopoints, finding the horizontal and vertical intercept (plug in 0 for x, then y and solve), oruse the slope to find another points.Note: As quick checks, re-

member that the sign ofslope (+,−) of the line de-termines whether or notthe line is increasing (+going up from left to right)or decreasing (− goingdown from left to right.Another check is to see ifthe line intersects the ver-tical axis on the positiveside or the negative. Thiscan be checked by lookingat wether or not b is posi-tive or negative.

Example: 2.4.4:

Graph the line using the given information. The first three will be the same infor-mation found in example 2.4.2.

1. We know the slope m = − 12 and b = 3. Our goal is to find two points on the

line. We may do this by plugging two x value into the equation y = − 12x + 3

found previously to find two y values. This will give us two points that lie onthe line. This however is more difficult than it need to be. We already haveb = 3, which give us the vertical intercept (0, 3). Us this point and the slopeto find the next point. The slope is also thought of as Rise

Run . Since the slopeis m = − 1

2 = −12 , we know to go down one step and two steps to the right.

Hence the other point is (0 + 2, 3− 1) = (2, 2). You may also just plot this inthe graph by using your pencil to move one space down and two spaces right.Once you draw these points, connect them to create a line. Remember a linecontinues on forever, so don’t stop your picture at the points.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

(0, 3)• (2, 2)

•x

y

2. We have the slope m = 34 , and a point on the line (2, 1).



We could do something similar in this problem since we have already solved forthe equation of the line y = 3

4x−12 . Notice it would be difficult to move three

spaces up and four spaces right starting at the point (0,− 12 ). Remember we

can use any point, so why not use the one given (2,1). Moving from this pointis much easier.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

(2, 1)•

(6, 4)•

x

y

3. We have two points on the line (4, 3) and (1, 2).Since we already have two points we can automatically graph the line by plot-ting the points and connecting them.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

(4, 3)•(1, 2)

•x

y

4. The equation 23x + 3y = 4. In this case we could write the equation in slope

intercept form, but there is no guarantee that the answer will have a easilygraph able intercept. It may be easier to plug in point to find nice points. Theeasiest points to try are x = 0, 1 or y = 0, 1. Since zero is the easiest, we willtry this first.If x = 0, then we have 2

3 (0) + 3y = 4. So y = 43 . This gives us the point (0, 4

3 )which is not a difficult point to graph if you know 4

3 is one and a third.If y = 0, then we have 2

3x+ 3(0) = 4. So x = 32 · 4 = 6. This gives us the point

(6, 0), which is also easy to plot. Hence



−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

(6, 0)•

(0, 43 )•

x

y

Some special examples lines which have a special slope are the vertical and horizontallines.

1. The vertical line is a line that goes straight up and down. Since the slope is how fast aline is going up or down, it is unclear what the slope would be in this situation. Thishappens when the slope formula m = y1−y2

x1−x2has a division by 0. This would make

the slope undefined. All vertical lines have a equation of x =constant, because it isconsistent in the x variable. Look at the example below to see that the x value neverchanges.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

(−4, 1) •

(−4, 4) •

(−4,−2)•

(−4,−4)•

x

y

2. In the horizontal line case, the line is neither going up or down. Since the slope sayshow fast its going up or down, the slope is zero. If you have a horizontal line, youmust write it as y =constant. This is because the y variable does not change. Youmay see this demonstrated below.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

(−4, 3)•

(−6, 3)•

(1, 3)•

(3, 3)•

x

y


Math 103 The Point-Slope Form of the Equation of a Line

2.5 The Point-Slope Form of the Equation of a Line

Goals:1. Use the point-slope

form to write equa-tions of a line.

2. Model data with lin-ear functions andmake predictions.

3. Find slopes andequations of paralleland perpendicularlines.

We have already discussed the point slope form of an equation and how it relates to theslope intercept form. We won’t discuss it further. If you need refer back to the previoussection or 2.5 in your book.

2.5.1 Model Data With Linear Functions and Make Predictions.

Linear functions are one of the most common ways to represent data, and make predictionson that data. One that you may have heard of is called linear regression. What happens ifyou have a function which models the data? You should be able to estimate future values.You could predict population, profit, or even cost of goods.

Example: 2.5.1:

In April 2001, all 250 public schools in Hawaii were closed because the teachers wenton strike. According to the Chicago Tribune (April 18, 2001), the teacher’s unionasserted that Hawaii’s high cost of living and low salaries resulted in high turnoverand made recruiting qualified educators from the mainland difficult. The averagesalary of teachers, s, depends on the cost of living, c. What this means is that ifthe state has a cost of living at 100, then a candy bar that costs $1 there wouldcost $1.33 in Hawaii. You’ve been hired to look over this issue. You are asked tofigure out whether or not the teachers in Hawaii have a right to strike because oflow salaries and whether Hawaii’s teachers should be paid more after considering thecost of living in Hawaii.

State Cost of living, c Salary, sMichigan 94.1 $48,711Pennsylvania 98.8 $48,457New York 109.4 $49,686California 114.9 $46,326Oregon 100.9 $43,789Washington 104.6 $38,530Nevada 103.2 $42,528Colorado 103.3 $38,157Alaska 125 $48,275Arizona 99.3 $34,582Hawaii 133 $40,416

The linear regression line for the data without Hawaii is given by theequation

s = $170.397c+ $25952.8

1. What is the input variable?The variable c or cost of living.

2. What is the output variable?The variable s or salary.

3. Use the equation above to determine the salary of a Hawaii teacher given a costliving of 133. Round your answer to the nearest dollar.s = $170.397(133) + $25952.8 = $48615.601 ≈ $48616

4. There was an offer from the government that salaries can be increased by 8%.Based on your findings from c, what are you going to tell the teachers of Hawaiiwith regards to their salaries. Should they go on a strike? With 8% raise theteacher would make $40416(1.08) = $43649. This is almost $5000 less than theregression model, so they should strike.



The average salary of teacher in Hawaii in 2018 is about $53,000. The inflation ratein Hawaii is 2.16% a year. If teacher pay only kept up with inflation, then theywould get a 2.16% increase every year for the past 17 years to his 2018. This meansthat they should get paid $48616(1.0216)17 = $69912.4. This is without getting araise. All this means is that the same pay back in 2001 is worth $69912.4 currently.Teachers are short by about $17,000 compared to their pay in 2001.

Example: 2.5.2:

The graph below is a scatterplot which represents the attendance rate compared tothe scores of the class.

5 10 15 20 25 30

20

40

60

80

100

•

•

•••

• •

•

•

•

•

•

•

••

Pass

x

y

1. Find the equation of the red line if you know the red line goes through thepoint (0, 41) and (10, 53.5)m = 53.5−41

10−0 = 12.510 = 125

100 = 54 . Hence plugging the slope in to the slope

intercept form we have, y = 54x+b. We could plug in a point to find b, however

notice that (0, 41) is the vertical intercept. Hence y = 54x+ 41.

2. About how many days do you need to have a good chance to pass?Since passing is a 70% and is the output, we can plug in 70 in and get

70 = 54x+ 41

54x = 29

x = 29 · 45

= 1165

≈ 23.2

If you have more than 23 days, then you are more likely to pass than if youhave less than or equal to 23 days attended.

2.5.2 Parallel and Perpendicular Lines.

We will first define what parallel and perpendicular lines are, in order to find properties thatthese pairs of lines must follow.



Definition 2.5.1: Two lines are parallel if they never intersect. That is two lines that nevertouch, are parallel.Definition 2.5.2: Two lines are perpendicular if the point at which they intersect forms aright angle (90◦) between the two lines.

Example: 2.5.3:

Based on the picture, determine if the lines are parallel, perpendicular, or neither.

1.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

x

y

The lines appear to be Parallel.

2.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

x

y

The lines appear to be Perpendicular.

3.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

x

y

The lines appear to be neither.



4.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

x

y

The lines appear to be neither.

5.

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

x

y

The lines appear to be perpendicular.

If you look at these examples carefully you will notice a few properties of parallel andperpendicular lines.

1. Parallel lines go up or down at the same rate. This is saying that their slope is thesame. However this is not the only requirement. In example 2.5.3 part 4, the lines havethe same slope, but they both touch the vertical axis (0, b) at the same point. Thelines are not parallel because they touch everywhere. We need them to have differentvertical intercepts, or different b values.

Theorem 2.5.1:

If the lines are not vertical or horizontal, then two lines are parallel if they havethe same slope but different vertical intercepts.

y = mx+ b and y = mx+ c b 6= c.

If they are vertical or horizontal, then they are parallel if the constant that thex or y is equal to is different in the two equations.

2. Perpendicular lines form 90◦ angles where they touch. This happens when they haveslopes that are negative reciprocals of each other (the negative flip of each other).Notice in example 2.5.3 part 5 that they have the same vertical intercept but are stillperpendicular.



Theorem 2.5.2:

If the lines are not vertical or horizontal, then two lines are perpendicular ifthey have slope which are negative reciprocals of each other.

y = m

nx+ b and y = − n

mx+ c.

If they are vertical or horizontal, then they are perpendicular if one is a verticalline and the other is a horizontal line.

Example: 2.5.4:

For each line, find a line that is perpendicular and one that is parallel to the onegiven.

1. y = 3x+ 2The line y = 3x + 4 is parallel since it has the same slope and b is different.The line y = − 1

3x+ 2 is perpendicular since it has a negative reciprocal slopefrom the original.

2. y = 32x−

34

The line y = 32x + 4 is parallel since it has the same slope and b is different.

The line y = − 23x+ 2 is perpendicular since it has a negative reciprocal slope

from the original.

3. x = 3A parallel line to x = 3 is any line x =constant as long as its not equal to 3,so x = 5 is parallel. A perpendicular line is any line y =constant, so y = 3 isperpendicular.

Systems of Lin-ear Equations

Math 103 Systems of Linear Equations in Two Variables

3 Systems of Linear Equations

Goals:1. Determine whether

an ordered pair is asolution of a systemof linear equations.

2. Solve systems oflinear equations bygraphing.

3. Solve systems oflinear equations bysubstitution.

4. Solve systems of lin-ear equations by ad-dition.

5. Select the most ef-ficient method forsolving a system oflinear equations.

6. Identify systemsthat do not haveexactly one ordered-pair solution.

A system of linear equations is a list of linear equations. The goal is to figure out the valuesof the variable that will make the equation of each linear equation in the system true. Forexample in a simple system

y = x+ 1

y = −x+ 2

the value x = 0 and y = 1 fails (plug in to check). The goal is to find all such point thatwill make each equation true. If you are ever given a point/s and they ask if it is a solutionto the system, then plug the values into both equations and see if it works. If it fails in oneequation, the point is not a solution.

3.1 Systems of Linear Equations in Two Variables

We have already discussed the goal of a system of linear equations, but what is a systemof linear equations. Linear equations are lines, and a system of linear equations is a list oflines. Think about the ways that lines can be drawn.

1. Touch at a point

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

•x

y

2. Do not touch

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

x

y



3. Lie on each other.

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

(4,4)

x

y

A system of linear equations is asking, ”where do the lines touch”. Because there areonly three possible situations, your answer will either be

1. They touch at the point (x, y), so the solution is (x, y).

2. They do not touch. They are parallel, so there is no solution.

3. They lie on top of each other, so there are infinitely many solutions.We will calculate these situations in two ways

Instructions: 3.1.1:Substitution Method for Solving a System of Linear Equations:

1. Pick a variable in one of the equations. (The easiest one)

2. Solve for that variable.

3. Plug the solution into the other equation. (The equation you did not use.)

4. After plugging in, solve for the other variable.

(a) You get a numerical value for the variable. Continue with the process.(b) You get all the variables to cancel and you get a false statement like 0 = 2,

then the lines are parallel and there are no solutions.(c) You get all the variables to cancel and you get a true statement like 0 = 0,

then the lines are on top of each other and there are infinitely many solutions.

5. Take the number and plug back into either equation to get a value for the firstvariable chosen.

6. Write your solution as a point.

Instructions: 3.1.2:Addition/Elimination Method for Solving a System of Linear Equations:

1. Line up the variables, the equals sign, and the constants by moving them aroundalgebraically.


3. Make the constant in front of the variable chosen the same but different sign ineach equation. Do this by multiplying one equation by a constant to both sides.

4. Rewrite the new equation and the one you did not multiply by a constant under



each other.

5. Add straight down.

6. Solve for the remaining variable

(a) You get a numerical value for the variable. Continue with the process.(b) You get all the variables to cancel and you get a false statement like 0 = 2,

then the lines are parallel and there are no solutions.(c) You get all the variables to cancel and you get a true statement like 0 = 0,

then the lines are on top of each other and there are infinitely many solutions.

7. Take the number and plug back into either equation to get a value for the firstvariable chosen.

8. Write your solution as a point.

Example: 3.1.1:

In the graph, blue is always the first equation and red is always the second equation.

1. Find the solution of the system,

0 = 4 + x− 2y

7x = 4y + 12

Substitution method:

x = 2y − 4 using the first

7(2y − 4) = 4y + 12 Plugging into the second

14y − 28 = 4y + 12

10y = 40

y = 4

0 = 4 + x− 2(4) Plug y = 4 into either one

x = 4

(4, 4)

Addition method:

x− 2y = −4

7x− 4y = 12

−2x+ 4y = 8 Multiply the first by -2

5x = 20 Add to the second

x = 4



(4)− 2y = −4 Plug x=4 into either one

−2y = −8

y = 4

(4, 4)

To graph follow section 2.4.3.

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

(4,4)

x

y

This says that the lines intersect at the point (4,4). The solution is (4,4).


0 = −84 + 24x+ 21y

27 = −3y + 157 x


y = 57x− 9 Using the second

0 = −84 + 24x+ 21(57x− 9) Plug into the first

0 = −84 + 24x+ 15x− 189

0 = 39x− 273

39x = 273

x = 7

27 = −3y + 157 (7) Plug x = 7 into either one

27 = −3y + 15

−3y = 12

y = −4



So we have (7,−4).Addition method:

0 = −27− 3y + 157 x

0 = −84 + 21y + 24x

0 = −189− 21y + 15x multiply the first by 7.

0 = −273 + 39x Add to the second

39x = 273

x = 7

27 = −3y + 157 (7) Plug x = 7 into either one

27 = −3y + 15

−3y = 12

y = −4

(7,−4)

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

(7,-4)

x

y

The lines touch at the point (7,−4). Hence the solution of the system is (7,−4).


−1− 14y + 3

8x = 0

2y − 3x = −4.


y = 32x− 2 using the second



−1− 14(3

2x− 2) + 38x = 0

−1− 38x+ 1

2 + 38x = 0

−1 + 12 = 0

−12 = 0 which is false, so no solution.

Addition method:

0 = −1− 14y + 3

8x

0 = 4 + 2y − 3x

0 = −8− 2y + 3x multiplying the first by 8.

0 = −4 Add to the 2nd

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

x

y

The lines are parallel, they do not touch and hence there is no solution.


6 + 3y + 2x = 2x

0 = −y − 2


y = −2 using the second

6 + 3(−2) + 2x = 2x Plug into the first



0 = 0

true =⇒ Infinitely many solutions

Addition method:

0 = 3y + 6

0 = −y − 2

0 = −3y − 6 multiply the second by 3

0 = 0 add to the first

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

x

y

They lie on top of each other, so there are infinitely many solutions.


Math 103 Problem Solving and Applications Using Systems of Equations

3.2 Problem Solving and Applications Using Systems of Equations

Goals:1. Solve problems us-

ing systems of equa-tions.

2. Use functions tomodel revenue, cost,and profit, andperform a breakeven analysis.

We will use the methods to solve systems of linear equations in a more realistic setting. Yes,this does require word problems. Now what you should do when solving word problemsis write out the important information, and get rid of all other unnecessary wording. Forexample, if you know there are 5 cats, you can just write 5 cats instead of there are 5 cats.We also want to know what we need to find. These are the first things we should do.

Example: 3.2.1:

One of a super markets most profitable items is drinks. During one week they solda total of 5032 drinks worth $ 9116. If all they sold was water at $1.50 and soda at$1.99, how much of each drink did they sell?We know that we want to find the number of drinks sold.

1. Let n = number of waters sold and m = number of sodas sold.

We also need to write the things we know.

1. We know that we sold a total of 5032 drinks.

2. Water cost $1.50.

3. Soda cost $1.99.

4. The total sale is $ 9116.

Using this information we can get

1. number of waters sold+number of sodas sold=n+m=5032

2. $1.50 · n+ $1.99 ·m = $9116

We have two linear equations, so we can solve using the addition/elimination orsubstitution methods.

n = 5032−m

$1.50 · (5032−m) + $1.99 ·m = $9116

$7548− $1.5m+ $1.99 ·m = $9116

$.49 ·m = $9166− $7548

m = 1568.49

m = 156849

100

m = 1568 · 10049

m = 32 · 100

m = 3200

Plugging back in we get

n = 5032− 3200

= 1832

We should check our answer by plugging back into the other equation $1.50 · 1832 +$1.99·3200 plug into your calculator and you will get $ 9116. This says that the super



market sold 1832 units of water and 3200 units of soda. As a picture the two functionscreated in the “using this information” step touch at the point (m,n)=(3200,1832).

1,000 2,000 3,000 4,000 5,000

500

1,000

1,500

2,000• (3200,1832)

m

n

Figure 3: The red line is function that represents the number of drinks sold, whilethe blue line is the function that represents the sale in dollars.

We may also use systems of linear equations to solve problems involving personal finances.

Example: 3.2.2:

You invested $5000 in the stock market. Part of your investment increased by 7%,but the rest fell by 10%. You had a total loss of $211. How much money was investedin each rate?We want to know how much money was invested in the two different rates.

1. Let n = money invested in the 7% rate and m = money invested in the 10%loss.


1. We know that we invested $5000.

2. n gained 7%.

3. m losed 10%.

4. The total loss was $211.


1. money invested in the 7% rate + money invested in the 10% loss=n+m=$5000

2. 7% of n+(-10%) of m=loss of $211 or .07(n)− .1(m) = −$211.

We have two linear equations, so we can solve using the addition/elimination orsubstitution methods.

n = $5000−m

.07($5000−m)− .1(m) = −$211

$350− .07m− .1m = −$211



−.17m = −$561

m = −$561−.17

m = $56117

100

m = $561 · 10017

m = $33 · 100

m = $3300


n = $5000− $3300

= $1700

We should check our answer by plugging back into the other equation .07($1700) −.1($3300) plug into your calculator and you will get -$ 211. This says that we invested$1700 in a stock that gained 7% while investing $3300 in stocks that lost 10%. As apicture the two functions created in the “using this information” step touch at thepoint (m,n)=(3300,1700).

1,000 2,000 3,000 4,000 5,000

500

1,000

1,500

2,000

•(3300,1700)

m

n

Figure 4: The red line is function that represents dollars invested, while the blue lineis the function that represents the loss of investment.

Another way that we use a system of linear equations is to model break even points ofbusinesses. You would like to know when your revenue will equal your cost. This will alsotell you after how many items sold will you be profitable.

Theorem 3.2.1:

1. The Revenue function is given by R(x) =price per unit sold (x)

2. The Cost Function is given by C(x)=fixed cost+cost per unit(x)In both cases x represents the number of units sold/made. This is assumingthat you only make product that you will sell.



Example: 3.2.3:

A firm producing Blu-rays finds that its fixed cost is $40,000 per title and has apressing cost of $2.00 per Blu-ray. If the company sells the Blu-rays at $20 a piece,what is the break even point of the company?We want to know break even point of the company

1. Let n = units created/sold.


1. The initial cost is $40,000

2. Every Blu-ray cost $2.00 to make.


1. R(x) = $20.00(n)

2. C(x) = $40, 000 + $2.00(n)

We have two linear equations, where we want to know how many Blu-rays we needto sell in order to have a break even (revenue=cost orR(x) = C(x).)

$40, 000 + $2.00(n) = C(x) = R(x) = $20.00(n)

$40, 000 = $18.00n

n = 40, 00018

n = 2222.2


R(x) = $20(

40, 00018

)= $44, 444.4

This says that the revenue and cost will be the same when we sell 2222.2 Blu-rays orhave a revenue of $44, 444.4. Since we cannot sell or have a repeated decimal dollaramount, we should round and say that we need to sell at least 2223 Blu-rays or makeat least $44,445. After this amount the company will be profitable.

Some other examples have to do with motion. You may think of a constant force that iseither being used with or against your motion. Think about a constant stream of air beingpushed against you as you run or with you (against your back) as your run.

Theorem 3.2.2:

Constant force applied to a constant rate gives you a new rate depending on theforces direction.

1. Rate of motion+Rate of force=x+ y, if the force is supporting the motion.

2. Rate of motion-Rate of force=x− y, if the force is against the motion.

As always (rate)(time)=distance.



Example: 3.2.4:

A canoe can travel 22 miles in 3 hours with a current but against the current theycan cover half the distance in 2 hours. How fast is the boat in still water?We want to know how fast is the boat in still water.

1. Let n = average rate of the canoe and m= average rate of the water.


1. Travel 22 miles in 3 hours.

2. 12 of 22 in 2 hours.


1. 3 hours (Rate of motion+Rate of force)=distance=3(n+m) = 22

2. 2 hours (Rate of motion-Rate of force)=distance2(n−m) = 11

We have two linear equations, that both have distance

3(n+m) = 22

3n+ 3m = 22

6n+ 6m = 44

2(n−m) = 11

2n− 2m = 11

6n− 6m = 33

6n+ 6m = 44

+ 6n− 6m = 33

12n = 77

n = 7712


3 · 7712 + 3m = 22

774 + 3m = 22

3m = 22− 774

3m = 88− 774

m = 1112

We should check our answer by plugging back into the other equation 2 · 7712 − 2 · 11

12plug into your calculator and you will get 11. This says that rate of the rate of the



canoe is 7712 miles per hour in still water, and the water is traveling at a rate of 11

12miles per hour. As a picture the two functions created in the “using this information”step touch at the point (m,n)=( 11

12 ,7712 ).

2 4 6 8 10

2

4

6

8

10

• ( 1122 ,

7712 )

m

n

Figure 5: The red line is function that represents the current of the water supportingthe canoe, while the blue line is the function that represents the current of the wateragainst the canoe.

Lets do a more relatable example and interesting problem.Example: 3.2.5:

On January 13, 2018 a false ballistic missile alert was issued. This alert was givenin a time where there was increased tension between North Korea and the Unitedstated. The assumption by many on the island was a nuclear missile strike fromNorth Korea was headed towards the island. A nuclear missile travels at a speed ofabout 15000 miles per hour. Lets say we have a anti ballistics missile (ABM) the cantravel at 9000 miles per hour. We also know the distance between north Korea andHawaii is 4,661 miles. Assume that it takes us 10 minutes to get an alert and fire theABM. Answer the following,

1. How long will each missile be in the air?

2. How far away will the interception be?

3. Is it a safe distance away if the nuclear fall out is around 45 miles (largestnuclear bomb in history)?

1. How long will each missile be in the air? We want to know how long eachmissile will be in the air.

(a) Let n = how long in hours the nuclear missile is in the air and m = howlong in hours the ABM is in the air.


(a) A nuclear missile travels at 15000 miles per hour.(b) Our ABB travels at 9000 miles per hour.(c) It takes us 10 minutes to fire the ABM.(d) The distance between North Korea and Hawaii is 4661.




(a) speed of nuke (n) + speed of ABM (m)=4661=15000(n)+9000(m) = 4661(b) n − 1

6 = m since 10 minutes is 16 of an hour. (All our units must be the

same.)

We have two linear equations, so we can solve using the addition/eliminationor substitution methods.

n− 16 = m

15000(n) + 9000(n− 16) = 4661

15000n+ 9000n− 1500) = 4661

24, 000n = 6161

n = 616124000


m = 616124000 −

16

= 216124000

We should check our answer by plugging back into the other equation15000( 6161

24000 ) + 9000( 216124000 ) plug into your calculator and you will get 4661.

This says that the nuclear missile will be in the air for 616124000 hours and the

ABM will be in the air for 216124000 hours. As a picture the two functions created

in the “using this information” step touch at the point (m,n)=( 2161

24000 ,6161

24000).

2. How far away will the interception be?The missile will be intercepted at a distance of m(9000) = 2161

24000 ·9000 = 810.375miles.

3. Is it a safe distance away if the nuclear fall out is around 45 miles (largestnuclear bomb in history)?This is a safe distance, assuming all the items that we did. For example;we assumed that we would be able to know that a missile was coming andlaunch a ABM in 10 minutes, we assumed the speed of the ABM and nuclearmissile, missile don’t travel in a straight line so the distance from North Koreato Hawaii will change (They go up and travel in a arc down to the target.),and we assumed that we could hit the incoming threat. Just as a tid bit ofinformation, the following picture represents the actual impact that the largestnuclear missile in history would have on our island.



Figure 6: The concentric circles represent fireball(3.97 miles), air blast fatal (20.3miles), thermal radiation (45.8 miles), and air blast injury (57 miles).


Math 103 Systems of Linear Equations in Three Variables

3.3 Systems of Linear Equations in Three Variables

Goals:1. Verify the solution

of a system of linearequations in threevariables.

2. Solve systems oflinear equations inthree variables.

3. Identify inconsistentand dependent sys-tems.

4. Solve problems us-ing systems in threevariables.

The main idea of systems of linear equations still holds true in the three variable case. Youwant to know where the graphs of these equations touch all at the same spot. Three linearvariable equations (ax+ by+ cz = r) are 3 dimensional planes. Hence we want to see wherethe three planes all touch at the same spot. It happens to be the case that in order to solvea system of three variable, we need to have 3 equations. This means that to check if a pointrepresents a solution of a system of three variable, you need to plug in the point into eachequation and see if you get a true statement. If it fails in any equation, then it is not asolution. For example is (x, y, z) = (−1, 2,−2) a solution to

x+ 2y − 3z = 9

2x− y + 2z = −8

−x+ 3y − 4z = 15.

This will be left up to the reader to check that this is actually a solution.

3.3.1 Cases of Systems of 3 Equations With 3 Variables

A system of 3 equations with 3 variables has more situations than a system of 2 equationswith two variable. We will now give a definition for two different types of systems as wellas demonstrate the different cases.Definition 3.3.1: A system of linear equations is called consistent if it has one solution anddependent if it has infinitely many solutions.Definition 3.3.2: A system of linear equations is called inconsistent if it has no solution.

The following systems illustrate the different cases of an inconsistent system of 3 equa-tions and 3 variables.

−2020

−5

510

20

y

z

−2020

−5

520

y

z

−2020

−5

520

y

z

Notice that in each case the planes do not all touch in a single spot. The first does nottouch at all. The second blue plane touch both the red and green plane, but the green andred do not touch each other. Finally, the third has all planes touching each other, but nonetouch in the same spot at the same time.

The next illustrations will demonstrate the types of consistent/dependent systems.

−2020

−5

5

−20

20y

z−20

20−5

5

−1

1y

z

−2020

−5

5

−200

200y

z



Similarly there are three cases. The first is where the three planes intercept at a singlepoint (consistent system). The second is when all three planes are exactly the same and lieon top of each other (dependent system). The third is where the planes intersect on a line(dependent system). For this class, we will be able to determine if we have case 1, but wewon’t be able to differentiate between case 2 and 3. This requires calc 3. What is nice aboutcase 2 and 3 is that they touch at infinitely many points. So similar to the two equation twovariable case, we will have a true statement and you may just state that there are infinitelymany solution, hence the system is dependent.

3.3.2 Solving Systems of 3 Equations With 3 Variables

The procedure of solving a system of 3 equations with 3 variables is very similar to theprocedure for solving a 2 equation system with 2 variables.

Instructions: 3.3.1:Substitution Method for Solving a System of 3 Linear Equations with 3 Variables:


2. Solve for that variable.

3. Plug the solution into the other equations. (The equations you did not use.)

4. Now we have a system of 2 equation in two variables. Solve like you did previously.

Instructions: 3.3.2:Addition/Elimination Method for Solving a System of 3 Linear Equations with 3

Variables:

1. Line up the variables, the equals sign, and the constants by moving them aroundalgebraically.

2. Pick two pairs of equations and a variable single variable that you are going toeliminate in the pairs. (The easiest one)

3. Make the constant in front of the variable chosen the same but different sign ineach equation. Do this by multiplying one equation by a constant to both sides.Do this for each pair.

4. Rewrite the new equation and the one you did not multiply by a constant undereach other.

5. Add straight down.

6. You have now eliminated one variable and have a 2 equation system of 2 variables.Solve like you did previously.

Note: If you happen topick two equations thatgive you a true statementafter the first eliminationor substitution method,then you must check theother pair to see if it is de-pendent of inconsistent. Ifthe other pair is inconsis-tent, then the whole thingis as well. If the otherpair gives you either a truestatement or an equationof 2 variables, then thesystem is dependent. Ifyou get true statement atthe end then it is depen-dent.

In order to conclude the case of a dependent system, you must use all the equations. Itis possible that the two equations you picked are dependent with each other, but the threeequations are not dependent. You can imagine two equations that lie on top of each otherbut the third is parallel. All three equations will never touch at the same point. Let’s nowdo a few examples.



Example: 3.3.1:

1. Solve the system of equations.

x+ 3y + 5z = 20

y − 4z = −16

3x− 2y + 9z = 36


y = −16 + 4z

x+ 3(−16 + 4z) + 5z = 20 plug in to first

x− 48 + 12z + 5z = 20

x+ 17z = 68

3x− 2(−16 + 4z) + 9z = 36 plug in to third

3x+ 32− 8z + 9z = 4

3x+ z = 4

Now we have two equations x+ 17z = 68 and 3x+ z = 4. We may use either ofthe methods used previously. I can see that there is a z which is easily solvedfor in 3x+ z = 72, so I will stick with substitution.

z = −3x

x+ 17(4− 3x) = 68

x+ 68− 51x = 68

−50x = 0

x = 0

3(0) + z = 4 Plug x = 0 in

z = 4

Plug x = 0 and z = 4 into any of the originals.

y − 4(4) = −16

y − 16 = −16

y = 0

This system is consistent and the graphs touch at (0, 0, 4).



Addition/Elimination:It is already lined up, but for the sake of ease I will throw in 0x into the second.

x+ 3y + 5z = 20

0x+ y − 4z = −16

3x− 2y + 9z = 36

I will use equations 2 and 3,

0x+ 2y − 8z = −32 multiply by 2

+ 3x− 2y + 9z = 36

3x+ z = 4

as well as 1 and 2

x+ 3y + 5z = 20

+ 0x− 3y + 12z = 48 multiply by −3

x+ 17z = 68

to get 3x + z = 4 and x + 17z = 68. For the sake of consistency I will solveusing the addition method.

−51x− 17z = −68 multiply by −17

+ x+ 17z = 68

−51x = 0

x = 0

Plugging back in the way we did in the substitution method will yield the sameresult.

2. Solve the system of equations.

2x− 4y + 3z = 17

x+ 2y − z = 0

4x− y − z = 6

Addition/Elimination:I think the substitution method is pretty straight forward, so I will do this one



in the addition method. I choose my pairs as

x+ 2y − z = 0

4x− y − z = 6

and

2x− 4y + 3z = 17

x+ 2y − z = 0

Eliminating x in both pairs to get

−4x− 8y + 4z = 0 multiply by −4

+ 4x− y − z = 6

−9y + 3z = 6

and

2x− 4y + 3z = 17

+ − 2x− 4y + 2z = 0 multiply by −2.

−8y + 5z = 17

So we have −9y + 3z = 6 and −8y + 5z = 17. Picking z, multiply the first by 5 andthe second by −3 to get

−45y + 15z = 30 multiply by 5

+ 24y − 15z = −51 multiply by −3

−21y = −21

y = 1

Plug in to −8y + 5z = 17 to get

−8(1) + 5z = 17

5z = 25

z = 5

Finally plug back into one of the originals.

x+ 2(1)− 5 = 0

x− 3 = 0

x = 3

The system is consistent, and the graphs touch at (3, 1, 5).

We now need to have some examples of problems that have no solution or infinitely manysolutions.



Example: 3.3.2:

1. Solve the system of linear equations.

3x+ 4y + 5z = 8

x− 2y + 3z = −6

2x− 4y + 6z = 8

Addition/Elimination Method:I will try to eliminate x, so my pairs will be

3x+ 4y + 5z = 8

x− 2y + 3z = −6

x− 2y + 3z = −6

2x− 4y + 6z = 8

We will get

3x+ 4y + 5z = 8

+ − 3x+ 6y − 9z = 18 multiply by −3

10y − 4z = 26

−2x+ 4y − 6z = 12 multiply by −2

+ 2x− 4y + 6z = 8

0 = 20

Since the graphs of x− 2y+ 3z = −6 and 2x− 4y+ 6z = 8 never touch, it isn’tpossible for all three to touch in the same spot. Hence no solution.

2. Solve the system of linear equations. infinite

x+ 2y + z = 4

3x− 4y + z = 4

6x− 8y + 2z = 8

Addition/Elimination Method:



I will try to eliminate z, so my pairs will be

x+ 2y + z = 4

3x− 4y + z = 4

Modifying the second equation we have

3x− 4y + z = 4

6x− 8y + 2z = 8

Using the pairs directly we get

−x− 2y − z = −4 multiply by −1

+ 3x− 4y + z = 4

2x− 6y = 0

and using the third equation and the modified second equation we get

−6x+ 8y − 2z = −8 multiply by −2

+ 6x− 8y + 2z = 8

0 = 0

Since one pair gives me a 2 variable equation and the other gives me a truestatement, the system is dependent and has infinity many solutions.

Inequalities andProblem Solving

Math 103 Solving Linear Inequalities

4 Inequalities and Problem Solving

Goals:1. Solve linear inequal-

ities.

2. Recognize inequali-ties with no solutionor all real numbersas solutions.

3. Solve applied prob-lems using linear in-equalities.

The nice part about this sections is that we have essential already did the material inchapter 1 and 2. Inequalities are in chapter one and linearity is in chapter 2. We will nowbe combining inequalities and linear functions to get results. Again the most important partis your ability to read to problem. If you can do this, this section will be easy.

4.1 Solving Linear Inequalities

Solving linear inequalities is exactly like solving for a variable. Every property that youused previously will work in this case. There are only two differences between solving withan equal sign and an inequality.

Theorem 4.1.1:

Things you must remember when solving inequalities.

1. If you divide or multiply by a negative number, the inequality switches direc-tion.

2. You cannot multiply or divide both sides by a variable unless you know it isalways positive or negative. This is because we do not know if the value of thevariable is positive or negative. It may change the direction of the inequality.

Example: 4.1.1:

Solve the inequality 3x+ 2 ≤ 2x+ 5.What if I just made pretend that the ≤ was an equal sign, then solve. We wouldhave

3x+ 2 = 2x+ 5

x+ 2 = 5

x = 3

Now replace the equal sign with the ≤. This is the solution. x ≤ 3. Notice if we didthe same procedure but left the ≤ symbol, we would get the same thing

3x+ 2 ≤ 2x+ 5

x+ 2 ≤ 5

x ≤ 3

This solution says, any value of the variable x that is smaller than or equal to 3

−5−4−3−2−1 0 1 2 3 4 5

•x

will make the inequality 3x+ 2 ≤ 2x+ 5 true.



−8 −6 −4 −2 2 4 6 8

−12−10−8−6−4−2

2468

1012 •

x

y

Figure 7: The red is represents 2x + 5 and the blue represents 3x + 2. The greenrepresents the x value of 3.

Notice that for all values of x that are smaller than 3, the equation 3x+ 2 is smallerthan or equal to 2x+ 5. Hence 3x+ 2 ≤ 2x+ 5.

4.1.1 Special Cases

There are inequalities that have all real numbers as solutions and some that have no solutions.Notice that the concept is very similar to solving a system of linear equations. This timewe are trying to see when the lines are either above or below each other.

Example: 4.1.2:

Solve the linear inequality.23x+ 5 ≤ 2

3x+ 3

This question is asking when the line 23x+ 5 is below or equal to the line 2

3x+ 3. Ifyou remember lines really well you don’t need to do anything to solve this. Whatcan you say about these lines? These lines are parallel where the first has verticalintercept (0, 5) and the other has vertical intercept (0, 3). They will never touch, andsince the first is above the second, this linear inequality has no solution. Let nowdemonstrate this algebraically.

23x+ 5 ≤ 2

3x+ 3

5 ≤ 3

Similarly to a system of linear equations, if the end result is false, then there is nosolution.



−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

x

y

Notice that there is no x value that we may plug in to make the blue line lower thanthe red line. Hence no solution.

The second special case that was mentioned is when all real numbers are solutions. Let’suse the same lines, but switch the inequality.

Example: 4.1.3:

Solve the linear inequality.23x+ 5 ≥ 2

3x+ 3

This question is asking when the line 23x + 5 is above or equal to the line 2

3x + 3.Similarly to the equations above, these lines are parallel to each other where the firstintersects the vertical axis at (0, 5), and the other intersects the vertical axis at (0, 3).Notice that since these lines are parallel the line 2

3x+ 5 is always above 23x+ 3. So

all x values satisfy this inequality. Again we will do this algebraically.

23x+ 5 ≥ 2

3x+ 3

5 ≥ 3

Similarly to a system of linear equations, if the end result is true then there isinfinitely many solutions, or all real values of x are solutions.

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

x

y

Notice that every x value that we may plug in to make the blue line above the redline.



4.1.2 Application of Linear Inequalities

In this section we will be discussing the applications of linear inequalities. We can use thelinear inequalities to determine useful information in many different types of problems. Firstwrite the linear equations and determine the inequality that is in between them using thedescriptive words in the problem. We will discuss a few examples.

Example: 4.1.4:

You are offered two jobs from two different companies. One has a base pay of $35,000a year with a salary increase of $500 every year. The other job starts at $50,000 ayear with a $100 increase every year.

1. For how long will the first job be paid less?The equations for the jobs salary are given by s = $35, 000 + $500t and s =$50, 000 + $100t. We want to know when salary one is less than salary two, or

Salary one ≤ Salary two

$35, 000 + $500t ≤ $50, 000 + $100t.

Hence

$35, 000 + $500t ≤ $50, 000 + $100t

$400t ≤ $15, 000

t ≤ 15, 000400

t ≤ 37.5

This says that salary one will be less than salary 2 for 37 years.

2. If you retire at 60 and started work at 20, how many years will you have at ahigher pay?You will have higher pay for 3 years?

3. Which job is pays better in the long run?Clearly job 2 is better. You will be making more money for 37 years of the job.

5 10 15 20 25 30 35 40 45 50

1

2

3

4

5

6

7

8·104

•

x

y


Math 103 Compound Inequalities

4.2 Compound Inequalities

Goals:1. Find the intersec-

tion of two sets.

2. Solve compound in-equalities involvingand.

3. Find the union oftwo sets

4. Solve compound in-equalities involvingor.

A compound inequality is the joining of multiple inequalities using the words “and” (∩) or“or” (∪). The word “and” asks for the values that make both the inequalities true, while“or” asks for the values that make one, the other, or both of the inequalities true.

Example: 4.2.1:

1. What values make the statement x < 2 or x > 5 true?Notice that these inequalities are represented in a picture as,

−8 −6 −4 −2 0 2 4 6 8

◦ ◦x

To write the blue line in interval notation we write, (−∞, 2). To write the redline in interval notation we write (5,∞). Since the problem is asking for thevalues that will make one, the other, or both true (or), we write (−∞, 2)∪(5,∞)as the answer.

2. What values make the statement x ≥ 2 and x < 5 true?Notice that these inequalities are represented in a picture as,

−8 −6 −4 −2 0 2 4 6 8

•◦

x

To write the blue line in interval notation we write, [2,∞). To write the redline in interval notation we write (−∞, 5). Since the problem is asking for thevalues that will both true (and), we write [2,∞)∩(−∞, 5) as the answer. Sinceand represents the overlap, we can simplify this to [2,∞) ∩ (−∞, 5) = [2, 5).You can see in the picture that the to lines would over lap on the interval [2, 5).This tells you that the values of x that will satisfy both of the inequalities atthe same time are the values in [2, 5).

To calculate more complex inequalities we will follow the procedures from section 4.1.The only difference is the way that we answer the question in the form of a sentence. Wewill begin with the “and” compound inequalities and then work on the “or” compoundinequalities.

4.2.1 And Compound Inequalities

Example: 4.2.2:

Solve 5(x− 2) > 15 and x−64 ≤ −2.

Solve each inequality separately, then graph and find the solution.

1)

5(x− 2) > 15

5x− 10 > 15

5x > 25

x > 5

2)x− 6

4 ≤ −2

x− 6 ≤ −8

x ≤ −2



−8 −6 −4 −2 0 2 4 6 8

◦◦x

Since we want the solutions that satisfy both at the same time (and), there is nosolution. This can be written as a set called the empty set ∅.

We can continue to solve these inequalities in this manner, or we can write it in a differentway so that we do not need to draw the picture. At the same time lets see a different wayof writing a compound linear inequality problem.

Example: 4.2.3:

Let p(n) = 4n+ 5 and h(n) = −4n− 5. Find all values of n for which p(n) ≤ 5 andh(n) ≤ 2.The statements p(n) = 4n+ 5 and p(n) ≤ 5 give us

5 ≥ p(n) = 4n+ 5

as long as the inequality point to the same value it does not matter if you place the5 on the left or the right. Similarly, h(n) = 3n− 4 and h(n) ≤ 2 give us

2 ≥ h(n) = −4n− 5.

Notice that if we multiply both sides of 2 ≥ h(n) = −4n− 5 by negative one to get

−2 ≤ 4n+ 5.

This is the same function on the right hand side as the inequality 5 ≥ 4n + 5. Wemay combine these inequalities to form the inequality

−2 ≤ 4n+ 5 ≤ 5.

Solving this by applying operations to the left, the right, and the middle.

−2 ≤ 4n+ 5 ≤ 5

−7 ≤ 4n ≤ 0

−74 ≤n ≤ 0

Writing this solution as an interval we get [−74 , 0]. The values of n in-between − 7

4and 0 will satisfy both of the inequalities at the same time.

This method is tedious, especially if the functions are not the same. It may be easier todo the graphical method. The situation that makes this method easier is if the problem isalready given as a ≤ function ≤ b or a < function < b where a and b are numbers.

4.2.2 Or Compound Inequalities

For “or” compound inequalities you need to solve both separately and graph. Rememberthe “or” statements are the elements in one, the other, or both. Consider this when writingthe solution.

Example: 4.2.4:

1. Solve: 3(2x+4)2 ≥ 5 or 3x+ 1 < 2x+ 5.



(a)

3(2x+ 4)2 ≥ 5

3(2x+ 4) ≥ 10

2x+ 4 ≥ 103

2x ≥ 103 − 4

2x ≥ 103 −

123

x ≥ − 23 · 2

x ≥ −13

(b)

3x+ 1 < 2x+ 5

x+ 1 < +5

x < 4

−8 −6 −4 −2 0 2 4 6 8

•◦

x

The intervals are written as (−∞, 4) ∪ [− 13 ,∞). Either pictorially or through

the intervals, notice that if we pick values that satisfy one, the other, or both,then we get all value of R. So all real numbers (−∞,∞) are solutions thatsatisfy one, the other, or both inequalities.

2. 3p+ 2 ≤ 5 or 5p− 7 ≥ 8

(a)

3p+ 2 ≤ 5

3p ≤ 3

p ≤ 1

(b)

5p− 7 ≥ 8

5p ≥ 15

p ≥ 3

−8 −6 −4 −2 0 2 4 6 8

• • p

The intervals are written as the union (−∞, 1] ∪ [3,∞). There is no way tosimplify this, so the solution that satisfies, one, the other, or both inequalitiesare the values of p in (−∞, 1] ∪ [3,∞).


Math 103 Equations and Inequalities Involving Absolute Value

4.3 Equations and Inequalities Involving Absolute Value

Goals:1. Solve absolute value

equations.

2. Solve absolute valueinequalities of theform |u| < c .

3. Solve absolute valueinequalities of theform |u| > c .

4. Recognize absolutevalue inequalitieswith no solution orall real numbers assolutions.

5. Solve problems us-ing absolute valueinequalities.

If you remember from section 1.2.2 the mathematical definition of the absolute value, thenthis section will make more sense. The definition is the piece wise function

f(x) = |x| ={x x ≥ 0−x x < 0

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−6−5−4−3−2−1

123456

x

y

The definition tells us to take the inside of the absolute value and do two things. In onecase keep the inside of the absolute value the same. In the second case take the inside of theabsolute value and multiply the whole thing by a negative 1. Knowing this definition willallow you to solve both equations and inequalities that have the absolute value in them.

4.3.1 Equations With Absolute Value

When solving equations using the absolute value we will first remove the absolute value,then use the definition of the absolute value. This will give us two cases. In both cases wemust now solve for the variable.

Example: 4.3.1:

Solve the equation

1. |3x| = 9

3x = 9

x = 3

or

−(3x) = 9

x = −3

2. | − 4 + 5x| = 16



−4 + 5x = 16

5x = 20

x = 4

or

−(−4 + 5x) = 16

4− 5x = 16

−5x = 12

x = −125

3. |v + 8| − 5 = 2

(v + 8)− 5 = 2

v + 3 = 2

v = −1

or

−(v + 8)− 5 = 2

−v − 8− 5 = 2

−v = 15

v = −15

4. 3| − 8x|+ 8 = 80

3(−8x) + 8 = 80

−24x = 72

x = −3

or

−3(−8x) + 8 = 80

24x = 72

x = 3



5. |7p+4|8 = 3

(7p+ 4)8 = 3

7p+ 4 = 24

7p = 20

p = 207

or

−(7p+ 4)8 = 3

−7p− 4 = 24

−7p = 28

p = −4

6. 2− 5|5m− 5| = −73

2− 5(5m− 5) = −73

−5(5m− 5) = −75

−25m+ 25 = −75

−25m = −100

m = 4

or

2 + 5(5m− 5) = −73

5(5m− 5) = −75

25m− 25 = −75

25m = −50

m = −2

7. 3|3− 5r| − 3 = 18



3(3− 5r)− 3 = 18

3(3− 5r) = 21

3− 5r = 7

−5r = 4

r = −45

or

−3(3− 5r)− 3 = 18

−3(3− 5r) = 21

3− 5r = −7

−5r = −10

r = 2

8. 5|9− 5n| − 7 = 38

5(9− 5n)− 7 = 38

5(9− 5n) = 45

9− 5n = 9

−5n = 0

n = 0

or

−5(9− 5n)− 7 = 38

−5(9− 5n) = 45

9− 5n = −9

−5n = −18

n = 185

4.3.2 Inequalites With Absolute Value

Now that we have a clear understanding of how the absolute value function works, we canimplement the inequality. Lets take some of the problems done previously and see what



would happen if we had an inequality instead.Example: 4.3.2:

1) |3x| ≤ 9

3x ≤ 9

x ≤ 3

or

−3x ≤ 9

x ≥ −3

We want the values of the inputthat make the absolute value of 3xhave a height below or equal tonine.

−10−8 −6 −4 −2 2 4 6 8 10

123456789

10• •

x

y

Remember that “or” inmathematics represents one theother or both. Hence this happenswhen the input is in [−3, 3]. Onething to notice is that if we combinethe two inequalities found in thesolution we get −3 ≤ x ≤ 3.

2) | − 4 + 5x| > 16

−4 + 5x > 16

5x > 20

x > 4

or

−(−4 + 5x) > 16

4− 5x > 16

−5x > 12

x < −125

We want the values of the inputthat make the absolute value of−4 + 5x have a height abovesixteen.

−10−8 −6 −4 −2 2 4 6 8 10

−8

−4

4

8

12

16

20◦ ◦

x

y

We want where the absolute valueof −4 + 5x is above 16. Notice thatthis happens when x < − 12

5 orx > 4. These two inequalitiescannot be combined, so our answergiven by for all values in(−∞,− 12

5 ) ∪ (4,∞) the function| − 4 + 5x| is above 16.

These examples lead to a way of calculating these types of problems.Theorem 4.3.1:

1. If the absolute value is given by |f(x)| < c, then we may solve it using −c <f(x) < c. This also works for ≤. To use this, first get the absolute value byitself on the left hand side and ensure that the right hand side is a constant.Then write in the form −c < f(x) < c. To solve this we do our standardalgebra but to the left, the middle, and the right.

2. If the absolute value is given by |f(x)| > c, then we solve using the definition



of absolute value. That is use f(x) > c or −f(x) > c. This also works for ≥.Using the numbers in example 4.3.2 part 2 as a reference, ensure that you donot write your answer as − 12

5 > x < 4 or 4 > x < − 125 . You must write the

statement given in example 4.3.2.

Lets try some more complicated examples that involve more than one step. Then in thelast two problems we will discuss some special cases where we can get the answer directlywithout any algebra.

Example: 4.3.3:

1) 5− 12 |3p− 5| ≤ 3

5− 12 |3p− 5| ≤ 3

−12 |3p− 5| ≤ −2

|3p− 5| ≥ 4

So we use the second case intheorem 4.3.1

3p− 5 ≥ 4

3p ≥ 9

p ≥ 3

or

−(3p− 5) ≥ 4

3p− 5 ≤ −4

3p ≤ 1

p ≤ 13

−10−8 −6 −4 −2 2 4 6 8 10

−8−6−4−2

2468

• •p

y

The function 5− 12 |3p− 5| is below

or equal to 3 for all inputs in(−∞, 1

3 ] ∪ [3,∞).

2)∣∣ 3n+1

2∣∣ ≤ 5 ∣∣∣∣3n+ 1

2

∣∣∣∣ ≤ 5

−5 ≤ 3n+ 12 ≤ 5

−10 ≤ 3n+ 1 ≤ 10

−11 ≤ 3n ≤ 9

−113 ≤ n ≤ 3

−10−8 −6 −4 −2 2 4 6 8 10

−8−6−4−2

2468

• •

n

y

The function∣∣ 3n+1

2∣∣ is below or

equal to 5 for all inputs in theinterval [− 11

3 , 3].



3) | 43x− 5| − 3 ≥ −4Notice that if we solve this we get| 43x− 5| ≥ −1. This is asking whenthe absolute value function | 43x− 5|is above or equal to −1. This isalways true since the absolute valuefunction is always greater than orequal to zero and consequentlyalways above −1. This tells us thatall real numbers are solutions tothis inequality. We write this as,the function | 43x− 5| − 3 is above orequal to 4 for all inputs on(−∞,∞).

−10−8 −6 −4 −2 2 4 6 8 10

−8−6−4−2

2468

x

y

4) −|4r − 3| − 5 > 1Similarly, notice that if we solvethis we get |4r − 3| < −6. This askswhen the absolute value function|4r − 3| is below −6. This should beclear that the absolute valuefunction can never be negative, so itwill never be below −6. This is asituation where we have nosolution. No input will let thefunction −|4r − 3| − 5 be above 1.

−10−8 −6 −4 −2 2 4 6 8 10

−8−6−4−2

2468

r

y

This gives us a theorem about the special cases of inequalities with absolute values.Theorem 4.3.2:

1. The inequality |f(x)| < c, where c is a negative number, has no solution. Thisalso holds for ≤.

2. The inequality |f(x)| > c, where c is a negative number, is true for all realnumbers for which f(x) is defined. This also holds for ≥.


Math 103 Linear Inequalities in Two Variables

4.4 Linear Inequalities in Two Variables

Goals:1. Graph a linear in-

equality in two vari-ables.

2. Use mathematicalmodels involvinglinear inequalities.

3. Graph a system oflinear inequalities.

Note: Notice that thebook does this a little dif-ferently. If you have ques-tions about their versionplease ask. You may do iteither way.

Similarly to all the inequalities above, we must first learn how to read the linear inequalitiesin two variables. This represents an inequality that has two different variables. For example;x and y or m and n.

Example: 4.4.1:

Lets do some simple examples.

1) y > 2x− 5First draw the graph of y = 2x+ 5and we will determine what theproblem is asking.

−10−8 −6 −4 −2 2 4 6 8 10

−8−6−4−2

2468

x

y

Notice that the shaded blue area isthe solutions to the question. Thisis because the problem is asking for,

y>2x− 5

the outputs/heights that are greaterthan/above the line 2x− 5. Alsonotice that the line is dottedbecause we are not allowing theheights that are equal to the linesince we do not have ≤ or ≥.

2) y ≤ − 74x+ 2

Again graph y = − 74x+ 2,

−10−8 −6 −4 −2 2 4 6 8 10

−8−6−4−2

2468

x

y

Similarly for this problem,

y≤−74x+ 2

the outputs/heights that are lessthan or equal to/below or equal tothe line − 7

4x+ 2. This time the lineis solid because we are allowing theheights that are equal to the line.

Note: In the special cases,the problem x > c, for ex-ample, asks us for all theinput/x values that arelarger than a certain num-ber c. This will happen tothe right of the number c.

−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

Instructions: 4.4.1:Solving Linear Inequalities With Two Variables:

1. Solve for the output. Get the output to one side. It does not matter which sidebut conventionally it is solved for on the left hand side.

2. Graph the line or lines (if there are more than one) using a solid line if ≥ or ≤and a dotted line if < or >. If you need refer back to section 2.4.3.Special cases:

(a) If you end up with x > or ≥ a constant c then shade the right hand side ofthe line x = c.

(b) If you end up with x < or ≤ a constant c then shade the left hand side ofthe line x = c.

(c) If you end up with y > or ≥ a constant c then shade above the line y = c.(d) If you end up with y < or ≤ a constant c then shade below the line y = c.



Read what the question is asking and shade the section that is being asked in theproblem.

3. If there is one line drawn then we are done and the shaded portion is our solution.If there is more than one line drawn, the solution is the overlap of the shadedportions.

4.4.1 Systems of Linear Inequalities

Let’s try some problems that require more than one line to be drawn. This is called a Systemof linear inequality. The procedure to solve these are the same as before.

Example: 4.4.2:

Graph and determine the solution set of the system,

y > −x− 1

y ≤ 35x− 5

1) y + x+ 1 > 0a) First solve for the output.

y + x+ 1 > 0

y > −x− 1

b) Graph

−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

2) 53 (y + 5) ≤ x

a) Solve for y

53(y + 5) ≤ x

(y + 5) ≤ 35x

y ≤ 35x− 5

b) Graph

−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

c) Of course you would graph these two together, but for the sake of showing whatwe are doing we will now combine the graphs.

−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

The solution set is the polygon that is shaded green and blue (the overlap).These solutions will satisfy both of the inequalities. That is, any point in that



section has a height strictly below the line −x − 1 (the one going down) andbelow or equal to the line 3

5x− 5 (the one going up).

Another possibility that can occur is the case with no solution. This happens when thereis no overlap in the shading of the solution sets. An example of this is

y ≥ 2x+ 1

y ≤ 2x− 1

These lines are parallel and give us the graph,

−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

Notice that the shaded area does not overlap. Therefore, there is no solution that satisfiesthis system.

Lastly let us do a system which has three lines and one of the special forms stated above.Example: 4.4.3:

Graph the solution set of the system

x− y < 3

−3 ≤ x < 3

y < 3

1. We only need to solve for the first one. Everything else is done.

x− y < 3

−y < 3− x

y > x− 3

2. Graph all three to get

−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

y = x− 3

−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

−3 ≤ x < 3

−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

y = 3

3. Overlapping all three we get,



−10−8−6−4−2 2 4 6 8 10

−8−6−4−2

2468

x

y

The middle trapezoid is the solution set that satisfies all three requirements,strictly above the green line, between inputs −3 and 3, allowing three to be aninput, and strictly below the line y = 3.

Polynomials,Polynomial Func-tions, and Equa-tions

Math 103 Introduction to Polynomial Functions

5 Polynomials, Polynomial Functions, and Equa-tions

Remember from before that every line is a function, but not every function is a line. Many ofthe equations found in this section will not be lines, but will be functions. Similarly though,every polynomial is a function, but not all functions are polynomials. If we think of it as aset we have the property

lines ⊂ polynomials ⊂ functions

We are trying to build up your knowledge of functions by giving you more and more generaltypes in the hopes of building your understanding of functions.

5.1 Introduction to Polynomial Functions

Goals:1. Use the vocabulary

of polynomials.

2. Evaluate polynomialfunctions.

3. Determine end be-havior.

4. Add polynomials.

5. Subtract polynomi-als.

Definition 5.1.1: A polynomial can be written in the form

cnxn + cn−1x

n−1 + . . .+ c2x2 + c1x+ c.

This is called the standard form. Each of the constants cn, . . . , c are real numbers andthe exponents are whole numbers. Polynomials can have subtraction. This isn’t explicitlyshown, however it is possible since the constants are real numbers. This allows for negativenumbers to appear.Definition 5.1.2: The degree of a polynomial with one variable is the largest exponent ofthe polynomial. The degree of a constant is 0 since cx0 = c. However since 0x0 = 0x1 =0x2 = . . . = 0xn we do not have a largest exponent for the constant term 0. Therefore theconstant term 0 has no defined degree.Definition 5.1.3: The terms are the constants, variables with their corresponding exponents,and the combinations of variables and constats that are being added or subtracted together.

Remember that your ability to read math is the key to being able to understand math,so read and understand the definitions.

Using the definitions above determine if the following are polynomials or not.Example: 5.1.1:

Determine if the functions are polynomials. If yes, what are the terms and the degreeof the polynomial?

1) f(x) = 3

This is a polynomial since it is ofthe form 3x0. So its degree is 0 andits only term is 3.

2) g(p) = p2 + 3p7 + 3This is a polynomial since it is ofthe form 3p7 + p2 + 3. So its degreeis 7 and has 3 terms 3p7, p2, and 3 .

3) p(m) = 2m2 +m+ 3 +m−2

This isn’t a polynomial since weneed positive exponents. The −2exponent does not follow thedefinition, so p(m) isn’t apolynomial.

4) h(n) = en2 + πn+√

2

This is a polynomial. Even thoughthe constants are irrational, theyare still real. All the exponents arewhole numbers so h(n) satisfies thedefinition of a polynomial. It hasdegree 2 and the terms are en2, πn,and√

2.



5) g(t) = et

This isn’t a polynomial.Polynomials do not allow variablesto be in the exponent. This howeveris a function, and is called anexponential function.

6) f(θ) = 43θ + 4

This is a polynomial. This is a line,and all lines are polynomials. It hasdegree 1 and the terms are 4

3θ and4.

7) g(γ) = γ23 + γ

This isn’t a polynomial. It has afractional exponent. This isn’tallowed.

8) f(x) = x1000000000

This is a polynomial. Though it hasa very large degree, it follows thedefinition of a polynomial. It has awhole number exponent and realcoefficient 1. Its degree is1000000000 and its only term isx1000000000

We also allow polynomials to have more than one variable. An example of this would be43x

2y + 3x4y2 + 6x+ y5. The only difference in this case is the degree.Definition 5.1.4: The degree of a polynomial with more than one variable is the largestsum of the degrees of the terms.

For the example 43x

2y1 + 3x4y2 + 6x1 + y5, we have the sums as 2+1=3, 4+2=6, 1 and5. So the degree is 6 since it is the largest.

5.1.1 Adding, Subtracting, and Evaluating Polynomials

Note: Remember thatadding or subtracting liketerms is like thinking ofthe variables as names orunits. We cannot add2 cars and 2 cats to get4 . If youneed a refresher go back tosection 1.2.4.

When adding and subtracting polynomials together we need to be careful. We may onlycombine like terms. Remember these are terms that have the exact same variable. Somenotation that will help you read the mathematics are

• (f + g)(x) = f(x) + g(x)= add the functions f and g together.

• (f − g)(x) = f(x)− g(x)= subtract the functions f and g together.

Example: 5.1.2:

Simplify the following expressions.1. Let f(x) = x2 + 3

2x+ 4 and g(x) = 4x+ 53x

2 + 4x4.Find (f + g)(x)

(f + g)(x) = f(x) + g(x)

=(x2 + 3

2x+ 4)

+(

4x+ 53x

2 + 4x4)

= (4x4) +(x2 + 5

3x2)

+(

32x+ 4x

)+ 4

= 4x4 + 33x

2 + 53x

2 + 32x+ 8

2x+ 4

= 4x4 + 83x

2 + 112 x+ 4



2. Let f(x) = x2 + 32x+ 4 and g(x) = 4x+ 5

3x2 + 4x4.

Find (f − g)(x)

(f − g)(x) = f(x)− g(x)

=(x2 + 3

2x+ 4)−(

4x+ 53x

2 + 4x4)

= x2 + 32x+ 4− 4x− 5

3x2 − 4x4

= (−4x4) +(x2 − 5

3x2)

+(

32x− 4x

)+ 4

= −4x4 + 33x

2 − 53x

2 + 32x−

82x+ 4

= −4x4 − 23x

2 − 52x+ 4

3. Let p(m,n) = 3m2n− 4mn3 + 3 and q(m,n) = −2m2n3 +m2n+ 5.Find (p+ q)(m,n).

(p+ q)(m,n) = p(m,n) + q(m,n)

= (3m2n− 4mn3 + 3) + (−2m2n3 +m2n+ 5)

= −2m2n3+3m2n+m2n− 4mn3 + 3 + 5

= −2m2n3+4m2n− 4mn3 + 8

The only terms that have the exact same variable is the part in green. Thereforeeverything else is left alone, and we add the green terms together.

Evaluating polynomials is done in the exact same way as before. If you need a reviewon the instructions, go to section 1.2.4. Lets use the same examples that we have alreadysolved for so that we can do this problem in two different ways.

Example: 5.1.3:

Let f(x) = x2 + 32x+ 4 and g(x) = 4x+ 5

3x2 + 4x4

Find (f + g)(6) and (f − g)(6). We will do this by evaluating each part separatelyas well as the end result of the algebra we did in the previous examples.

1. (f + g)(6)



(a) Since (f + g)(6) = f(6) + g(6) wecan evaluate f and g separately,then add them together.

f(6) = (6)2 + 32(6) + 4

= 36 + 9 + 4

= 49

and

g(6) = 4(6) + 53(6)2 + 4(6)4

= 24 + 53(36) + 1296

= 24 + 60 + 5184

= 5268.

So, f(6) + g(6) = 49 + 5268 =5317.

(b) We also have done the algebra toget that (f+g)(x) = 4x4 + 8

3x2 +

112 x+4. We can plug directly into

this to get the answer.

(f + g)(6)

= 4(6)4 + 83(6)2 + 11

2 (6) + 4

= 4(1296) + 83(36) + 11

2 (6) + 4

= 5184 + 96 + 33 + 4

= 5317

Notice that this is the same solu-tion as the one found earlier.

2. (f − g)(6)

(a) Since (f − g)(6) = f(6)− g(6) wecan evaluate f and g separately,then add them together.

f(6) = (6)2 + 32(6) + 4

= 36 + 9 + 4

= 49

and

g(6) = −4(6)− 53(6)2 − 4(6)4

= −24− 53(36)− 5184

= −24− 60− 5184

= −5268.

So, f(6) − g(6) = 49 − 5268 =−5219.

(b) Similarly, we will plug 6 into (f−g)(x) = −4x4 − 2

3x2 − 5

2x+ 4.

(f − g)(6)

= −4(6)4 − 23(6)2 − 5

2(6) + 4

= −4(1296)− 23(36)− 5

2(6) + 4

= −5184− 24− 15 + 4

= −5219

Which is the same answer

5.1.2 Tail Ends of Polynomials

In this section we will get a general idea of what happens to polynomial as we go further outto the right and to the left. In math we sometimes call this the tail ends of the graphs. Sincethere are two parts that make up polynomials, the degree and the coefficients. These two



parts will determine what the function does in the tail end. When drawing these graphs,make sure that you draw them smooth with no edges. All polynomials are smooth andcontinuous (no breaks).

Instructions: 5.1.1:To graph the tail ends of the function you must have the equation of the polynomial in standard form. Then look

at the degree and the leading coefficient (number attached to the variable with the largest exponent) to determine oneof the four following cases.

1. If the degree is evenand the leading co-efficient is positive,then the graph tendsup on both sides.

x

y

2. If the degree is evenand the leading co-efficient is negative,then the graph tendsdown on both sides.

x

y

3. If the degree is oddand the leading co-efficient is positive,then the graph tendsup on the right anddown on the left.

x

y

4. If the degree is oddand the leading co-efficient is negative,then the graph tendsup on the left anddown on the right.

x

y

Example: 5.1.4:

Determine the how the polynomial functions act in their respective tail ends.

1) f(x) = 4x3 + 3x+ 2

This polynomial is written instandard and has a positive leadingterm with an odd degree. This saysthat the graph will tend down onthe left and up on the right.

2) g(m) = −(3m2 +m) + 1

This is not written in standard formyet. We need to distribute thenegative to get −3m2 −m+ 1. Thishas a negative leading coefficientand an even degree. So the graphtends down on both sides.

3) p(n) = x100000000 + 2

This is written in standard formand has a positive leadingcoefficient and an even degree. So ittends up on both sides.

4) f(t) = 43 t

3 − 4t4

This polynomial is in standard formeven if the terms are in the wrongorder. Remember the degree is thelargest power, so it has an evendegree 4 with negative leadingcoefficient. This says that the graphwill tend down on both sides.

5) h(p) = 0p17 − 4p3

This polynomial is actually −4p3,and it has odd degree 3 with anegative leading coefficient. Thismeans that the graph will tend upto the left and down to the right.

6) f(x) = −πx3 + 3

Even thought the leading coefficientis π, it is negative with a odd degreeof three. Hence the graph tends upto the left and down to the right.

If you are getting confused in this section, the odds are that you do not understand allthe words that are being used. Ensure that you know all the definitions of the words beforeyou start problems. This way when you read a problem you will understand what you arereading.


Math 103 Multiplication of Polynomials

5.2 Multiplication of Polynomials

Goals:1. Multiply monomi-

als.

2. Multiply monomialand polynomial.

3. Multiply polynomi-als when neither is amonomial.

4. Use FOIL in polyno-mial multiplication.

5. Square binomials.

6. Multiply the sumand difference of twoterms.

7. Find the product offunctions.

8. Use polynomial mul-tiplication to evalu-ate functions.

Books normally teach multiplication of polynomials using easy cases and then building upto more complicated problems. This less however, will teach multiplication of polynomialsin a general way. This will allow you to do any type of multiplication, regardless of if it has1,2, or n different terms.

5.2.1 General Multiplication

Instructions: 5.2.1:The General Distributive Property:

To multiply two expressions that have one or more terms we will follow the procedure

represented below.

(cnxn + cn−1xn−1 + . . .+ c2x

2 + c1x+ c)(cnxn + cn−1xn−1 + . . .+ c2x

2 + c1x+ c)

What this says is that we take the first term (red) and multiply it to all terms inthe second expression. Then do the same for the second term (green). We continuethis until you use all the terms in the first expression. Then simplify using your algebrarules.

Many of us have learned FOIL. FOIL is a special case of this rule that only works when thetwo expressions have two terms. This rule will always work. Even if the expressions are notfunction, or if they are not polynomials.

Example: 5.2.1:

Solve the given multiplication problems1. 6v(2v + 3)

6v(2v + 3) = 6v(2v) + 6v(3)

= 12v2 + 18v

2. −y2(−8x2 − 6xy − y2)

−y2(−8x2 − 6xy − y2) = (−y2)(−8x2)− (−y2)(6xy)− (−y2)(y2)

= 8x2y2 + 6xy3 + y4

3. (2n+ 2)(6n+ 1)

(2n+ 2)(6n+ 1) = 2n(6n) + 2n(1) + 2(6n) + 2(1)



= 12n2 + 2n+ 12n+ 2

= 12n2 + 14n+ 2

4. (6a− 6)(−2a2 − 4a− 8)

= 6a(−2a2)− 6a(4a)− 6a(8) + (−6)(−2a2)− (−6)(4a)− (−6)(8)

= −12a3 − 24a2 − 48a+ 12a2 + 24a+ 48

= −12a3 − 12a2 − 24a+ 48

5. (−3m2 − 2mn− 8n2)(8m2 + 4mn+ n2)

This is a long problem which will be broken into parts to fit in the book.

(−3m2)(8m2) + (−3m2)(4mn) + (−3m2)(n2) 1st term mult

+(−2mn)(8m2) + (−2mn)(4mn) + (−2mn)(n2) 2nd term mult

+(−8n2)(8m2) + (−8n2)(4mn) + (−8n2)(n2) 3rd term mult

= −24m3 − 12m3n− 3m2n2 evaluate red line

−16m3n− 8m2n2 − 2mn3 evaluate green line

−64m2n2 − 32mn3 − 8n3 evaluate blue line

= −24m3 − 28m3n− 75m2n2 − 34mn3 − 8n3

6. f(n) = n2 + 6n− 4 and g(n) = 2n− 4.Find (fg)(n) = f(n) · g(n).

(fg)(n) = f(n) · g(n)

= (n2 + 6n− 4)(2n− 4)

= (n2)(2n)− (n2)(4) + (6n)(2n)− (6n)(4) + (−4)(2n)− (−4)(4)

= 2n3 − 4n2 + 12n2 − 24n− 8n+ 16

= 2n3 + 8n2 − 32n+ 16

7. (πr2 + 1)(2πr + 3)(r − 1)



(πr2 + 1)(2πr + 3)(r − 1)

= [(πr2)(2πr) + (πr2)(3) + (1)(2πr) + (1)(3)](r − 1)

= [2π2r3 + 3πr2 + 2πr + 3](r − 1)

= (2π2r3)r − 2π2r3(1) + (3πr2)(r)− 3πr2(1) + (2πr)(r)− 2πr(1) + 3r − 3(1)

= 2π2r4 − 2π2r3 + 3πr3 − 3πr2 + 2πr2 − 2πr + 3r − 3

= 2π2r4 + (−2π2 + 3π)r3 − πr2 + (−2π + 3)r − 3

8.√

2x2(√

2x+√

2)

√2x2(√

2x+√

2) = (√

2x2)(√

2x) + (√

2x2)(√

2)

= 2x3 + 2x2

9. (x+ 1)(x− 1)− x2 + 1

(x+ 1)(x− 1)− x2 + 1 = (x)(x)− (x)(1) + (1)(x)− (1)(1)− x2 + 1

= x2 − x+ x− 1− x2 + 1

= 0

10. (x+ 1)(x+ 1)

(x+ 1)(x+ 1) = (x)(x) + (x)(1) + (1)(x) + (1)(1)

= x2 + 2x+ 1

11. (f(x) + g(x))(f(x)− g(x))

(f(x) + g(x))(f(x)− g(x)) = (f(x))(f(x))− f(x)g(x) + g(x)f(x)− (g(x))(g(x))

= (f(x))2 − (g(x))2



12. (f(x) + g(x))(f(x) + g(x))

(f(x) + g(x))(f(x) + g(x)) = (f(x))(f(x)) + f(x)g(x) + g(x)f(x) + (g(x))(g(x))

= (f(x))2 + 2f(x)g(x) + (g(x))2

Note: Factoring and dis-tribution are opposites ofeach other. When you dothe distribution you getthe equation that you willneed to factor. The itemsthat you are multiplyingtogether during distribu-tion are the factors of theend result of distribution.

The last four problems demonstrate some special cases that will help us when we learnhow to factor.

Theorem 5.2.1:

1. (A+B)2 = A2 + 2AB +B2

2. (A−B)2 = A2 − 2AB +B2

3. (A+B)(A−B) = (A−B)(A+B) = A2 −B2

Lets try some of these special cases. These are as stated “special”. Therefore this willnot always work. They will only work in these three cases. However the way that we workedout multiplication earlier will always work. The special cases are there so that if you see it,then it will be faster. If you don’t do the multiplication as you normally would, and youwill get the same answer.

Example: 5.2.2:

Evaluate the following multiplications.

1. (x+ 3)(x− 3)

Notice that this follow special case 3. So we square the first term x and squarethe second term 3 and subtract it from x2.

(x+ 3)(x− 3) = x2 − 32

= x2 − 9

2. (9x+ 5)(9x+ 5)

Notice that this follow special case 1. So

(9x+ 5)(9x+ 5)

= (9x)2 + 2(9x)(5) + 52

= 81x2 + 90x+ 25

3. (2x+ 3 + 5y)(2x+ 3− 5y)

This follows special case 3. It may not look like it, but lets rewrite it to see



that it does.

(2x+ 3 + 5y)(2x+ 3− 5y) = [(2x+ 3) + 5y][(2x+ 3)− 5y]

= (2x+ 3)2 − 52

= 4x2 + 12x+ 9− 25

= 4x2 + 12x− 16

4. (2x+ 3− 5y)2

Similarly, this may not look like it, but it follows special case 2.

(2x+ 3− 5y)2 = [(2x+ 3)− 5y][(2x+ 3)− 5y]

= (2x+ 3)2 − 2(2x+ 3)(5y) + (5y)2

= 4x2 + 12x+ 9− 20xy − 30y + 25y2

= 4x2 − 20xy + 12x− 30y + 25y2 + 9


Math 103 Greatest Common Factors and Factor by Grouping

5.3 Greatest Common Factors and Factor by Grouping

Goals:1. Factor out the great-

est common factorof a polynomial.

2. Factor out a com-mon factor with anegative coefficient.

3. Factor by grouping.

Factoring is one of the most useful skills in algebra and will be used for the remainder ofyour math career. The goal of factoring is to write a number or algebraic expression as aproduct. The factors are the elements that are being multiplied together to get the product.For example; 6 = 2(3) has factors of 2 and 3 to get the product of 6. When doing factoringproblems, if you do not have an end result which is a multiplication problem, then you didnot factor.

5.3.1 Greatest Common Factor (GCF)

Definition 5.3.1: A factor of a number or algebraic expression is a number or algebraicexpression which divides it evenly (no remainder).

A = BC

A,B, and C can be numbers or algebraic expressions.The GCF is the most simple example of factoring. To understand this look at the words

in the acronym GCF. Greatest (biggest/largest) Common ( shared by) Factor (numberor algebraic expression which divides evenly). This is saying find the largest number oralgebraic expression which divides/shared by all terms. An easy example is find the GCFof 18, 24, and 6. Notice that the largest number that divides 18, 24, and 6 is 6. So 6 is theGCF.

Instructions: 5.3.1:GCF Factoring:

1. Find the GCF of the constants.

2. Find the GCF of the variables or algebraic expressions.

3. Write it as a multiplication of the GCFs and some other term so that the productis the original.

(GCF)(GCF)( ) = original

Example: 5.3.1:

Factor the following expressions.

1) −8x7 + 24x6 + 12x5

A 4 can be factored out of all theconstants, and a x5 can be pulledout of all the variables. So

−8x7 + 24x6 + 12x5

= −2x2(4x5) + 6x(4x5) + 3(4x5)

= 4x5(−2x2 + 6x+ 3)

Notice that if we found the GCFthat there is noting left to factorout of −2x2 + 6x+ 3. This iscompletely factored. Also noticethat if we did the distribution wewould get the original polynomialback.

2) −3πk3 + 15πk2 − 6πk

We can pull out 3π and k from theconstants and variables respectively.

− 3πk3 + 15πk2 − 6πk

= −k2(3πk) + 5k(3πk)− 2(3πk)

= (3πk)(−k2 + 5k − 2)

Again check that if we do thedistribution that we get the originalpolynomial −3πk3 + 15πk2 − 6πkback. The book says to factor outthe negative if the leadingcoefficient is negative. This is usefulbecause it makes things cleaner, butit isn’t necessary.



3) 32n3m3 + 28nm2 − 20n2m3

Off hand I cannot see the largest number that divides, 32, 28, and 20. I doknow that 2 divides all but I don’t know if this is the largest. So lets starthere. I also know that I can pull out a n and a m2 out.

32n3m3 + 28nm2 − 20n2m3

= 16n2m(2nm2) + 14(2nm2)− 10nm(2nm2)

= (2nm2)(16n2m+ 14− 10mn)

Notice that we can still pull out a 2 from all the constants. So

= (2nm2)(2)(8n2m+ 7− 5mn)

= 4nm2(8n2m+ 7− 5mn)

Now there is nothing else that we can factor out. This is now completelyfactored. Similarly check to see if we get the original back after distribution.

4) 12r2s+ 4rs− 13

Notice that there are no constants that divide all the constanst of the terms,nor are there variables that divide all the variables. This expression is notfactorable using the GCF method. This is expression is not factorable at all,but before you conclude this you must try some other methods which we willlearn later.

5) 4p3(4p+ 1) + 3(4p+ 1)

There are no constants that candivide both 4 and 3 except 1. Thealgebraic expression part is moreinteresting. Notice that in bothterms they have a 4p+ 1. Hence wecan factor this out of both.

4p3(4p+ 1) + 3(4p+ 1)

= (4p+ 1)(4p3 + 3)

This is essentially the last part ofall the previous problems, butsometimes factoring can berepresented in this way.

6) 9(x+ 4) + 2x(2x+ 8)

This one is a little tricky. Thesecond terms 2x+ 8 can be factoredas 2(x+ 4). Substituting this backin we get 9(x+ 4) + 4x(x+ 4). Nowwe have a (x+ 4) in common so

9(x+ 4) + 4x(x+ 4)

= (x+ 4)(9 + 4x)

The last two examples lead to a different type of factoring called factoring by grouping.This however is the same as factoring by GCF. They just give it a special name. The goalis to group terms together so that you can perform GCF factoring on the smaller groups,then factor using GCF on the remaining terms.

Instructions: 5.3.2:Factoring by Grouping:

1. Group terms that have a GCF. Can be grouped by two or more.



2. Factor out the GCF from each group.

3. Factor out the GCF from the new terms. There may not be one. If this happenstry another grouping.

If there ends up being no GCF in the last step for all groupings, then the polynomialis not factorable.

Example: 5.3.2:

Factor the following expressions

1. 24p3 + 15p2 − 56p− 35

24p3 + 15p2 − 56p− 35 = (24p3 + 15p2) + (−56p− 35)

= 3p2(8p+ 5) + (−7)(8p+ 5)

= (8p+ 5)(3p2 − 7)

2. 24r3 − 64r2 − 21r + 56

24r3 − 64r2 − 21r + 56 = (24r3 − 64r2) + (−21r + 56)

= 8r2(3r − 8) + (−7)(3r − 8)

= (3r − 8)(8r2 − 7)

3. 56xw + 49xk2 − 24yw − 21yk2

56xw + 49xk2 − 24yw − 21yk2 = (56xw + 49xk2) + (−24yw − 21yk2)

= 7x(8w + 7k2) + (−7y)(4w − 3k2)

Notice that this did not work, so try a different grouping.

56xw + 49xk2 − 24yw − 21yk2 = (−21yk2 + 49xk2) + (−24yw + 56xw)

= −7k2(3y − 7x) + (−8w)(3y − 7x)

= (3y − 7x)(−7k2 − 8w)

Some people may get (3y − 7x)(−7k2 − 8w) = (7x + 3y)(7k2 +8w) depending on what GCF they used. Both are correct.

4. 42mc+ 36md− 7n2c− 6n2d



42mc+ 36md− 7n2c− 6n2d = (42mc+ 36md) + (−7n2c− 6n2d)

= 6m(7c+ 6d) + n2(−7c− 6d)

Notice that (−7c − 6d) is the negative version of (7c + 6d), so factor out anegative one.

6m(7c+ 6d) + n2(−7c− 6d) = 6m(7c+ 6d)− n2(7c+ 6d)

= (7c+ 6d)(6m− n2)

5. 3x2 − 8y + 12x− 2xy

We need to rearrange this one since the first two terms have no common factors.

3x2 − 8y + 12x− 2xy = (3x2 + 12x) + (−2xy − 8y)

= 3x(x+ 4) + (−2y)(x+ 4)

= (x+ 4)(3x− 2y)

6. ax2 + 3ax− 11a+ bx2 + 3bx− 11b

This is an example of a grouping of three. Notice that the first three termshave an a in them and the last three terms have a b in them. We will groupthem accordingly.

ax2 + 3ax− 11a+ bx2 + 3bx− 11b = (ax2 + 3ax− 11a) + (bx2 + 3bx− 11b)

= a(x2 + 3x− 11) + b(x2 + 3x− 11)

= (x2 + 3x− 11)(a+ b)

This is a very useful type of factoring that we can use later on if you do not understandhow to factor trinomials. There is a systematic way to factor which involves factoring bygrouping, so keep this method in mind.


Math 103 Factoring Trinomials

5.4 Factoring Trinomials

Goals:1. Factor a trinomial

whose leading coeffi-cient is 1.

2. Factor using a sub-stitution.

3. Factor a trinomialwhose leading coeffi-cient is not 1.

4. Factor trinomials bygrouping.

We will now learn more ways to factor. The factor using GCF method is a very straightforward simple method of factoring that we have seen does not always work. Some of thesenew methods are easier than others but may take more time than the more difficult ones.The more proficient you are at the essentially guess and check method, the faster it is.

5.4.1 Factor by Guess and Check

In my opinion, this method is the most difficult way of factoring. However, if you learn itwell, then you will be extremely fast at factoring. This is by far the fastest method whenyou understand how to do it.

Instructions: 5.4.1:Factor by Guess and Check: To factor a polynomial of the form ax2 + bx+ c,

1. Factor out the GCF if there are any. If not, move on to step 2.

2. Find all the pair factors of c. For example; 9 has pair factors of 3 and 3, 9 and1, −3 and −3, and lastly −9 and −1. These are all factors since if we multiplythem together we get 9.

3. Find all the positive pair factors of a. If a is negative, factor out the negative onefrom the polynomial so that you may use only the positive factors of a.

4. Multiply one pair from step 2 (x, y) and one pair from step 3 (m,n) together.Then add the products together.

xm+ yn and xn+ ym

Check to see if this number is b. If it is you are done. If not try a new pair.

5. The factors of a go in the box, while the factors of c go into the triangle.

(2x+4)(2x+4)

Double check that when you do the distribution you get the original back.

If you are planning on doing more mathematics I highly recommend learning how to dofactoring this way. The more that you do these types of problems the better you will becomeat them. You will build something called mathematical intuition which will allow you to dothese problems without writing all the steps down. Until then write everything down. It isvery difficult to keep all the information required to factor in your head.

Example: 5.4.1:

Factor the trinomials.



1) b2 + 8b+ 7

1) There is no GCF.2) The pair factors of 7, are 7 and

1, and −7 and −1.3) The pair factors of 1 are 1 and

14) We have

(−7)(1) + (−1)(1) = −8. Thisdoes not work, so try anotherpair. Hence(7)(1) + (1)(1) = 8.

5) The factored form is(b+ 7)(b+ 1). Do thedistribution to check that weget b2 + 8b+ 7.

2) n2 − 11n+ 10

1) There is no GCF.2) The pair factors of 10 are 10

and 1, 5 and 2, −10 and −1,and −5 and −2.

3) The pair factors of 1 are 1 and1.

4) Notice that(−10)(1) + (−1)(1) = −11

5) The factored form is(n− 10)(n− 1).

3) 4p2 + 4p− 8

1) Notice that we can factor out a 4. So we have 4(p2 + p− 2).2) Since b is now -2, the factors are −2 and 1 as well as 2 and −13) We now have a is 1. So the factors are 1 and 1.4) The sum of 2(1) + (−1)(1) = 1.5) The factored form is 4(p+ 2)(p− 1).

4) 3m2 − 2m− 5

1) There is no GCF.2) The factors of −5 are −5 and 1, as well as 5 and −1.3) The factors of 3 are 3 and 1.4) This has 4 possible options. Let’s make a table to keep organized.

Factors of −5 mult to 3 and 15, −1 5(3) + (−1)(1) = 145, −1 5(1) + (−1)(3) = 2−5, 1 −5(3) + (1)(1) = −14−5, 1 −5(1) + (1)(3) = −2

5) The last one is the one we want, so the factored form is (3m− 5)(m+ 1).Notice that if we do the distribution that the −5 multiplies to the 1mand the 3m multiplies to the 1. This is what we did in the table.



5) 5n2 + 19n+ 12

1) There is no GCF.2) The factors of 12 are (12,1), (6,2), (4,3), and the negative versions

respectively.3) The factors of 5 are (5,1).4) Again, make a table. There are 12 different possibilities. Since b is

positive we need only check the positive factors of c. This is because ifwe add two negative numbers, we will get a negative number. Knowingthis we won’t need to check as much. Since b is positive and c is positive,we need both factors of c to be positive.

Factors of 12 mult to 5 and 112, 1 12(5) + 1(1) = 6112, 1 12(1) + 1(5) = 17−12, −1 Not both positive−12, −1 Not both positive

6, 2 6(5) + 2(1) = 326, 2 6(1) + 2(5) = 16−6, −2 Not both positive−6, −2 Not both positive

4, 3 4(5) + 3(1) = 234, 3 4(1) + 3(5) = 19−4, −3 Not both positive−4, −3 Not both positive

5) I wrote all the possibilities to show that we only needed to do half thework because we knew that all the cases labeled “not both positive” didnot work. The factored form uses the red. So (5n+ 4)(n+ 3)

5.4.2 Factor by Grouping

This is a very systematic way of factoring. It always works, but takes more time comparedto a person who is proficient using the guess and check method.

Instructions: 5.4.2:Factor by Grouping: To factor a polynomial of the form ax2 + bx+ c,

1. Factor out the GCF. This isn’t required, but makes the problem much easier.

2. Take a and c and multiply them together.

3. Find all the factors of ac.

4. Add these numbers together to get b.

5. Split bx into two parts using the factors of ac. bx = mx+ nx

6. Factor out the GCF from the first two terms, then the GCF from the last twoterms.

7. Factor by grouping.

Lets use the same examples above to show that we can do them in this fashion.



Example: 5.4.2:

1) b2 + 8b+ 7

1) No GCF.2) 1(7)=73) The factors are 1 and 7 as well

as −1 and −7.4) If we add 1 and 7 together we

get 8.5) b2 + 8b+ 7 = b2 + 7b+ b+ 76) b2 +7b+b+7 = b(b+7)+(b+7)7) (b+ 7)(b+ 1) is the factored

form of b2 + 8b+ 7.

2) n2 − 11n+ 10

1) No GCF.2) 1(10)=103) The factors are 10 and 1, −10

and −1, 5 and 2, and lastly −5and −2.

4) If we add −10 and −1 togetherwe get −11.

5) n2−11n+10 = n2−10n−n+106) n2 − 10n− n+ 10 =

n(n− 10)− (n− 10)7) (n− 10)(n− 1) is the factored

form of n2 − 11n+ 10.

3) 4p2 + 4p− 8

1) Factor out a 4 to get 4(p2 + p− 2)2) 1(-2)=-23) The factors are −2 and 1 as well as −1 and 2.4) If we add −1 and 2 together we get 1.5) 4(p2 + p− 2) = 4(p2 − p+ 2p− 2)6) 4(p2 − p+ 2p− 2) = 4[p(p− 1) + 2(p− 1)]7) 4(p− 1)(p+ 2) is the factored form of 4(p2 + p− 2).

4) 3m2 − 2m− 5

1) No GCF.2) 3(-5)=-153) The factors are −15 and 1, −1 and 15, −3 and 5, and lastly −5 and 3.4) If we add −5 and 3 together we get −2.5) 3m2 − 2m− 5 = 3m2 − 5m+ 3m− 56) 3m2 − 5m+ 3m− 5 = m(3m− 5) + (3m− 5)7) (3m− 5)(m+ 1) is the factored form of 3m2 − 2m− 5.



5) 5n2 + 19n+ 12

1) No GCF.2) 5(12)=603) The factors are 60 and 1, 30 and 2, 10 and 6, 15 and 4, and all the

negative versions.4) If we add 15 and 4 together we get 19.5) 5n2 + 19n+ 12 = 5n2 + 15n+ 4n+ 126) 5n2 + 15n+ 4n+ 12 = 5n(n+ 3) + 4(n+ 3)7) (n+ 3)(5n+ 4) is the factored form of 5n2 + 19n+ 12.

5.4.3 Factor Using the Quadratic Equation

We will do one more type of factoring using the quadratic equation. The quadratic equationis a way to solve for the roots or zeros of a polynomial. These are the input values thatmake the polynomial zero when we plug the value in.

Theorem 5.4.1:

The quadratic formula solves for the input values that make the polynomial ax2 +bx+ c = 0. The quadratic formula is given as

−b±√b2 − 4ac

2a

Instructions: 5.4.3:Quadratic Formula Factoring:

1. Factor out the GCF.

2. Solve for the roots of the polynomial using the quadratic equation.

3. Set the roots equal to the input. input = root1 and input = root2.

4. Use algebra to make the right hand side zero.

5. These are your factors. Write them as a product.

This method works because of a advanced mathematical theorem that says.Theorem 5.4.2:

Every non-zero, single-variable, degree n polynomial with complex coefficients has,counted with multiplicity, exactly n complex roots. All the roots of the polynomialcan be found in C the complex numbers.Another way to state this theorem is that every n degree polynomial can be factoredinto n linear factors. That is, let f(x) be a n degree polynomial. Then,

f(x) = a(x− c1)(x− c2) . . . (x− cn)

where ci ∈ C for all i.

You will learn how to find all the roots of a polynomial in future math classes. There ishowever a discussion on complex numbers in a later section.



Example: 5.4.3:

Factor the expression completely.

1) b2 + 8b+ 7

1) No GCF.2)

b = −8±√

82 − 4(1)(7)2(1)

= −8±√

64− 282

= −8±√

362

= −8± 62

3) b = −8+62 and b = −8−6

2

4)

b = −8 + 62

= −22

= −1

=⇒ b+ 1 = 0

and

b = −8− 62

= −142

= −7

=⇒ b+ 7 = 0

5) The factored form ofb2 + 8b+ 7 is (b+ 7)(b+ 1).

2) n2 − 11n+ 10

1) No GCF.2)

= −(−11)±√

(−11)2 − 4(10)2(1)

= 11±√

121− 402

= 11±√

812

= 11± 92

3) n = 11+92 and n = 11−9

2

4)

n = 11 + 92

= 202

= 10

=⇒ n− 10 = 0

and

n = 11− 92

= 22

= 1

=⇒ n− 1 = 0

5) The factored form ofn2 − 11n+ 10 is(n− 10)(n− 1).



3) 4p2 + 4p− 8

1) 4(p2 + p− 2)2)

= −1±√

12 − 4(1)(−2)2(1)

= −1±√

1 + 82

= −1±√

92

= −1± 32

3) p = −1+32 and p = −1−3

2

4)

p = −1 + 32

= 22

= 1

=⇒ p− 1 = 0

and

p = −1− 32

= −42

= −2

=⇒ p+ 2 = 0

5) The factored form of4(p2 + p− 2) is 4(p+ 2)(p− 1).

4) 3m2 − 2m− 5

1) No GCF.2)

= 2±√

(−2)2 − 4(3)(−5)2(3)

= 2±√

4 + 606

= 2±√

646

= 2± 86

3) m = 2+86 and m = 2−8

6

4)

m = 2 + 86

= 106

= 53

=⇒ 3m− 5 = 0

and

m = 2− 86

= −66

= −1

=⇒ m+ 1 = 0

5) The factored form of3m2 − 2m− 5 is(m+ 1)(3m− 5).



5) 5n2 + 19n+ 12

1) No GCF.2)

= −19±√

(19)2 − 4(5)(12)2(5)

= −19±√

361− 24010

= −19±√

12110

= −19± 1110

3) n = −19+1110 and n = −19−11

10

4)

n = −19 + 1110

= −810

= −45

=⇒ 5n+ 4 = 0

and

n = −19− 1110

= −3010

= −3

=⇒ n+ 3 = 0

5) The factored form of 5n2 + 19n+ 12 is (n+ 3)(5n+ 4).

Another type of factoring appears when there are more than one variable. This typehowever follows the same methods that we have already used to factor. The difference isthat there are two variable. This polynomial is of the form

ax2 + bxy + cy2.

For example factor x2−4xy−21y2. Remember that the two end numbers multiply togetherto get cy2. So we know that it looks like (2x+4y)(2x+4y). Notice that −7 and 3 have



a product of −21 and a sum of −4. So

(x− 7y)(x+ 3y).

This is just like factoring x2−4x−21. We just have the y variable on the right hand sideof each parenthesis. If you cannot see this, use the methods that we were doing previouslyto solve this equation. Just remember the additional variable in the right hand side of eachparenthesis.

5.4.4 Factor by Substitution

This last type of factoring is a way to factor when the powers are not two and one. This isan application for the methods that we have been using to factor.

Instructions: 5.4.4:Factor by Substitution:

1. Check that they polynomial is of the form

axn + bxn2 + c

That is the middle exponent is half of the degree. Ex: 2x6 + 3x3 + 5 or 5m8 +3m4 + 9.

2. Substitute using u = xn2 .

3. Now the polynomial will look like au2 + bu+ c.

4. Factor like normal.

5. Substitute back xn2 for each u.

6. Check that the result is factored completely.

Example: 5.4.4:


1. 2n10 + 3n5 − 9

(a) 2n10 + 3n5 − 9 is of the right form. The middle exponent is half of thedegree.

(b) Let u = n5. So u2 = (n5)2 = n10. Hence

2n10 + 3n5 − 9 = 2u2 + 3u− 9

(c) I will factor using guess and check. We know that the factors of −9 are 1,−3 and 3, and −1 and 9. The factors of 2 are just 2 and 1. Notice that3(2) + (−3)(1) = 3 So the factored form is

(2u− 3)(u+ 3).

(d) Plugging back in we get,

(2u− 3)(u+ 3) = (2x5 − 3)(x5 + 3).

(e) This is factored completely.



2. −2a4 + 13a2 − 20

(a) −2a4 + 13a2 − 20 is in the correct form.(b) We should factor out the negative attached to the 2 first. So −(2a4 −

13a2 + 20). Now let u = a2 and u2 = a4. Hence

−(2a4 − 13a2 + 20) = −(2u2 − 13u+ 20)

(c) I will factor using grouping. Multiply a and c, 2(20)=40. Notice that −5and −8 have a product of 40 and a sum of −13. So

−(2u2 − 13u+ 20) = −(2u2 − 8u− 5u+ 20)

= −[(2u)(u− 4) + (−5)(u− 4)]

= −(u− 4)(2u− 5)

(d) This is actually −(u− 4)(2u− 5) = −(a2 − 4)(2a2 − 5).(e) Notice thought that we can factor a2 − 4 as (a− 2)(a+ 2). So

−(a2 − 4)(2a2 − 5) = −(a− 2)(a+ 2)(2a2 − 5)

.

You may be asking “how will I know when we can continue factoring?”. These arespecial cases which we will discuss in the next section. We have already discussed these inthe distribution section. Like before, if you cannot see it, then you need to try to factor eachpart you will find that you won’t be able to do it. Along with trial and error, mathematicalintuition also plays a factor in how fast you will notice if something is factorable. It is notthe case that your mathematical intuition will always be correct, but it will allow you todetermine if you should put more effort into trying to factor something. For example; mymathematical intuition says that a2 − 4 should be factorable, while 2a2 − 5 is not. So I willspend more time trying to factor a2 − 4 since I believe that it can be done. Mathematicalintuition comes from the understanding of the math, not repetition of problems. The moreyou understand the math that you do the more accurate and useful your intuition willbecome.


Math 103 Factoring Special Forms

5.5 Factoring Special Forms

Goals:1. Factor the difference

of two squares.

2. Factor perfectsquare trinomials.

3. Use grouping to ob-tain the difference oftwo squares.

4. Factor the sum ordifference of twocubes.

We have already discussed many of these special forms in section 5.2. These special formsare not required to know if you want to factor. The methods we learned in section 5.3 andsection 5.4 will work without knowing what is in this section. This section is created toteach special forms so that you do not have to go through those long tedious processed. Ifyou are able to spot these special forms, then we can factor in a few seconds.

5.5.1 Difference of Squares

Definition 5.5.1: A polynomial of the form A2 − B2 is of the form of a difference of twosquares, and can be factored as

A2 −B2 = (A+B)(A−B).

Example: 5.5.1:

Factor completely1) 4x2 − 25

4x2 − 25 = (2x)2 − 52

= (2x+ 5)(2x− 5)

(2x+ 5)(2x− 5) is the factoredform of 4x2 − 25.

2) a2b2 − 16

a2b2 − 16 = (ab)2 − 42

= (ab+ 4)(ab− 4)

(ab+ 4)(ab− 4) is the factored formof a2b2 − 16.

3) m4 − n4

m4 − n4

= (m2)2 − (n2)2

= (m2 + n2)(m2 − n2)

= (m2 + n2)(m+ n)(m− n)

(m2 + n2)(m+ n)(m− n) is thefactored form of m4 − n4.

4) −64a2 + 925b

2

− 64a2 + 925b

2

= −(8a)2 +(

35b)2

=(

35b)2− (8a)2

=(

35b+ 8a

)(35b− 8a

)( 3

5b+ 8a) ( 3

5b− 8a)

is the factoredform of −64a2 + 9

25b2.

5) 16m2 − (3n+ 2y)2

16m2 − (3n+ 2y)2 = (4m)2 − (3n+ 2y)2

= (4m+ (3n+ 2y))(4m− (3n+ 2y))

(4m+ (3n+ 2y))(4m− (3n+ 2y)) is the factored form of 16m2 − (3n+ 2y)2.



6) (p+ 4q)2 − 25r2

(p+ 4q)2 − 25r2 = (p+ 4q)2 − (5r)2

= ((p+ 4q) + 5r)((p+ 4q)− (5r))

((p+ 4q) + 5r)((p+ 4q)− (5r)) is the factored form of (p+ 4q)2 − 25r2.

Remember the last example in section 5.4. The reason that it was know that a2− 4 wasfactorable was because it is a difference of squares. Make sure that you factor your problemscompletely.

Lets use this in more complicated types of problems. These problems will need to befactored more than once. You will need to make groupings so that we can get a differenceof squares

Example: 5.5.2:

Factor completely.

1) x4 − x2 − 6x− 9

x4 − x2 − 6x− 9 = x4 − (x2 + 6x+ 9)

= (x2)2 − (x+ 3)2

= (x2 + (x+ 3))(x2 − (x+ 3))

(x2 + (x+ 3))(x2 − (x+ 3)) is the factored form of x4 − x2 − 6x− 9.

2) x2 − 12x+ 12y − y2

x2 − 12x+ 12y − y2 = x2 − y2 − 12x+ 12y

= (x+ y)(x− y) + (−12)(x− y)

= (x− y)(x+ y − 12)

(x− y)(x+ y − 12) is the factored form of x2 − 12x+ 12y − y2.



3) x3 + 5x2 − 9x− 45

x3 + 5x2 − 9x− 45 = (x3 + 5x2) + (−9x− 45)

= x2(x+ 5) + (−9)(x+ 5)

= (x+ 5)(x2 − 9)

= (x+ 5)(x2 − (3)2)

= (x+ 5)(x+ 3)(x− 3)

(x+ 5)(x+ 3)(x− 3) is the factored form of x3 + 5x2 − 9x− 45.

5.5.2 Perfect Square Trinomials

Definition 5.5.2: A perfect square trinomial is a polynomial that is of the form A2+2AB+B2

or A2 − 2AB +B2. These polynomials factor as

A2 + 2AB +B2 = (A+B)2

A2 − 2AB +B2 = (A−B)2

These problems are ones that we have done in section 5.2. The difference is that we willstart in the standard form and convert to factored form. Not the other way around.

Example: 5.5.3:


1) p2 + 2p+ 1

p2 + 2p+ 1 = (p)2 + 2p(1) + (1)2

= (p+ 1)2

(p+ 1)2 is the factored form ofp2 + 2p+ 1.

2) y2 − 20y + 100

y2 − 20y + 100

= (y)2 − 2(y)(10) + (10)2

= (y − 10)2

(y − 10)2 is the factored form ofy2 − 20y + 100.

3) 9x2 + 24x+ 16

9x2 + 24x+ 16

= (3x)2 + 2(3x)(4) + 42

= (3x+ 4)2

(3x+ 4)2 is the factored form of9x2 + 24x+ 16.

4) 16t2 − 40t+ 25

16t2 − 40t+ 25

= (4t)2 − 2(4t)(5) + (5)2

= (4t− 5)2

(4t− 5)2 is the factored form of16t2 − 40t+ 25.



5) x3 + 6x2y + 9xy2

x3 + 6x2y + 9xy2 = x(x2 + 6xy + 9y2)

= x((x)2 + 2(x)(3y) + (3y)2)

= x(x+ 3y)2

x(x+ 3y)2 is the factored form of x3 + 6x2y2 + 9xy3.

6) 4k3w + 20k2w2 + 25kw3

4k3w + 20k2w2 + 25kw3

= kw(4k2 + 20kw + 25w2)

= kw((2k)2 + 2(2k)(5w) + (5w)2)

= kw(2k + 5w)2

kw(2k + 5w)2 is the factored form of 4k3w + 20k2w2 + 25kw3.

5.5.3 Sum and Difference of Cubes

This is another simple to use form of factoring. Just like a difference of squares, the differenceof cubes looks very similar.Definition 5.5.3: A polynomial of the form A3 − B3 is of the form of a difference of twosquares, and can be factored as

A3 −B3 = (A−B)(A2 +AB +B2).

Definition 5.5.4: A polynomial of the form A3 + B3 is of the form of a difference of twosquares, and can be factored as

A3 +B3 = (A+B)(A2 −AB +B2).

Being able to see that the numbers or variables can be written as ( )3 is a skill thatcan be built. This is the reason why you do not want to use a calculator until you are ableto understand and see these properties. Calculators should be used to check your work. Inorder to build your math abilities, you must perform the math without a calculator.

Example: 5.5.4:

Factor the following.1. x3 + 125

x3 + 125 = (x)3 + (5)3

= (x+ 5)(x2 − 5x+ 52)

= (x+ 5)(x2 − 5x+ 25)



(x+ 5)(x2 − 5x+ 25) is the factored form of x3 + 125.

2. −a3 − 8

−a3 − 8 = −(a3 + 23)

= −(a+ 2)(a2 − 2a+ 4)

−(a+ 2)(a2 − 2a+ 4) is the factored form of −a3 − 8.

3. 8a3 + 125

8a3 + 125

= (2a)3 + 53

= (2a+ 5)((2a)2 − 2a(5) + 52)

= (2a+ 5)(4a2 − 10a+ 25)

(2a + 5)(4a2 − 10a + 25) is the factored form of 8a3 + 125.

4. 250m4 + 128m

250m4 + 128m = 2m(125m3 + 64)

= 2m((5m)3 + 43)

= 2m(5m+ 4)((5m)2 − 5m(4) + 42)

= 2m(5m+ 4)(25m2 − 20m+ 16)

2m(5m+ 4)(25m2 − 20m+ 16) is the factored form of 150m4 + 128m.

5. 1000yx3 + 64y4

1000yx3 + 64y4 = y(1000x3 + 64y3)

= y((10x)3 + (4y)3)

= y(10x+ 4y)((10x)2 − 10x(4y) + (4y)2)



= y(10x+ 4y)(100x2 − 40xy + 16y2)

= 2y(5x+ 2y)(100x2 − 40xy + 16y2)

= 2y(4)(5x+ 2y)(25x2 − 10xy + 4y2)

= 8y(5x+ 2y)(25x2 − 10xy + 4y2)

8y(5x+ 2y)(25x2 − 10xy + 4y2) is the factored form of 1000yx3 + 64y4.

6. 23x

3y3 − 23m

3n3

23x

3y3 − 23m

3n3 = 23(x3y3 −m3n3)

= 23((xy)3 − (mn)3)

= 23(xy −mn)((xy)2 + (xy)(mn) + (mn)2)

= 23(xy −mn)(x2y2 + xymn+m2n2)

23 (xy −mn)(x2y2 + xymn+m2n2) is the factored form of 2

3x3y3 − 2

3m3n3.


Math 103 A General Factoring Strategy

5.6 A General Factoring Strategy

Goals:1. Use a general strat-

egy for factoringpolynomials.

In this section we will be going over factoring problems that are not in any specific section.That is, we won’t know the type of factoring, for example; a difference of squares, grouping,or guess and check type of factoring problem. This section is the best section to use todetermine if you are truly able to factor.

5.6.1 Random Factoring Problems

Instructions: 5.6.1:General Factoring Instructions:

1. Factor out the GCF.

2. Determine the type of factoring. These are ordered.

(a) If there are two terms, try to see if the polynomial is ai. Difference of squares, A2 −B2 = (A+B)(A−B).ii. Sum of cubes, A3 +B3 = (A+B)(A2 −AB +B2).

iii. Difference of cubes, A3 +B3 = (A−B)(A2 +AB +B2).(b) If there are three terms, try to see if the polynomial is a

i. Perfect squares (+), A2 + 2AB +B2 = (A+B)2

ii. Perfect squares (−), A2 − 2AB +B2 = (A−B)2

(c) If there are four or more terms, try to see if you cani. Factor by grouping.

(d) If all else fails, use any factor method from section 5.4. You may have tosubstitute.

Working out general problems is the most difficult, but will test whether you are ableto perform on an exam. On an exam, you may not have instructions that say “factor by

”. You will have to know how to factor the problem with no help.Example: 5.6.1:

1) x3 − 5x2 − x+ 5

This has four terms, so we will factor by grouping.

x3 − 5x2 − x+ 5 = (x3 − 5x2) + (−x+ 5)

= x2(x− 5)− (x− 5)

= (x− 5)(x2 − 1)

We are not done since x2 − 1 is a difference of squares. Hence

(x− 5)(x2 − 1) = (x− 5)(x− 1)(x+ 1)

(x− 5)(x− 1)(x+ 1) is the factored form of x3 − 5x2 − x+ 5.



2) a4 − 13a2 + 40

Since we do not have a degree of 2, we need to do a substitution where u = a2.

a4 − 13a2 + 40 = u2 − 13u+ 40

= u2 − 5u− 8u+ 40

= u(u− 5)− 8(u− 5)

= (u− 5)(u− 8)

= (a2 − 5)(a2 − 8)

(a2 − 5)(a2 − 8) is the factored form of a4 − 13a2 + 40.

3) m6 − 4m2

This is a difference of squares.

m6 − 4m2 = (m3)2 − (2m)2

= (m3 + 2m)(m3 − 2m)

= m(m2 + 2)(m3 − 2m)

= m2(m2 + 2)(m2 − 2)

m2(m2 + 2)(m2 − 2) is the factored form of m6 − 4m2.You could have also done this one by factoring out m2 first. Then using thedifference of squares on m4 − 4.

4) 2n2 − 13n+ 20

We will have to factor using the methods in section 5.4.

2n2 − 13n+ 20 = 2n2 − 5n− 8n+ 20

= n(2n− 5)− 4(2n− 5)

= (2n− 5)(n− 4)

(2n− 5)(n− 4) is the factored form of 2n2 − 13n+ 20.

5) −64x3 + 125

This is a difference of cubes. It will look like it after the first step.

−64x3 + 125 = −(4x)3 + (5)3

= 53 − (4x)3

= (5− 4x)(52 + 5(4x) + (4x)2)

= (5− 4x)(25 + 20x+ 16x2)

(5− 4x)(25 + 20x+ 16x2) is the factored form of −64x3 + 125.



6) 4p3 − p2 − 4p+ 1

4p3 − p2 − 4p+ 1 = (4p3 − p2) + (−4p+ 1)

= p2(4p− 1)− (4p− 1)

= (4p− 1)(p2 − 1)

We are not done since p2 − 1 is a difference of squares.

(4p− 1)(p2 − 1) = (4p− 1)(p+ 1)(p− 1)

(4p− 1)(p+ 1)(p− 1) is the factored form of 4p3 − p2 − 4p+ 1.

7) x2 + 21x+ 100

We will try using guess and check. The factors of 100 are 100 and 1, 10 and10, 5 and 20, 50 and 2, and lastly 25 and 4. Since 100 is positive and the bterm is positive, we only need to consider the positive factors of c. Notice thatnone of these factors will have a sum of 21. Hence x2 + 21x+ 100 is notfactorable.

8) 9b2 + 12b+ 4

This is a perfect square.

9b2 + 12b+ 4 = (3b)2 + 2(3b)2 + 22

= (3b+ 2)2

I originally thought that I would have to use the factor method. I did notnotice that it was a perfect square. This would have worked but would havetake more time. So (3b+ 2)2 is the factored form of 9b2 + 12b+ 4.

9) y5 − 4y3 − 8y2 + 32

Since there are four terms, we will factor by grouping.

y5 − 4y3 − 8y2 + 32 = (y5 − 4y3) + (−8y2 + 32)

= y3(y2 − 4)− 8(y2 − 4)

= (y2 − 4)(y3 − 8)

We are not done. We know that y2 − 4 is a difference of squares, and y3 − 8 isa perfect cube. Hence

(y2 − 4)(y3 − 8) = (y + 2)(y − 2)(y − 2)(y2 + 2y + 4)

(y + 2)(y − 2)2(y2 + 2y + 4) is the factored form of y5 − 4y3 − 8y2 + 32.



10) (a+ b)3 + c3

(a+ b)3 + c3 = (a+ b+ c)((a+ b)2 − c(a+ b) + c2)

(a+ b+ c)((a+ b)2 − c(a+ b) + c2) is the factored form of (a+ b)3 + c3.

11) 3x3y3 + 6x2y3 + 9xy4

Factor out the GCF.

3x3y3 + 6x2y3 + 9xy4 = 3xy3(x2 + 2x+ 3y)

You will notice that this is factored completely. Thus 3xy3(x2 + 2x+ 3y) isthe factored form of 3x3y3 + 6x2y3 + 9xy4.

QuadraticEquations andFunctions

Math 103 The Square Root Property and Completing the Square

6 Quadratic Equations and Functions

Quadratic functions in general look like

x

y

x

y

They have the ability to be thinner or thicker as well as start further to the left or theright.

In this section we will dive more deeply into quadratic equations to see many more vari-ations of what we have seen previously. Some may be easier, some more difficult. However,we already have all the tools necessary to go through this entire chapter.

6.1 The Square Root Property and Completing the Square

Goals:1. Solve quadratic

equations usingthe square rootproperty.

2. Complete the squareof a binomial.

3. Solve quadraticequations by com-pleting the square.

4. Solve problems us-ing the square rootproperty.

The square root property is a fancy term used to say that |x| =√x2. As we have seen in

definition 1.2.2, the absolute value |x| is given by the piecewise function,

f(x) = |x| ={x x ≥ 0−x x < 0

Notice that the we take the inside of the absolute value either leave it alone or multiply itby a negative one. This property is what they are calling the square root property whenwritten in the form

√u2.

Theorem 6.1.1:

Let u be an algebraic expression and d be in R. If u2 = d, then

u2 = d

√u2 = |u| =

√d

u = ±√d

In general, to solve these types of problems we use the following.Theorem 6.1.2:

Let u be an algebraic expression and d be in R. If un = d, then



1. If n is even,

un = d

n√un = ± n

√d

u = ± n√d

2. If n is odd,

un = d

n√un = n

√d

u = n√d

When evaluating the square root ensure that it is simplified before you state your answer.This requires some algebra.

Remember, the square root is asking for a number that you can multiply to itself to getthe inside of the square root. So

√9 = 3 since 3 · 3 = 9. This is traditional way of thinking

about the square root, but what happens if you have a number√

90. The real concept is tobreak down the interior of the root into factors. Then, depending on the n in the n

√find a number repeated n times. This is simple for the square root, since n = 2. Lastly pullout one number from the grouping to the outside of the square root. In the example

√90,

we have√

90 =√

9 · 10

=√

3 · 3 · 5 · 2

= 3√

5 · 2

= 3√

10

This idea works for any n√ and even works for the

√9 case. Notice

√9 =

√3 · 3 =

3√

1 = 3.Note:This only works ifthe inside of the squareroot is purely multiplica-tion. If it is not it doesnot work. For example;√

9 + x =√

3 · 3 + x thisdoes not equal 3

√1 + x or

3√x.

Instructions: 6.1.1:Simplifying nth Roots: Consider n

√a.

1. Break down a into its prime factors.

a = p1 · p2 . . . · pm.

n√a = n

√p1 · p2 . . . · pm

2. Look for n of the same number being multiplied.

3. Write that number times the remains square root.

4. If there are more than one number that repeats n times, repeat 3 until there arenone left.

We will now use this property and the square root property to solve quadratic equations.To do this we must isolate the variable with the exponent of 2. Then use the square rootproperty.



Example: 6.1.1:

Solve the quadratic equations.

1) 5x2 − 10 = 0

5x2 − 10 = 0

5x2 = 10

x2 = 2√x2 =

√2

x = ±√

2

So when we plug in 2 or −2 into5x2 − 10 we get 0.

2) 4p2 = 7

4p2 = 7

p2 = 74√

p2 =√

74

p = ±√

72

So when we plug in√

72 or −

√7

2 into4p2 we get 7.

3) 23b

2 = 18

23b

2 = 18

b2 = 18 · 32

√b2 =

√27

b = ±3√

3

So when we plug in 3√

3 or −3√

3into 2

3b2 we get 18.

4) 5a2 + 10 = 0

5a2 + 10 = 0

5a2 = −10

a2 = −2√a2 =

√−2

a = ±√−2

However we have not been allowedto use

√−numbers. We will discuss

this further below.

Previously we were not allowed to use even roots of negative numbers. This does notmake sense. However, mathematicians have come up with a way to use this very commonproperty to make a new algebraic space called the complex plane. Essentially,

√−1 = i.

Using this we may do many things that were previously inaccessible to use. In mathematicsthis is called Complex Analysis. Using this property, we have the solution to number 5 as,

a = ±√−2

a = ±√−1√

2

a = ±i√

2

We should check to see that this solution is correct by plugging back in and seeing if we getthe correct output.



5(i√

2)2 + 10 = 5(√−1)2(

√2)2 + 10

= 5 · (−1) · (2) + 10

= −10 + 10

= 0

5(−i√

2)2 + 10 = 5(√−1)2(−

√2)2 + 10

= 5 · (−1) · (2) + 10

= −10 + 10

= 0

When we plug in i√

2 or −i√

2 into 5a2 + 10 we get 0.This is for problems that look like u2 = c. If the problem is of the form

√u = c we have

different rules.

Instructions: 6.1.2:u2 = c VS

√u = c: Let u and c ∈ R.

1. If u2 = c, then use the square root property.

2. If√u = c,

(a) If c is positive, then square both sides and solve.(b) If c is negative, then there is no solution. This is because in order for

√u

to be a function, we restrict the values that it can take to only the positivevalues. Ex:

√x = 2 =⇒ x = 4. This property holds for any even nth root.

In theorem 6.1.1 we say that u is an algebraic expression. So what would happen if wehad a problem which did not end up as x2 =number, but rather (expression)2=number.This comes up a lot when we try to complete the square.

6.1.1 Complete the Square

The goal of completing the square is to factor ax2+bx+c so that we may solve ax2+bx+c = 0even though ax2 + bx + c is un-factorable using out current methods. Now completing thesquare will work even if ax2 + bx + c is factorable, but it will take more time than theprevious methods.

Theorem 6.1.3:

To complete the square with the quadratic x2 + bx+ c we need to add and subtract(b2)2 to the original. We will have

x2 + bx+(b

2

)2−(b

2

)2+ c.

Notice that x2 + bx+(b2)2 is a perfect square. Hence we have(

x+(b

2

))2−(b

2

)2+ c.



Notice that this is called completing the square because the end result is a perfect square.We started with a polynomial which was not a perfect square and completed it so that itwas. If you do not get this result, then you did not do it correctly.

Instructions: 6.1.3:Complete the Square:

0. Either bring everything to one side, or bring the variables to one side and theconstant to the other. Either works.

1. When solving, check if you can factor using other methods.

2. If not factor out the leading coefficient so there is no constant attached to thesquared variable.

3. Take the new b and divide it by 2, then square it.

4. Add and subtract this to the original or part 2 if part 2 was done. Make sure thatit is inside the parenthesis if you factored.

5. Factor the perfect square, and combine the constant terms.

6. If asked to solve, then solve using theorem 6.1.1.

Example: 6.1.2:

Complete the square of the following expressions.

1) y2 + 3y

1) N/A2) N/A

3)( 3

2)2

4) y2 + 3y +( 3

2)2 −

( 32)2

5)(y + 3

2)2 − 9

4

2) x2 + 5x+ 4

1) This is factorable. It factors as(x+ 4)(x+ 1). However lets dothis using complete the square.We will show later that thiswill give us the same answer ina solving problem.

2) N/A

3)( 5

2)2

4) x2 + 5x+( 5

2)2 −

( 52)2 + 4

5)(x+ 5

2)2 − 9

4

3) 2p2 + 4p

1) N/A2) 2(p2 + 2p)

3)( 2

2)2 = 1

4) 2(p2 + 2p+ 1− 1)5) 2[(p+ 1)2 − 1] = 2(p+ 1)2 − 2

4) 3w2 + 4w + 6

1) N/A2) 3(w2 + 4

3w + 2)

3)( 4

6)2 =

( 23)2

4) 3(w2 + 43w +

( 23)2 −

( 23)2 + 2)

5) 3[(w + 2

3)2 + 14

9 ] =3(w + 2

3)2 + 14

3

Lets use this technique to solve quadratic equations.



Example: 6.1.3:

Solve the following.

1. x2 + 5x+ 4 = 0

x2 + 5x+ 4 = 0

(x+ 4)(x+ 1) = 0

x = −4 or x = −1

x2 + 5x+ 4 = 0(x+ 5

2

)2− 9

4 = 0

(x+ 5

2

)2= 9

4√(x+ 5

2

)2=√

94

x+ 52 = ±3

2

x = −52 ±

32

So x = − 52 + 3

2 = − 22 = −1 or

x = − 52 −

32 = − 8

2 = −4.

We get the same answer, but the former factoring methods are much eas-ier to solve. Hence when we plug in −4 or −1 into x2 + 5x + 4 we get 0.

2. b2 + 2b = −20

b2 + 2b+ 20 = 0

b2 + 2b+(

22

)2−(

22

)2+ 20 = 0

(b+ 1)2 + 19 = 0

(b+ 1)2 = −19√(b+ 1)2 =

√−19

b+ 1 = ±i√

19

b = −1± i√

19

When we plug in −1 − i√

19 or −1 + i√

19 into b2 + 2b we get −20.

3. k2 − 4k + 1 = −5



k2 − 4k = −6

k2 − 4k +(−42

)2−(−42

)2= −6

(k − 2)2 − 4 = −6

(k − 2)2 = −2√(k − 2)2 =

√−2

k − 2 = ±i√

2

k = 2± i√

2

When we plug in 2 + i√

2 or 2 − i√

2 into k2 − 4k + 1 we get −5.

4. n2 = 18n+ 40

n2 = 18n+ 40

n2 − 18n = 40

n2 − 18n+(−18

2

)2−(−18

2

)2= 40

(n− 9)2 − 81 = 40

(n− 9)2 = 121√(n− 9)2 =

√121

n− 9 = ±11

n = 9± 11

When we plug in 20 or −2 into n2 − 18n we get 40.

5. 9n2 + 79 = −18n

9n2 + 18n = −79

9(n2 + 2n) = −79

9(n2 + 2n+(

22

)2−(

22

)2) = −79



9[(n+ 1)2 − 1] = −79

9(n+ 1)2 − 9 = −79

9(n+ 1)2 = −70

(n+ 1)2 = −709√

(n+ 1)2 =√−70

9

n+ 1 = ±i√

703

n = −1± i√

703

When we plug in −1 + i√

703 or −1 − i

√703 into 9n2 + 18n we get −79.

6. 8y2 + 16y = 42

8y2 + 16y = 42

8(y2 + 2y) = 42

8(y2 + 2y + 1− 1) = 42

8[(y + 1)2 − 1] = 42

8(y + 1)2 − 8 = 42

8(y + 1)2 = 50

(y + 1)2 = 508√

(y + 1)2 =√

254

y + 1 = ±52

y = −1± 52

When we plug in −1 + 52 = 3

2 or −1− 52 = − 7

2 into 8y2 + 16y we get 42.


Math 103 The Quadratic Equations and The Quadratic Formula

6.2 The Quadratic Equations and The Quadratic Formula

Goals:1. Solve quadratic

equations.

2. Solve quadraticequations using thequadratic formula.

3. Use the discrimi-nant to determinethe number andtype of solutions.

4. Determine the mostefficient method touse when solving aquadratic equation.

5. Write quadraticequations fromsolutions.

6. Use the quadraticformula to solveproblems.

Definition 6.2.1: A quadratic equation in x is an equation of the form

ax2 + bx+ c = 0.

where a, b and c are real numbers, with a 6= 0.

Theorem 6.2.1:

Let A and B be algebraic expressions. If AB = 0, then either A = 0 or B = 0

These two ideas will allow us to solve quadratic equations. We will do this by factoringor using the quadratic equation.

6.2.1 Quadratic Equation

Instructions: 6.2.1:Solving Quadratic Equations:

1. Write the equation in the form ax2 + bx+ c.

2. Factor completely.

3. Apply theorem 6.2.1 by setting each non-constant factor equal to zero.

4. Solve these new equations.

5. Check the solutions by plugging back into the original equation.

I tend to learn material that will work for all the cases that I could possibly work with. Thismeans learning how to solve more general equations that may not be quadratic. The moregeneral instructions that work for all polynomials

Instructions: 6.2.2:Solving Polynomial Equations:

1. Set one side equal to zero.

2. Factor completely.

3. Set each non-constant factor equal to zero, by theorem 6.2.1.

4. Solve each new equation.

5. Check the solutions by plugging back into the original equation.

Note: You may not di-vide by a variable. Forexample, solving 2x2 = xcannot be done by divid-ing both sides by x to get2x = 1. So x = 1

2 .If we did it properly wewould have x(2x− 1) = 0.And the solutions wouldbe x =0 or 1

2 . Dividingby a variable will removepossible solutions.

Notice that these instructions are nearly identical. The only difference is that we did notrestrict ourselves to a polynomial of degree 2. When factoring, do not use the quadraticformula method of factoring. If you do you will have the solutions already. Factoring wasmeant to be used to solve these equations in a easier way than the quadratic formula.

Example: 6.2.1:

Solve the equations.

1. 3x+ 36 = 0



3x+ 36 = 0

3(x+ 12) = 0

=⇒ x+ 12 = 0

x = −12

When we plug x = −12 into 3x + 36 we get zero.This is another way of solving this linear equation.

2. 2c3 − 16c2 − 18c = 0

2c3 − 16c2 − 18c = 0

2c(c− 9)(c+ 1) = 0

So

2c=0 c=0c− 9=0 c=9c+ 1=0 c=-1

When we plug c = 0, c = 9, or c = −1 into 2c3 − 16c2 − 18c we get zero.

3. b2 + 18b = 24b+ 40

b2 + 18b = 24b+ 40

b2 + 18b− 24b− 40 = 0

b2 − 6b− 40 = 0

(b− 10)(b+ 4) = 0

So

b− 10 = 0

b = 10

or

b+ 4 = 0

b = −4

When we plug b = 10 or b = −4 into b2 − 6b − 40 we get zero.



4. −8n2 − 15n+ 2 = 0

−8n2 − 15n+ 2 = 0

−(8n2 + 15n− 2) = 0

−(8n− 1)(n+ 2) = 0

So

8n− 1 = 0

8n = 1

n =18

or

n+ 2 = 0

n = −2

When we plug n = 18 or n = −2 into −8n2 − 15n+ 2 we get zero.

5. m3 − 3m2 = m− 3

m3 − 3m2 = m− 3

m3 − 3m2 −m+ 3 = 0

(m3 − 3m2)− (m− 3) = 0

m2(m− 3)− (m− 3) = 0

(m− 3)(m2 − 1) = 0

So

m− 3 = 0

m = 3

or

m2 − 1 = 0

m2 = 1

m = ±√

1

m = ±1

When we plug n = 3, n = 1, or n = −1 into m3 − 3m2 −m + 3 we get zero.



6. 50y2 − 72 = 0

50y2 − 72 = 0

2(25y2 − 36) = 0

2((5y)2 − 62) = 0

2(5y − 6)(5y + 6) = 0

So

5y − 6 = 0

5y = 6

y = 65

or

5y + 6 = 0

5y = −6

y = −65

When we plug y = 65 or y = − 6

5 into 50y2 − 72 we get zero.

7. (x− 4)2 = 9

(x− 4)2 = 9

(x− 4)2 − 9 = 0

x2 − 8x+ 16− 9 = 0

x2 − 8x+ 7 = 0

(x− 7)(x− 1) = 0

So

x− 7 = 0

x = 7

or

x− 1 = 0

x = 1



When we plug x = 7 or x = 1 into (x − 4)2 − 9 we get zero.

8. t3 − 5t2 − 4t+ 20 = 0

t3 − 5t2 − 4t+ 20 = 0

(t3 − 5t2)− (4t− 20) = 0

t2(t− 5)− 4(t− 5) = 0

(t− 5)(t2 − 4) = 0

So

t− 5 = 0

t = 5

or

t2 − 4 = 0

t2 = 4

t = ±√

4

t = ±2

When we plug t = 7, t = 2, or t = −2 into t3 − 5t2 − 4t+ 20 we get zero.

6.2.2 Quadratic Formula

This section is essentially a review. We have already seen the quadratic equation in sec-tion 5.4. In case you forgot, the quadratic formula solves for the input values that make thepolynomial ax2 + bx+ c = 0. The quadratic formula is given as

−b±√b2 − 4ac

2a .

Now that we know how to complete the square we can prove the quadratic equation.

Proof. The solutions to ax2 + bx+ c = 0 are given by

−b±√b2 − 4ac

2a .



ax2 + bx+ c = 0

=⇒ x2 + b

ax+ c

a= 0

=⇒ x2 + b

ax = − c

a

=⇒ x2 + b

ax+

(b

2a

)2−(b

2a

)2= − c

a

=⇒(x+ b

2a

)2− b2

4a2 = − ca

=⇒(x+ b

2a

)2= − c

a+ b2

4a2

=⇒(x+ b

2a

)2= −4ac

4a2 + b2

4a2

=⇒(x+ b

2a

)2= −4ac+ b2

4a2

=⇒ x+ b

2a = ±√b2 − 4ac

4a2

=⇒ x = − b

2a ±√b2 − 4ac

2a

=⇒ x = −b±√b2 − 4ac

2a

We now know that the quadratic equation is true. There us a special part of the quadraticequation called the discriminant. The discriminant is used to determine the number ofsolutions to the quadratic equation before solving it. The discriminant is the part of thequadratic equation in the square root.

Theorem 6.2.2:

Consider the quadratic equation ax2 + bx+ c = 0. The discriminant is defined to beb2 − 4ac. If

1. b2 − 4ac > 0 then the quadratic equation has 2 real solutions.

2. b2 − 4ac = 0 then the quadratic equation has 1 real solution that repeats.

3. b2−4ac < 0 then the quadratic equation has no real solutions. It has 2 complexsolutions.

Example: 6.2.2:

Before solving the quadratic equations, determine if the solutions are real or complex,and the number of solutions. Then solve using the quadratic equation.



1) 2x2 + 3x+ 5 = 0The discriminate is32 − 4(2)(5) = 9− 40 = −31. Hencewe will have 2 complex solutions.

x = −3±√

32 − 4(2)(5)2(2)

= −3±√−31

4

= −3± i√

314

When we plug in −3+i√

314 or

−3−i√

314 into 2x2 + 3x+ 5 we get 0.

2) 3y2 = −5y + 2The polynomial can be written as3y2 + 5y − 2 = 0. Hence, thediscriminate52 − 4(3)(−2) = 25 + 24 = 49 tellsus that there are two real solutions.

x = −5±√

52 − 4(3)(−2)2(3)

= −5±√

496

= −5± 76

When we plug in −5+76 = 2

6 = 13 or

−5−76 = −12

6 = −2 into 3y2 + 5y − 2we get 0.

3) −w2 + 4w = 4The polynomial can be written as−w2 + 4w − 4 = 0. Hence, thediscriminate42 − 4(−1)(−4) = 16− 16 = 0 tellsus that is only one solution, whichrepeats twice.

x = −4±√

42 − 4(−1)(−4)2(−1)

= −4±√

0−2

= −4± 0−2

= 2

This is also a perfect square. Whenwe plug in 2 into −w2 + 4w − 4 weget 0.

4) −6p2 − 3p+ 3 = 0We can first divide through by a −3to make the quadratic formulaeasier. Hence we have2p2 + p− 1 = 0. The discriminant is12 − 4(2)(−1) = 9. Hence we have 2real solutions.

x = −1±√

12 − 4(2)(−1)2(2)

= −1±√

94

= −1± 34

When we plug in −1 or 12 into

−6p2 − 3p+ 3 we get 0.

When solving quadratic equation there will be a fastest way or most efficient way to solvethem. When in doubt, the quadratic equation will always work. However, it can and manytimes is the most cumbersome way of handling quadratic equations. In math there are twoways to handle problems. The brute force method, or the finesse method. The quadraticequation would fall under the brute force category, while factoring would fall under thefinesse category. The better you are using finesse, the faster and easier the mathematics willbe.



Form Method Example

ax2 + bx + c = 0 and thequadratic equation can befactored easily.

Factor and use theo-rem 6.2.1. 3x2 + 8x+ 4 = 0

(3x+ 2)(x+ 2) = 0

3x+ 2 = 0 or x+ 2 = 0

x = −23 or x = −2

ax2 + c = 0 there is no bterm.

Use the square root prop-erty 3x2 + 5 = 0

3x2 = −5

x2 = −52

x = ±i√

52

u2 + c = 0 where u is analgebraic expression.

Use the square root prop-erty. (3x+ 5)2 − 3 = 0

(3x+ 5)2 = 3

3x+ 5 = ±√

3

3x = −5±√

3

x = −5±√

33

ax2 +bx+c = 0 where thequadratic equation is noteasily factored.

Here you have two op-tions. You may eithercomplete the square, thenuse the square root prop-erty or quadratic equa-tion.

We have done manyquadratic formula typeproblems, so there willnot be one in the ex-ample. Instead there isa complete the squareexample.

2x2 + 5x+ 1 = 0

x2 + 52x+ 1 = 0

x2 + 52x+

(54

)2−(

54

)2+ 1 = 0

(x+ 5

4

)2= 9

16

x+ 54 = ±3

4

x = −54 ±

34

x = −12 or x = −2

Let us now solve problems similar to the factoring by quadratic equation type of problems.What if we are given the solutions to a quadratic equation? Can we get the equation fromthis?

Example: 6.2.3:

Given the solutions, find the equation of the polynomial.

1. w = 2 with multiplicity 2.



w = 2

=⇒ w − 2 = 0

Hence the polynomial is

(w − 2)(w − 2) = 0

w2 − 4w + 4 = 0

The polynomial w2 − 4w + 4 = 0 has the solution 2 with multiplicity 2.

2. x = 1 and x = − 12

x = −1

=⇒ x+ 1 = 0

and

x = 12

=⇒ x− 12 = 0

Hence the polynomial is

(x+ 1)(x− 1

2

)= 0

x2 + 12x−

12 = 0

The polynomial x2 + 12x−

12 = 0 has solution 1 and − 1

2 .

3. y = −2 and y = 13

y = −2

=⇒ y + 2 = 0

and

y = 13

=⇒ y − 13 = 0



So the polynomial is

(y + 2)(y − 1

3

)= 0

y2 + 2y − 13y −

23 = 0

y2 + 53y −

23 = 0

We may also multiply the left and the right hand side by 3 to eliminate thefractions. Notice that we get the polynomial 3y2 + 5y − 2 = 0, which hassolutions 1

3 and −2.

4. p = −3±i√

314

p = −3 + i√

314

=⇒ 4p = −3 + i√

31

=⇒ 4p+ 3− i√

31 = 0

and

p = −3− i√

314

=⇒ 4p = −3− i√

31

=⇒ 4p+ 3 + i√

31 = 0

So the polynomial is

(4p+ 3− i√

31)(4p+ 3 + i√

31) = 0

16p2 + 12p+ 4pi√

31 + 12p+ 9 + 3i√

31− 4pi√

31− 3i√

31− i2(31) = 0

16p2 + 24p+ 9− (−1)(31) = 0

16p2 + 24p+ 40 = 0

8(2p2 + 3p+ 5) = 0

2p2 + 3p+ 5 = 0

The polynomial that give us solutions p = −3±i√

314 is 2p2 + 3p+ 5 = 0.

Many times real life examples will not follow the equation of a line. Therefore, we willneed to understand new functions and polynomials that will allow us more power to solvepractical problems.

Example: 6.2.4:

1. The hypotenuse of a right angled triangle is 20 cm. The difference between itsother two side is 4 cm. Find the length of the sides.Using the information we can obtain a picture of the triangle.



x− 4

x

20

Using this, we can use the pythagorean theorem a2 + b2 = c2 to obtain

(x− 4)2 + x2 = 202.

Hence

(x− 4)2 + x2 = 202

x2 − 8x+ 16 + x2 = 400

2x2 − 8x+ 16− 400 = 0

2x2 − 8x− 384 = 0

x2 − 4x− 192 = 0

x2 − 16x+ 12x− 192 = 0

x(x− 16) + 12(x− 16) = 0

(x+ 12)(x− 16) = 0

So x = 16 or x = −12. Since we cannot have a negative length, the length ofone side is 16. The other side can be found by plugging into x−4 since the lastside is 4 less than 16, or 12. Hence the lengths of the triangle are 16 and 12.

2. The area of a rectangular field is 2000 m2 and its perimeter is 180 m. Find thelength and width of the field.We know that for a rectangle

x

y

A = 2000 = x · y and P = 180 = 2x + 2y =⇒ x + y = 90. Using P we cansolve for one of the variable and plug into the equation of A.

90 = x+ y

90− y = x

x = −y + 90



Hence

2000 = (−y + 90) · y

2000 = −y2 + 90y

y2 − 90y + 2000 = 0

(y − 50)(y − 40) = 0

So y can equal 50 or 40.

(a) If y=50, then x+ 50 = 90 =⇒ x = 40.(b) If y=40, then x+ 40 = 90 =⇒ x = 50.

This says that the length of the sides of a rectangle that hasarea 2000m2 and perimeter or 180 m are 40 and 50 meters.

3. A ball is thrown upwards from a rooftop which is above from the ground.It will reach a maximum vertical height and then fall back to the ground.The height of the ball ”h” from the ground at time ”t” seconds is given by,h = −16t2 + 64t+ 80. How long will the ball take to hit the ground?This question is asking for what time will the ball be at height zero. Henceh = 0, and we want to solve for the t variable which will make −16t2 + 64t+ 80zero.

−16t2 + 64t+ 80 = 0

−16(t2 − 4t− 5) = 0

t2 − 4t− 5 = 0

(t− 5)(t+ 1) = 0

So t = 5 or t = −1. It does not make sense to say that the ball was at heightzero at −1 time, hence the only solution is t = 5. So five seconds after the ballis thrown, the ball will hit the ground.


Math 103 Quadratic Functions and Their Graphs

6.3 Quadratic Functions and Their Graphs

Goals:1. Recognize charac-

teristics of parabo-las.

2. Graph parabolas inthe form f(x) =a(x− h)2 + k.

3. Graph parabolas inthe form f(x) =ax2 + bx+ c.

4. Determine aquadratic func-tions minimum ormaximum value.

5. Solve problems in-volving a functionsminimum or maxi-mum value.

We have seen in the very beginning of this chapter that the quadratic equation

ax2 + bx+ c = 0

generally has a graph of the form

•(−4, 2) x

y

•(3,−3)

x

y

where the dot represents the vertex. Notice that the vertex is at the bottom when thea value is positive and at the top when the a value is negative. This is because the graphseither have a cup shaped graph opening up or down according to the sign of a. Quadraticequations however can be written in another form

a(x− h)2 + k a 6= 0

Proof.

ax2 + bx+ c = a

(x2 + b

ax

)+ c

= a

(x2 + b

ax+

(b

2a

)2−(b

2a

)2)

+ c

= a

(x+ b

2a

)2− a b

2

4a2 + c

= a

(x−

(− b

2a

))2+(− b

2

4a + c

)

Hence h = − b2 and k = − b2

4a + c, and this is written in the form a(x− h)2 + k.

This form is nice because it give you the vertex (h, k). The vertex is the lowest or thehighest point of the quadratic function and is known as either the maximum of the minimum.

Theorem 6.3.1:

Consider the quadratic equation

ax2 + bx+ c = a(x− h)2 + k

1. If a is positive, then the vertex (h, k) is the minimum.

2. If a is negative, then the vertex (h, k) is the maximum.



The goal is to find the maximum or minimum points of a quadratic equation as well asgraph them.There are two ways to graph a quadratic equation. It is dependent of the formthat they are written in.

Instructions: 6.3.1:Graph a(x− h)2 + k:

1. Determine if the graph opens up or down using a.

2. Find the vertex (h, k).

3. Find all the horizontal intercepts by plugging 0 into the output and solving.

4. Find all the vertical intercepts by plugging 0 into the input and solving. Use onlythe real solutions.

5. Plot the intercepts, vertex, and any additional points if necessary, and connectthem with a smooth bowl shaped curve.

Instructions: 6.3.2:Graph ax2 + bx+ c:

1. Determine if the graph opens up or down using a.

2. Find the vertex(− b

2a ,−b2

4a + c)

.

3. Find all the horizontal intercepts by plugging 0 into the output and solving.

4. When plugging 0 into the input we will always get c. So the vertical intercept is(0, c).

5. Plot the intercepts, vertex, and any additional points if necessary, and connectthem with a smooth bowl shaped curve.

If you prefer to memorize two different steps then you may do so. I prefer to memorizeas little as possible. Hence I would rather convert the quadratic equation ax2 + bx+ c intoa(x− h)2 + k by completing the square. Completing the square is something that you willneed to know anyway, so it will not hurt to have more practice. This will allow you toremember only the instructions of how to graph a(x− h)2 + k.

Example: 6.3.1:

Determine if the quadratic equation has a maximum or minimum, determine whatthe minimum or maximum is, and graph the function.

1. y = −x2 + 2x+ 1Since a = −1 this graph opens down and has a maximum. By completing thesquare we get

−x2 + 2x+ 1 = −(x2 − 2x) + 1

= −(x2 − 2x+ 1− 1) + 1

= −[(x− 1)2 − 1] + 1

= −(x− 1)2 + 2

This quadratic equation has its maximum at its vertex at (1, 2). Now we mustsolve for the intercepts.

a) Vertical intercept: y = −(0)2 + 2(0) + 1 = 1, so the vertical intercept is(0, 1).



b) Horizontal intercept:

−x2 + 2x+ 1 = 0

−(x2 − 2x) + 1 = 0

−(x2 − 2x+ 1− 1) + 1 = 0

−[(x− 1)2 − 1] + 1 = 0

−(x− 1)2 + 2 = 0

(x− 1)2 = 2

x− 1 = ±√

2

x = 1±√

2

The horizontal intercepts are (1−√

2, 0) and (1−√

2, 0).

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

••

•• x

y

2. y = 2x2 − 16x+ 33

Since a = 2 this graph opens up and has a minimum. We know that the vertexis(−−16

2·2 ,−(−16)2

4(2) + 33)

= (4, 1)

a) Vertical intercept: y = 2(0)2 − 16(0) + 33 = 33, so the vertical interceptis (0, 33).

b) Horizontal intercept: Notice that the discriminant is (−16)2 − 4(2)(33) =−8. So there is no solution, and the graph does not touch the horizontalaxis.

−2 2 4 6

5101520253035

•

•

x

y



3. (x− 3)2 + 1Since a = 1 this graph opens up and has a minimum at the vertex (3, 1).

a) Vertical intercept: y = (0− 3)2 + 1 = 9 + 1 = 10, so the vertical interceptis (0, 10).


(x− 3)2 + 1 = 0

(x− 3)2 = −1

x− 3 = ±√−1

x = 3± i

So there are no real solution and hence the graph does not touch thehorizontal axis.

−2 2 4 6

2

4

6

8

10

12

•

•

x

y

4. − 12 (x+ 2)2

Since a = − 12 this graph opens down and has a maximum at its vertex (−2, 0)

a) Vertical intercept: y = − 12 (0 + 2)2 = − 1

2 (2)2 = −2, so the vertical inter-cept is (0,−2).


−12(x+ 2)2 = 0

(x+ 2)2 = 0

x+ 2 = 0

x = −2

The horizontal intercepts are (−2, 0) and it has a multiplicity of 2.



−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

•

•

• x

y

Minimum and maximum values can be used in many application problems. You will learnmany more types of min max problems in calculus. For no we will only use the quadraticequations.

Example: 6.3.2:

1. The value of Jennifer’s stock portfolio is given by the functionv(t) = 50 + 73t − 3t2 , where v is the value of the portfolio in hundredsof dollars and t is the time in months. How much money did Jennifer startwith? When will the value of Jennifer’s portfolio be at a maximum?

To answer the first question we must know what her starting money is at time 0.So, by plugging t = 0 into the formula we will be able to answer this question.

v(0) = 50 + 73(0)− 3(0)2 = 50

Jeniffer started with $50.

The second question requires that we know where the vertex of the quadraticequation is. We also need to determine if it is a maximum or a minimum. Sincea = −3, the quadratic equation will have a maximum at its vertex.

(− 73

2(−3) ,−732

4(−3) + 50)

=(

736 ,

532912 + 50

)

=(

736 ,

592912

)

5 10 15 20

100

200

300

400

500 •Maximum

• Starting money x

y

2. Taro brand poi company has daily production costs of C(x) = 800−8x+0.25x2,where C is the total cost (in dollars) and x is the number of 20 pound bagsof poi produced. How many units should be produced every day to yield a



minimum cost?Similarly to above, we will need to find the vertex. Since the a value is 0.25,the quadratic equation will have a minimum at its vertex.

(− −8

2( 1

4) ,− (−8)2

4( 1

4) + 800

)=(

812,−64 + 800

)= (16, 736)

Notice this says that the company can make 16, 20lb bags of poi for $736. Thisis a cost of about $2.30 per pound.

10 20 30 40 50

100200300400500600700800900

1,000

•Minimum

x

y


Math 103 Equations Quadratic in Form

6.4 Equations Quadratic in Form

Goals:1. Solve equations that

are quadratic inform.

Equations that are quadratic in form are in the form

axn + bxn2 + c = 0.

These problems are similar to the factor by substitution problems in section 5.4.4. We willfollow the same procedure for factoring as in section 5.4.4, but we will now use are newfound knowledge of solving equations to solve equations that are quadratic in form.

Instructions: 6.4.1:Solving equations that are quadratic in form:

1. Bring everything to one side and check to see that the equation is of the form

axn + bxn2 + c = 0.

2. Substitute and then factor.

3. Set each factor equal to zero.

4. Substitute back and solve.

Example: 6.4.1:

Solve the following equations.

1. 2y4 + y2 − 15 = 0

2y4 + y2 − 15 = 0

2u2 + u− 15 = 0 u = y2 and u2 = y4

(2u− 5)(u+ 3) = 0

So

2u− 5 = 0 or u+ 3 = 0

2y2 − 5 = 0 or y2 + 3 = 0

y2 = 52 or y2 = −3

y = ±√

52 or y = ±i

√3

When we plug in√

52 , −

√52 , i√

3, or −i√

3 into 2y4 + y2 − 15 we get 0.

2. 7m− 41√m− 6 = 0



7m− 41√m− 6 = 0

7u2 − 41u− 6 = 0 u =√m and u2 = m

(7u+ 1)(u− 6) = 0

So

7u+ 1 = 0 or u− 6 = 0

7√m+ 1 = 0 or

√m− 6 = 0

√m = −1

7 or√m = 6

No solution or m = 36 By instructions: 6.1.2

When we plug in 36 into 7m − 41√m − 6 we get 0.

3. (w − 3)2 − 2(w − 3) = 15

(w − 3)2 − 2(w − 3) = 15

u2 − 2u− 15 = 0 u = (w − 3) and u2 = (w − 3)2

(u− 5)(u+ 3) = 0

So

u− 5 = 0 or u+ 3 = 0

(w − 3)− 5 = 0 or (w − 3) + 3 = 0

w − 8 = 0 or w = 0

w = 8 or w = 0

When we plug in 8 or 0 into (w − 3)2 − 2(w − 3) we get 15.

4. 2p 12 + p

14 − 1 = 0

2p 12 + p

14 − 1 = 0

2u2 + u− 1 = 0 u = p14 and u2 = p

12

(2u− 1)(u+ 1) = 0



So

2u− 1 = 0 or u+ 1 = 0

2p 14 − 1 = 0 or p

14 + 1 = 0

p14 = 1

2 or p14 = −1

p =(

12

)4or No solution By instructions: 6.1.2

p = 116 or No solution

When we plug in 116 into 2p 1

2 + p14 − 1 we get 0.

5. (x2 − 4x)2 + 7(x2 − 4x) + 12 = 0

(x2 − 4x)2 + 7(x2 − 4x) + 12 = 0

u2 + 7u+ 12 = 0 u = x2 − 4x and u2 = (x2 − 4x)2

(u+ 4)(u+ 3) = 0

So

u+ 4 = 0 or u+ 3 = 0

x2 − 4x+ 4 = 0 or x2 − 4x+ 3 = 0

(x− 2)2 = 0 or (x− 3)(x− 1) = 0

x = 2, 3 or 1

When we plug in 2, 3, or 1 into (x2 − 4x)2 + 7(x2 − 4x) + 12 we get 0.

6. m 25 −m 1

5 = 6

m25 −m 1

5 = 6

u2 − u− 6 = 0 u = m15 and u2 = m

25

(u− 3)(u+ 2) = 0



So

u− 3 = 0 or u+ 2 = 0

m15 − 3 = 0 or m

15 + 2 = 0

m15 = 3 or m

15 = −2

m = 35 = 243 or m = −32

When we plug in 243 or −32 into m25 − m

15 we get 6.

7. 1x2 − 6

x + 8 = 0

1x2 −

6x

+ 8 = 0

(1x

)2− 6 1

x+ 8 = 0

u2 − 6u+ 8 = 0 u = 1x and u2 =

( 1x

)2

(u− 4)(u− 2) = 0

So

u− 4 = 0 or u− 2 = 0

1x− 4 = 0 or 1

x− 2 = 0

1x

= 4 or 1x

= 2

1 = 4x or 1 = 2x

x = 14

or x = 12

When we plug in 14 or 1

2 into 1x2 − 6

x + 8 we get 0.

8. 1(x+1)2 − 5

x+1 + 4 = 0



1(x+ 1)2 −

5(x+ 1) + 4 = 0

(1

x+ 1

)2− 5 1

x+ 1 + 4 = 0

u2 − 5u+ 4 = 0 u = 1x+1 and u2 =

(1

x+1

)2

(u− 4)(u− 1) = 0

So

u− 4 = 0 or u− 1 = 0

1x+ 1 − 4 = 0 or 1

x+ 1 − 1 = 0

1x+ 1 = 4 or 1

x+ 1 = 1

1 = 4(x+ 1) or 1 = x+ 1

1 = 4x+ 4 or 1 = x+ 1

−3 = 4x or 0 = x

x = −34 or x = 0

When we plug in − 34 or 0 into 1

(x+1)2 − 5x+1 + 4 we get 0.


Math 103 Polynomial Inequalities

6.5 Polynomial Inequalities

Goals:1. Solve polynomial in-

equalities.

2. Solve problemsmodeled by polyno-mial inequalities.

We have already done a simple case of a polynomial inequality. This was done in section 4.1.Remember that linear equations are polynomials of the form y = mx + b, a first degreepolynomial. The general concept of general polynomial inequalities is the same as linearinequalities.

6.5.1 Polynomial Inequalities

Polynomial inequalities are of the form

f(x) = anxn + an−1x

n−1 + . . .+ a2x2 + a1x+ a0 > 0

where the ai are real numbers. Also we may substitute > with , <, ≥, or ≤. This is moreuseful than linear inequalities. If you learn this type of inequality, then you can automaticallysolve for linear inequalities.

Instructions: 6.5.1:Solving Polynomial Inequalities:

For each of the statements we may replace > with , <, ≥, or ≤. Let f(x) be apolynomial.

f(x) = anxn + an−1x

n−1 + . . .+ a2x2 + a1x+ a0

1. Write the inequality in the form f(x) > 0.

2. Solve f(x) = 0. These are called the zeros, roots, or boundary points. Letc1, . . . , cn be the roots of f(x).

3. Draw a number line broken into parts, labeling the roots. For the sake of ease,we will only plot c1, c2 and c3.

c1 c2 c3

x

4. Test one number in each of the intervals by picking a number in each interval andplugging it into f(x). We only care about if it is positive or negative, so it maybe easier to have f(x) factored.

(a) If f(x) is negative, then f(x) < 0 on that interval.(b) If f(x) is positive, then f(x) > 0 on that interval.

5. Write the solution set using interval notation.

When using ≥ or ≤, you must include the roots in your solution set.

Note: These graphs arehere to show that thesolutions found in exam-ple 6.5.1 are correct.

1) −p2 + 14p− 49

3 6 9 12 15

−10−8−6−4−2

246

•p

y

Example: 6.5.1:

Find all solutions of the following inequalities.

1. −p2 + 14p− 49 ≤ 0

We can rewrite this as

−p2 + 14p− 49 ≤ 0

p2 − 14p+ 49 ≥ 0



Now solve p2 − 14p+ 49 = 0

(p− 7)2 = 0

=⇒ p− 7 = 0

p = 7

−9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 8 97x

Notice that if we plug in any value into (p− 7)2 we will get a positive number.Also we have the symbol ≥. So 7 is also in the solution set. Hence we have

−9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 8 97x

(−∞,∞) is the solution to when p2 − 14p+ 49 is above or equal to 0.

2. x2 + 16x+ 24 > 6x

First, bring everything to the left hand side.

x2 + 16x+ 24 > 6x

x2 + 16x+ 24− 6x > 0

x2 + 10x+ 24 > 0

Now solve

x2 + 10x+ 24 = 0

(x+ 6)(x+ 4) = 0

=⇒ x = −6 or x = −4

−9 −8 −7 −5 −3 −2 −1 0 1 2 3 4 5 6 7 8 9−6 −4x

Pick the points −10, −5, and 10.

(−10 + 6)(−10 + 4)

positive

(−5 + 6)(−5 + 4)

negative

(10 + 6)(10 + 4)

positive

We want the polynomial to be greater than 0 or positive. Also, since we have>, the roots are not included in the solution set. So



−9 −8 −7 −5 −3 −2 −1 0 1 2 3 4 5 6 7 8 9−6 −4

◦ ◦

x

the solution set (−∞,−6) ∪ (−4,∞) is the solution to when x2 + 10x + 24 isabove 0.

2) −p2 + 14p− 49

−10−8−6−4−2 2 4−4−2

2468

10

• • x

y

Using the multiplicity, you may determine where the function is positive and negativeby testing only one point.

Theorem 6.5.1:

1. If the multiplicity of a root is even, then the output of the polynomial has thesame sign on the interval to the left and right of the root.

2. If the multiplicity of a root is odd, then the output of the polynomial hasdifferent signs on the interval to the left and right of the root.

You will find that this is much faster and easier. It will also show up again when graphinggeneral polynomials.

1) (m + 5)(m − 2)(m −1)(m+ 1)

−6 −4 −2 2 4 6

−100−50

50100150

• ••• m

y

2) (q + 8)2(q + 5)(q + 7)2

−9 −6

−10

−5

5

10

•••q

y

Example: 6.5.2:

1. (m+ 5)(m− 2)(m− 1)(m+ 1) < 0It is clear that the roots are −5, 2, 1, and −1. Also the multiplicity of each ofthese roots is 1. This says that the sign is different on the left and the right ofthe root. Testing one point which is easy, like m = 0, we get

(0 + 5)(0− 2)(0− 1)(0 + 1) = positive.

So on the interval that contains zero, the polynomial is positive. Since themultiplicity of the roots are odd, the sign will change as we pass over the root.

−9 −8 −7 −6 −4 −3 −2 0 3 4 5 6 7 8 9−5 21−1

+ − ++ −

x

So the solution set is given by the negative intervals without the roots.

−9 −8 −7 −6 −4 −3 −2 0 3 4 5 6 7 8 9−5 21−1

◦◦ ◦ ◦

x

So the solution set (−5,−1) ∪ (1, 2) is where (m+ 5)(m− 2)(m− 1)(m+ 1) isbelow 0.

2. (y + 8)2(y + 5)(y + 7)2 ≥ 0

This should also be clear that the solutions to (y + 8)2(y + 5)(y + 7)2 = 0 aregiven by −8, −5, and −7. However, in this case the multiplicity of −8 and −7is two, and the multiplicity of −5 is one. So when we pass over the roots −8and −7, the sign will stay the same, but will change over −5. Testing again aneasy number y = 0 we get

(0 + 8)2(0 + 5)(0 + 7)2 = positive.



−9 −6 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9−5−7−8

+−−−

x

So the solution set is given by the positive intervals including the roots.

−9 −6 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9−5−7−8

•••

x

So the solution set {−8} ∪ {−7} ∪ [−5,∞) is where (y + 8)2(y + 5)(y + 7)2 isabove or equal to 0.

Lets discuss a more interesting problem that requires some thinking. You will need tounderstand what the solutions of an inequality are asking for in order to solve this.

Example: 6.5.3:

Write a polynomial inequality with the solution: [−1, 2] ∪ [3,∞).

Since we know that single points are part of the solutions set, and we include the endpoint 3, we must have either ≥ or ≤. We also know that the polynomial has −1, −2,and 3 as roots. So we must have (x+ 1)n, (x− 2)m, and (x− 3)p in our polynomial.Lets pick ≥. If you choose ≤ the polynomial may change.

(x+ 1)n(x− 2)m(x− 3)p ≥ 0

We know that the polynomial must be greater than 0 on [−1, 2] and [3,∞) since it isin the solution set. Also every other interval must be negative since it is not in thesolution set. Hence

−9 −8 −7 −6 −5 −4 −3 −2 0 1 4 5 6 7 8 9−1 2 3

+−+−

x

This picture tells us information about the multiplicity of the roots. We know thatsince the sign changes after passing each root, the exponents must all be odd. Hencean example of a polynomial inequality that satisfies this solution set is

(x+ 1)(x− 2)(x− 3) ≥ 0

−2 2 4

−10−5

51015

• • • x

y

There are actually infinitely many solutions as long as the polynomial follow thisrule.

(x+ 1)n(x− 2)m(x− 3)p ≥ 0

Where n, m, and p are all odd.

Rational Ex-pressions, Func-tions, and Equa-tions

Math 103 Rational Expressions and Functions: Multiplying and Dividing

7 Rational Expressions, Functions, and Equations

What is a rational number? Remember a rational number is defined as {pq |p, and q ∈ Z. q 6=0}. This idea is what we will use to create rational functions.Definition 7.0.1: Rational functions are of the form

f(x)g(x) = anx

n + an−1xn−1 + . . .+ a2x

2 + a1x+ a0

bmxm + bm−1xm−1 + . . .+ b2x2 + b1x+ b0

where the ai and bj are real numbers, n and m are natural number, and g(x) 6= 0.Rational functions are in the form of a fraction where the numerator and the denominator

are polynomials. Just like in rational numbers or division, we may not divide by zero. Henceg(x) 6= 0.

7.1 Rational Expressions and Functions: Multiplying and Dividing

Goals:1. Evaluate rational

functions.

2. Find the domain ofa rational function.

3. Interpret informa-tion given by thegraph of a rationalfunction.

4. Simplify rational ex-pressions.

5. Multiply rationalexpressions.

6. Divide rational ex-pressions.

A basic example of a rational function is f(x) = 1x . This is given by the graph

−4 −2 2 4

−15

−10

−5

5

10

15

x

y

Note: In this book we willnot learn how to graphthese types of functions.It is just displayed so thatyou have a visual repre-sentation of what is hap-pening to the function asthe input approaches anumber where the rationalfunction will have a divi-sion by zero.

Notice that the division by zero happens when x = 0. Also notice that the graphapproaches y = 0 but never touches it. This is called a vertical asymptote (x = 0) andhorizontal asymptote (y = 0). These are very important part of rational functions. Ingeneral when determining the domain of a function we need to look for two restrictions

• Even roots of negative numbers.

• Division by zero.

Since a rational function has only natural numbers as power, we do not need to lookfor even roots of negative numbers. The example above has a restriction at x = 0, so thedomain of f(x) = 1

x is (−∞, 0)∪ (0,∞). We can see this in the graph. The function has nooutput for an input of x = 0.

In the following examples the functions will be graphed.Example: 7.1.1:

Find the roots and the vertical asymptotes of the rational functions.

1. f(x) = (x+4)(x+2)x2−1



The function is zero when

(x+ 4)(x+ 2)x2 − 1 = 0

=⇒ (x+ 4)(x+ 2) = 0

=⇒ x+ 4 = 0 or x+ 2 = 0

=⇒ x = −4 or x = −2.

The function has as vertical asymptote when the denominator is zero. Thishappens when

x2 − 1 = 0

x2 = 1

x = ±1

−4 −3 −2 −1 1 2 3 4

−15

−10

−5

5

10

15

x

y

Similarly the graph never touches the vertical asymptotes represented by thelines x = −1 or x = 1.

2. g(m) = (2m+3)(2m−1)(m+3)

The function is zero when

(2m+ 3)(2m− 1)(m+ 3) = 0

=⇒ 2m+ 3 = 0

=⇒ 2m = −3

=⇒ m = −32

The function has as vertical asymptote when the denominator is zero. This



happens when

(2m− 1)(m+ 3) = 0

2m− 1 = 0 or m+ 3 = 0

2m = 1 or m = −3

m = 12 or m = −3

−4 −3 −2 −1 1 2 3 4

−15

−10

−5

5

10

15

x

y

These examples illustrate the domain of a function. Notice that they are restricted bythe division by zero, as stated above. Lets now write out a domain of a rational function ininterval notation.

Example: 7.1.2:

What is the domain of f(x) = 2x2+3x+12x+1 .

This function can be rewritten as

2x2 + 3x+ 12x+ 1 = (2x+ 1)(x+ 1)

2x+ 1 .

This says that we have division by zero when

2x+ 1 = 0

2x = −1

x = −12

So the domain is given by (−∞,− 12 ) ∪ (− 1

2 ,∞).

You may be asking why we don’t simplify f(x) to

(2x+ 1)(x+ 1)2x+ 1 = x+ 1.

When we have a rational function we must be careful. We must always consider the originalfunction. The original function has a restriction of the domain at x = − 1

2 . The simplifiedversion x+ 1 does not. You may however say that these are the same when we restrict thedomain of x+ 1.

x+ 1 where x 6= 12

Pictorially, this looks like



−9 −6 −3 3 6 9

−15

−10

−5

5

10

15

◦ x

y

Figure 8: f(x) = 2x2+3x+12x+1

−9 −6 −3 3 6 9

−15

−10

−5

5

10

15

x

y

Figure 9: f(x) = x+ 1

Simplifying rational expressions requires factoring. Factor the numerator and denomi-nator so that you may cancel.

Example: 7.1.3:

Simplify the expression 6x2+13x+62x2+11x+12 .

Factor the numerator and denominator.

6x2 + 13x+ 62x2 + 11x+ 12 = (3x+ 2)(2x+ 3)

(x+ 4)(2x+ 3)

= 3x+ 2x+ 4 x 6= − 3

2

Don’t forget to restrict the domain.

Simplifying is very useful. Simplify, means to make simpler. We would like for theproblems to be simpler if at all possible. We will use simplifying in the coming topics.

7.1.1 Multiplying Rational Expressions

When multiplying rational expressions we want to

Instructions: 7.1.1:Multiplying Rational Expressions:

1. Factor the numerator and denominator of all fractions being multiplies.

2. Write as one fraction.

3. Simplify and write the restriction.

Example: 7.1.4:

1. m+13m−15 ·

8m−80m2−9m−10

1) m+13m−15 ·

8m−80m2−9m−10 = m+1

3(m−5) ·8(m−10)

(m−10)(m+1)

2) 8(m+1)(m−10)3(m−5)(m−10)(m+1)

3) 83(m−5) m 6= 10 or − 1.

2. 45x2

x−9 ·x2−5x−363x3+12x2



1) 45x2

x−9 ·x2−5x−363x3+12x2 = 45x2

x−9 ·(x−9)(x+4)

3x2(x+4)

2) 45x2(x−9)(x+4)3x2(x−9)(x+4)

3) No input of x 6= 9, 0, −4

Be careful to simplify as much as possible. What if we are given

(p2 + 1)(2p− 5)5− 2p

Is there anything that we can simplify? It may not seem like it, but we can simplify (2p−5)and 5− 2p. This is because we may factor out a −1 from the denominator to get

(p2 + 1)(2p− 5)5− 2p = (p2 + 1)(2p− 5)

−(−5 + 2p)

= (p2 + 1)(2p− 5)−(2p− 5)

= (p2 + 1)−1

= −(p2 + 1) p 6= 52

Make sure that you check to see if the factors are the same with opposite sign.

7.1.2 Dividing Rational Expressions

The nice part about dividing rational expressions is that the process for solving them isessentially the same as multiplying rational expressions.

Instructions: 7.1.2:Dividing Rational Expressions:

1. Two different ways of writing division.

(a) f(x)g(x) ÷

p(x)q(x) = f(x)

g(x) ·p(x)q(x)

(b)f(x)g(x)p(x)q(x)

= f(x)g(x) ·

p(x)q(x)

In either case, we multiply by the reciprocal of the fraction that we are using todivide.

Now the problem is a multiplication problem. So finish using the multiplication ofrational expressions instructions.

2) Factor the numerator and denominator of all fractions being multiplied.

3) Write as one fraction.

4) Simplify and write the restriction. The restrictions in this case are all parts thathave division by zero. You need to check where the numerator and denominatorof the divisor (what you are dividing by) is zero as well as where the denominatorof the dividend (what you are dividing) is zero



Example: 7.1.5:

1. 10b2+42b+366b2−2b−60 ÷

40b+483b2−13b+10

(a) 10b2+42b+366b2−2b−60 ÷

40b+483b2−13b+10 = 10b2+42b+36

6b2−2b−60 ·3b2−13b+10

40b+48

(b) 10b2+42b+366b2−2b−60 ·

3b2−13b+1040b+48 = 2(5b+6)(b+3)

2(3b−10)(b+3) ·(3b−10)(b−1)

8(5b+6)

(c) 2(5b+6)(b+3)(3b−10)(b−1)16(3b−10)(b+3)(5b+6)

(d) To get the restrictions, we need to checki. 6b2 − 2b− 60 = 0

6b2 − 2b− 60 = 0

2(3b− 10)(b+ 3) = 0

3b− 10 = 0 or b+ 3 = 0

3b = 10 or b = −3

b = 103 or b = −3

ii. 3b2 − 13b+ 10 = 0

3b2 − 13b+ 10 = 0

(3b− 10)(b− 1) = 0

3b− 10 = 0 or b− 1 = 0

3b = 10 or b = 1

b = 103 or b = 1

iii. 40b+ 48 = 0

40b+ 48 = 0

8(5b+ 6) = 0

5b = −6

b = −65

10b2+42b+366b2−2b−60 ÷

40b+483b2−13b+10 = 1

8 (b− 1) where b 6= 103 , −3, 1, − 6

5 .

2.(

3m2−25m−1827m+18

)(5m−3

5m2−33m+18

)(a)

(3m2−25m−18

27m+18

)(5m−3

5m2−33m+18

) =(

3m2−25m−1827m+18

)·(

5m2−33m+185m−3

)(b)

(3m2−25m−18

27m+18

)·(

5m2−33m+185m−3

)=(

(3m+2)(m−9)9(3m+2)

)·(

(5m−3)(m−6)5m−3

)



(c)(

(3m+2)(m−9)(5m−3)(m−6)9(3m+2)(5m−3)

)(d) To get the restrictions, we need to check

i. 27m+ 18 = 0

27m+ 18 = 0

27m = −18

m = −1827

= −23

You may also get this using the 3m+ 2 in the denominator.ii. 5m− 3 = 0

5m− 3 = 0

5m =3

m = 35

iii. 5m2 − 33m+ 18 = 0

5m2 − 33m+ 18 = 0

(5m− 3)(m− 6) = 0

5m− 3 = 0 or m− 6 = 0

m = 35 or m = 6

(3m2−25m−18

27m+18

)(5m−3

5m2−33m+18

) = 19 (m− 9)(m− 6) where m 6= − 18

27 ,35 , 6.


Math 103 Adding and Subtracting Rational Expressions

7.2 Adding and Subtracting Rational Expressions

Goals:1. Add rational expres-

sions with the samedenominator.

2. Subtract rationalexpressions with thesame denominator.

3. Find the least com-mon denominator.

4. Add and subtractrational expressionswith different de-nominators.

5. Add and subtractrational expressionswith opposite de-nominators.

Adding and subtracting rational expressions is the same as adding and subtracting rationalnumbers. What was the first step when adding a

b + cd? Find the LCD, then add the top

and keep the bottom. This idea is the same with rational expressions. So, before we canstart these problems we need to know how to find the LCD of fractions that have a algebraicexpression in the denominator. The reason to find the LCD is to be able to add two fractionstogether that do not have the same denomenator.

Instructions: 7.2.1:Combining Fractions With Algebraic Expressions in The Denominator:

1. Factor all the denominators of the fractions that you want to add/subtract to-gether.

2. Write a list of all the items that are in the denominator starting with the firstfraction. If the next fraction has the same item that need to be added to the list,then skip that item. Continue this process for all fractions.Ex: 1

x2 + 1x . Since the denominator of the first fraction is x ·x, the second fraction

has nothing in the denominator that is different. So the list is just x and x. Notx, x, and x.

3. For each fraction, multiply the top and bottom by this list, minus what is alreadyin the bottom.

4. Write as one fraction, by adding/subtracting all the tops and keeping the bottomthe same. Simplify

For now lets only solve these problems by writing it as one fraction. No simplifying.Example: 7.2.1:

Write as one fraction.

1. 1(x+1)(x−1) −

2xx

1) Done2) (x+ 1), (x− 1) and x.3)

1(x+ 1)(x− 1) ·

x

x− 2x

x· (x+ 1)(x− 1)

(x+ 1)(x− 1)

x

x(x+ 1)(x− 1) −2x(x+ 1)(x− 1)x(x+ 1)(x− 1)

4)x− 2x(x+ 1)(x− 1)x(x+ 1)(x− 1)

2. (w+2)w2 + 2w(w+1)

w(2w−3)

1) Done2) w, w, and (2w − 3).



3)

(w + 2)w2 · (2w − 3)

(2w − 3) + 2w(w + 1)w(2w − 3) ·

w

w

(w + 2)(2w − 3)w2(2w − 3) + 2w2(w + 1)

w2(2w − 3)

4)(w + 2)(2w − 3) + 2w2(w + 1)

w2(2w − 3)

3. g+2g2+2g+1 + 3

g

1) g2 + 2g + 1 = (g + 1)(g + 1).2) g + 1, g + 1, and g.3)

g + 2g2 + 2g + 1 ·

g

g+ 3g· (g + 1)2

(g + 1)2

g(g + 2)g(g2 + 2g + 1) + 3(g + 1)2

g(g + 1)2

4)g(g + 2) + 3(g2 + 2g + 1)

g(g2 + 2g + 1)

4. p2p2+3p−5 −

12p+5 + 2

1) 2p2 + 3p− 5 = (2p+ 5)(p− 1)

2) (2p+ 5), (p− 1), and 1.

3)

p

2p2 + 3p− 5 −1

2p+ 5 ·p− 1p− 1 + 2(2p+ 5)(p− 1)

(2p+ 5)(p− 1)

p

2p2 + 3p− 5 −p− 1

(2p+ 5)(p− 1) + 2(2p+ 5)(p− 1)(2p+ 5)(p− 1)

4)p− (p− 1) + 2(2p+ 5)(p− 1)

(2p+ 5)(p− 1)

Note: Don’t forget thatwhen we cancel we needto write the restrictions onthe domain to ensure thatwe have the same equa-tion. Ex: x

x = 1, forx 6= 0.

Now let us continue this process so that we can simplify the addition or subtraction. Wewill see that the simplified version of the rational expressions are much easier to evaluateand solve.



Example: 7.2.2:

Perform the given operation and simplify.

1. x2+3x−2(x+5)(x−2) + 4x+12

(x+5)(x−2)

x2 + 3x− 2(x+ 5)(x− 2) + 4x+ 12

(x+ 5)(x− 2) = x2 + 3x− 2 + 4x+ 12(x+ 5)(x− 2)

= x2 + 7x+ 10(x+ 5)(x− 2)

= (x+ 5)(x+ 2)(x+ 5)(x− 2)

= x+ 2x− 2 x 6= −2

2. 3p−5p2−25 −

2p+5

3p− 5p2 − 25 −

2p+ 5 = 3p− 5

(p− 5)(p+ 5) −2

p+ 5

= 3p− 5(p− 5)(p+ 5) −

2p+ 5 ·

p− 5p− 5

= 3p− 5(p− 5)(p+ 5) −

2(p− 5)(p+ 5)(p− 5)

= 3p− 5− 2(p− 5)(p− 5)(p+ 5)

= 3p− 5− 2p+ 10(p− 5)(p+ 5)

= p+ 5(p− 5)(p+ 5)

= 1p− 5 p 6= −5

3. m+12m−1 −

2m3m+4 + 4m2−11m−12

6m2+5m−4The factors of the denominators are 2m− 1, 3m+ 4, and (3m+ 4)(2m− 1). Sothe list is only given by 2m− 1 and 3m+ 4. Hence

m+ 12m− 1 −

2m3m+ 4 + 4m2 − 11m− 12

6m2 + 5m− 4

= m+ 12m− 1 ·

3m+ 43m+ 4 −

2m3m+ 4 ·

2m− 12m− 1 + 4m2 − 11m− 12

(2m− 1)(3m+ 4)



= (m+ 1)(3m+ 4)(2m− 1)(3m+ 4) −

(2m)(2m− 1)(3m+ 4)(2m− 1) + 4m2 − 11m− 12

(2m− 1)(3m+ 4)

= (m+ 1)(3m+ 4)− (2m)(2m− 1) + 4m2 − 11m− 12(2m− 1)(3m+ 4)

= 3m2 + 7m+ 4−4m2 + 2m+ 4m2−11m− 12(2m− 1)(3m+ 4)

= 3m2 − 2m− 8(2m− 1)(3m+ 4)

= (3m+ 4)(m− 2)(2m− 1)(3m+ 4)

= m− 22m− 1 m 6= − 4

3

4. 2q2+q−33q+5 · 9q+15

6q2+3q−9 + q2+33q+1

2q2 + q − 33q + 5 · 9q + 15

6q2 + 3q − 9 + q2 + 33q + 1

= (2q2 + q − 3)(9q + 15)(3q + 5)(6q2 + 3q − 9) + q2 + 3

3q + 1

= 18q3 + 30q2 + 9q2 + 15q − 27q − 4518q3 + 9q2 − 27q + 30q2 + 15q − 45 + q2 + 3

3q + 1

Notice how long this problem starts to get. One thing that many people forgetthat they can do is simplify at any point. We should simplify right from thebeginning. Notice

2q2 + q − 33q + 5 · 9q + 15

6q2 + 3q − 9 + q2 + 33q + 1

= 2q2 + q − 33q + 5 · 3(3q + 5)

3(2q2 + q − 3) + q2 + 33q + 1

= (2q2 + q − 3)(3q + 5)(3q + 5)(2q2 + q − 3) + q2 + 3

3q + 1

= 1 + q2 + 33q + 1

= 3q + 13q + 1 + q2 + 3

3q + 1



= 3q + 1 + q2 + 33q + 1

= q2 + 3q + 43q + 1

Again, the word simplify, by definition, means to make easier, or to make simpler.The goal is to make to problem easier to work with.


Math 103 Complicated Rational Expressions

7.3 Complicated Rational Expressions

Goals:1. Simplify complex ra-

tional expressions bymultiplying by 1.

2. Simplify complex ra-tional expressions bydividing.

Lets consider more complicated rational expressions. To do this lets look at a complicatedrational number. For example

34 + 2

31 + 4

3.

What do we need to do to solve this? First we need to remember that there are hiddenparenthesis that make the expression

34 + 2

31 + 4

3=( 3

4 + 23)(

1 + 43) .

This allow us to use PEMDAS to work the parenthesis parts first and separately.

1)34+2

3912 + 8

121712

2)

1+43

33 + 4

373

Now combine them to form the original rational number.( 34 + 2

3)(

1 + 43) =

171273

Simplify using your division of fractions rules.171273

= 1712 ·

37 = 17

28 .

Notice that when breaking this complicated problem into smaller parts it becomes a basicproblem that we have been doing since the beginning. The reason why we spend so muchtime learning fraction rules and power rules is because these rules apply when the problemshave algebraic expressions in them as well.Note: Many books like to

find the LCD of the nu-merator and the denomi-nator together. This canmake the problems verydifficult. For example 1

2 +13 + 1

4 + 16 . If we want

to find the LCD of thesefour numbers we could,but it would be more com-plicated than finding theLCD of 1

2 and 14 first,

then 13 and 1

6 next. Justlike in the example, breakthe problem into smallerparts.

Let’s apply the same idea to problems with algebraic expressions in the numerate andthe denominator.

Instructions: 7.3.1:Simplify Complicated Rational Expressions:

1. Break the problem down into parts. Top and bottom.

2. Do each part separately by finding their respective LDDs and perform the oper-ations.

3. Reform the original rational expression.

4. Use division of fraction rules and simplify.

Example: 7.3.1:

Perform the given operations.



1. 1− 1x2

1− 1x

1− 1x2

1− 1x

=x2

x2 − 1x2

xx −

1x

LCD is x2 and x.

=x2−1x2

x−1x

= x2 − 1x2 · x

x− 1

= x2 − 1x(x− 1)

= (x− 1)(x+ 1)x(x− 1)

= x+ 1x

x 6= 1

2.2

ab2− 3ab3 + 1

ab4

a2b+ab− 1

ab

2ab2 − 3

ab3 + 1ab

4a2b + ab− 1

ab

=2ab2 · bb −

3ab3 + 1

ab ·b2

b2

4a2b + ab · a2b

a2b −1ab ·

aa

LCDs are ab3 and a2b.

=2bab3 − 3

ab3 + b2

ab3

4a2b + a3b2

a2b −aa2b

=2b−3+b2

ab3

4+a3b2−aa2b

= 2b− 3 + b2

ab3 · a2b

4 + a3b2 − a

= 2b− 3 + b2

b2 · a

4 + a3b2 − 1 a 6= 0

= a(b+ 3)(b− 1)b2(4 + a3b2 − a) a 6= 0

Since the denominator is not factorable, we are done.

3.x−3x+3−

x+3x−3

x−3x+3 + x+3

x−3

x−3x+3 −

x+3x−3

x−3x+3 + x+3

x−3=

x−3x+3 ·

x−3x−3 −

x+3x−3 ·

x+3x+3

x−3x+3 ·

x−3x−3 + x+3

x−3 ·x+3x+3

LCD is (x+ 3)(x− 3) for both.

=(x−3)2

(x+3)(x−3) −(x+3)2

(x+3)(x−3)(x−3)2

(x+3)(x−3) + (x+3)2

(x+3)(x−3)



=(x−3)2−(x+3)2

(x+3)(x−3)(x−3)2+(x+3)2

(x+3)(x−3)

= (x− 3)2 − (x+ 3)2

(x+ 3)(x− 3) · (x+ 3)(x− 3)(x− 3)2 + (x+ 3)2

= (x− 3)2 − (x+ 3)2

(x− 3)2 + (x+ 3)2 x 6= 3,−3

= x2 − 6x+ 9− (x2 + 6x+ 9)x2 − 6x+ 9 + x2 + 6x+ 9 x 6= 3,−3

= −12x2x2 + 18 x 6= 3,−3

= − 6x2 + 9 x 6= 3,−3

4. m−2+2m−1

m+4m−2

In this example the rational expression is written in the form x−1. So first letsconvert it into the form that we have been working with. x−1 = 1

x1

m−2 + 2m−1

m+ 4m−2 =1m2 + 2 1

m1

m+ 4 1m2

Convert

=1m2 + 2 1

m1 · mmm · m2

m2 + 4 1m2

LCD is m2 for both.

=1m2 + 2m

m2

m3

m2 + 4m2

=1+2mm2

m3+4m2

= 1 + 2mm2 · m2

m3 + 4

1 + 2mm3 + 4 m 6= 0


Math 103 Division of Polynomials

7.4 Division of Polynomials

Goals:1. Divide a polynomial

by a monomial.

2. Use long division todivide by a poly-nomial containingmore than one term.

Lets say that the fraction f(x)g(x) is simplified. Can we still perform the division? Of course!

Just like before, think about this as if we were discussing rational numbers instead. Howwould we evaluate 32

5 ? We would write

65)

32302

What this tells us is that there are 6 groupings of 5 in 32 with a remainder of 2. Anotherway to say this is

32 = 6(5) + 3

This is called the Euclidean division algorithm.Theorem 7.4.1:

1. For Integers:Given two integers a and b with b 6= 0, there exists unique integers q and r suchthat

a = bq + r,

where 0 ≤ r < |b|.Another way that this is seen is

a

b= q + r

b.

Ex: 325 = 6 + 3

5 . 32 divided by five is 6 and three fifths.

2. For Polynomials:Given two univariate polynomials (polynomial with one variable) f(x) andg(x) 6= 0. There exists two polynomials q(x) and r(x), such that

f(x) = g(x)q(x) + r(x)

anddeg(r(x)) < deg(g(x)).

Another way that this is seen is

f(x)g(x) = q(x) + r(x)

g(x) .

Lets try do some long division using polynomials. The first type is the easiest case. Thishappens when the divisor is a monomial (single termed polynomial). To solve this we usethe power rules as before.Note: Don’t forget to dis-

tribute the division to allthe terms in the numera-tor. Do not cancel like

x+ 1x

= 1 + 1 = 2.

Writing these problems as1x · (x + 1) helps to elim-inate this common mis-take. Most people do notmess up the distributiveproperty.

Example: 7.4.1:

1. 3x3+x2+4x+32x2

3x3 + x2 + 4x+ 32x2 = 1

2x2 · (3x3 + x2 + 4x+ 3)

= 3x3

2x2 + x2

2x2 + 4x2x2 + 3

2x2



= 32x

3−2 + 12x

2−2 + 42x

1−2 + 32x

0−2

= 32x

1 + 12x

0 + 2x−1 + 32x−2

= 32x+ 1

2 + 2 · 1x

+ 32 ·

1x2

2. x3y2+5x2y+2xy2+3y3

3xy2

x3y2 + 5x2y + 2xy2 + 3y3

3xy2 = x3y2

3xy2 + 5x2y

3xy2 + 2xy2

3xy2 + 3y3

3xy2

= 13x

3−1y2−2 + 53x

2−1y1−2 + 23x

1−1y2−2 + 33x

0−1y3−2

= 13x

2y0 + 53x

1y−1 + 23x

0y0 + x−1y1

= 13x

2 + 53yx+ 2

3 + 1xy

Note: From thetheorem abovef(x) = 3x3 + x2 + 4x + 3and g(x) = 2x2. Fromthis we can see thatf(x)g(x) =

( 32x+ 1

2 + 2 · 1x

)+

32 ·

1x2 So q(x) =( 3

2x+ 12 + 2 · 1

x

)and

r(x) = 3.

Now calculating the long division for rational expressions with something other than aunivariate polynomial in the denominator is much more tedious.

Instructions: 7.4.1:Long Division of Polynomial: For f(x)

g(x) .

1. Divide the first term of f(x) by the first term of g(x). And write the quotientabove corresponding degree of f(x) in the long division symbol.

2. Multiply this to g(x) and write below f(x).

3. Subtract the product received in 2) to f(x). You may also think of this as switch-ing the sign of 2) and adding it to f(x) instead. We should have the first term off(x) cancel. If not then you divided wrong in 1).

4. Bring down the next term of f(x).

5. Repeat all steps for the new line until the remainder cannot be divided by g(x).This happens when deg(g(x)) < deg(new line). The new line is r(x).

Example: 7.4.2:

Perform the long division. What is f(x), g(x), q(x), and r(x).

1. x3+x2−1x−1

x− 1)

x3 + x2 − 1Write the division in the long divisionform.



x2

x− 1)

x3 + x2 − 1The first term of f(x) is x3 and thefirst term of g(x) is x. Divide to getx3

x = x2. Now write this above theterm of f(x) that has degree 2.

x2

x− 1)

x3 + x2 − 1− x3 + x2

2x2

Multiply x2 to g(x) to get x2(x− 1) =x3 − x2. Now subtract this from f(x).Notice that the first term cancels.

x2 + 2xx− 1

)x3 + x2 − 1

− x3 + x2

2x2

Bring down the next term of f(x).Since it is blank, there is nothing to do.So we not repeat the steps. 2x2

x = 2x.Write above the corresponding degreeof f(x).

x2 + 2xx− 1

)x3 + x2 − 1

− x3 + x2

2x2

− 2x2 + 2x

Multiply 2x(x − 1) = 2x2 − 2x andsubtract this from the new line.

x2 + 2xx− 1

)x3 + x2 − 1

− x3 + x2

2x2

− 2x2 + 2x2x− 1

Bring down the next term.

x2 + 2x+ 2x− 1

)x3 + x2 − 1

− x3 + x2

2x2

− 2x2 + 2x2x− 1

Divide 2xx = 2. Write this above the

corresponding degree of f(x).

x2 + 2x+ 2x− 1

)x3 + x2 − 1

− x3 + x2

2x2

− 2x2 + 2x2x− 1

− 2x+ 2

Multiply 2(x − 1) = 2x − 2. Subtractthis from the new line.

x2 + 2x+ 2x− 1

)x3 + x2 − 1

− x3 + x2

2x2

− 2x2 + 2x2x− 1

− 2x+ 21

Since the degree of the new line (0) isless than the degree of g(x) (1), we aredone

To finish answering the question we have

(a) f(x) = x3 + x2 − 1(b) g(x) = x− 1(c) q(x) = x2 + 2x+ 2



(d) r(x) = 1

The two ways to write this given theorem 7.4.1 are

f(x) = g(x)q(x) + r(x)x3 + x2 − 1 = (x− 1)(x2 + 2x+ 2) + 1

f(x)g(x) = q(x) + r(x)

g(x)x3 + x2 − 1

x− 1 = x2 + 2x+ 2 + 1x− 1

Our answer is the second version.

Notice that f(x) is our numerator/dividend, g(x) is the denominator/divisor, q(x) is thepolynomial above the long division symbol, and r(x) is whatever is leftover when we can nolonger divide.

For the new problems we will omit the play by play instructions for the sake of spaceand perform the long division.

Example: 7.4.3:

Perform the long division. What are the polynomials f , g, q, and r.

1. 4m4+m3+3m2+m+5m2+1

4m2 +m− 1m2 + 1

)4m4 +m3 + 3m2 +m+ 5

− 4m4 − 4m2

m3 −m2 +m−m3 −m

−m2 + 5m2 + 1

6

(a) f(m) = 4m4 +m3 + 3m2 +m+ 5(b) g(m) = m2 + 1(c) q(m) = 4m2 +m− 1(d) r(m) = 6

4m4+m3+3m2+m+5m2+1 = 4m2 + m − 1 +

6m2+1 .

2. 6p4+5p3+3p−53p2−2p

2p2 + 3p+ 23p2 − 2p

)6p4 + 5p3 + 3p− 5− 6p4 + 4p3

9p3

− 9p3 + 6p2

6p2 + 3p− 6p2 + 4p

7p− 5

(a) f(p) = 6p4 + 5p3 + 3p− 5(b) g(p) = 3p2 − 2p(c) q(p) = 2p2 + 3p+ 2(d) r(p) = 7p − 5 Since we have 7p

and the −5 left over.

6p4+5p3+3p−53p2−2p = 2p2 + 3p+ 2 + 7p−5

3p2−2p .

3. 2n3+7n2−4n+7n2+2n−1



2n + 3n2 + 2n− 1

)2n3 + 7n2 − 4n + 7

− 2n3 − 4n2 + 2n3n2 − 2n + 7− 3n2 − 6n + 3

− 8n+ 10

(a) f(n) = 2n3 + 7n2 − 4n+ 7(b) g(n) = n2 + 2n− 1(c) q(n) = 2n+ 3(d) r(n) = −8n+ 10

2n3+7n2−4n+7n2+2n−1 = 2n+ 3 + −8n+10

n2+2n−1 .

4. q4−13q−42q2−q−6

t2 + t + 7t2 − t− 6

)t4 − 13t− 42

− t4 + t3 + 6t2t3 + 6t2 − 13t− t3 + t2 + 6t

7t2 − 7t− 42− 7t2 + 7t+ 42

0

(a) f(t) = t4 − 13t− 42

(b) g(t) = t2 − t− 6

(c) q(t) = t2 + t+ 7

(d) r(t) = 0

t4−13t−42t2−t−6 = t2 + t+ 7 + 0.


Math 103 Synthetic Division and the Remainder Theorem

7.5 Synthetic Division and the Remainder Theorem

Goals:1. Divide polynomials

using synthetic divi-sion.

2. Evaluate a polyno-mial function usingthe Remainder The-orem.

3. Show that a num-ber is a solution of apolynomial equationusing the RemainderTheorem.

There is a special way to do polynomial division. This is called synthetic division. Be verywary. You may only use synthetic division when the denominator is of the form x− c.

Instructions: 7.5.1:Synthetic division:

1. Write the synthetic division of

anxn + an−1x

n−1 + . . .+ a1x+ a0

x− c.

For the sake of ease we will use 3x2+2x+4x+2 . Write this as

3 2 4

c = −2

Notice that we are only looking at the constant terms and the coefficient of thepolynomial.

2. Bring the coefficient of the first term down.

3 2 4

c = −2 ↓

3

3. Multiply this coefficient by c, and write below the next number to the right.

3 2 4

c = −2 −6

3

4. Add this column together.

3 2 4

c = −2 −6

3 −4

5. Repeat 3 and 4 until you get to the last constant.

3 2 4

c = −2 −6 8

3 −4 12

6. The numbers on the bottom, not including the last number to the far right,represent the coefficients of the quotient, q(x). The quotient is always one degreeless than f(x). The last number on the far right represents the remainder, r(x).

3x2 + 2x+ 4x+ 2 = 3x− 4 + 12

x+ 2



Example: 7.5.1:

Do the following division using synthetic division.

1. m2−7m−11m−8

1 − 7 − 118 8 8

1 1 − 3

m2−7m−11m−8 = m+ 1− 3

m−8

2. 2p2+7p−39p−7

2 7 − 397 14 147

2 21 108

2p2+7p−39p−7 = 2p+ 21 + 108

p−7

3. n3+7n2+14n+3n+2

1 7 14 3− 2 − 2 − 10 − 8

1 5 4 − 5

n3+7n2+14n+3n+2 = n2 + 5n+ 4− 5

n+2

4. q3−10q2+20q+26q−5

1 − 10 20 265 5 − 25 − 25

1 − 5 − 5 1

q3−10q2+20q+26q−5 = q2 − 5q − 5 + 1

q−5

5. 5m3−13m2+40m+18m−7

5 − 13 40 187 35 154 1358

5 22 194 1376

5m3−13m2+40m+18m−7 = 5m2 + 22m+ 194 + 1376

m−7

6. −5n2+n3+8n+41+n First rearrange so that it looks more familiar

−5n2 + n3 + 8n+ 41 + n

= n3 − 5n2 + 8n+ 4n+ 1



1 − 5 8 4− 1 − 1 6 − 14

1 − 6 14 − 10

−5n2+n3+8n+41+n = n2 − 6n+ 14− 10

n+1

This type of problem has a very nice property that has to do with the Euclidean divisionalgorithm, theorem 7.4.1. We know that

f(x) = g(x)q(x) + r(x).

In this case though g(x) = x− c, and r(x) will always be a constant. Call it r. Hence

f(x) = (x− c)q(x) + r.

What happens when we plug in c for x?

f(c) = (c− c)q(c) + r

f(c) = (0)q(c) + r

f(c) = r.

So we can evaluate f at c to be r. We call this the remainder theoremTheorem 7.5.1:

If a polynomial f(x) is divided by x− c, then the remainder of f(x)x−c is f(c).

This may seem like a waste of time until you consider a problem like 5 from example 7.5.1.Evaluating f(m) = 5m3−13m2 +40m+28 at 7 is fairly difficult. Since we have the syntheticdivision done we can conclude that

f(m) = (m− 7)(5m2 + 22m+ 194) + 1376.

Sof(7) = 1376.

Lets try some problems with larger exponents to see this.Example: 7.5.2:

Use synthetic division and the remainder theorem (theorem 7.5.1) to find f(c).

1. f(x) = 4x6 + 5x5 + 3x4 + 3x2 + 4 at x = 3.

4 5 3 0 3 0 43 12 51 162 486 1467 4401

4 17 54 162 489 1467 4405

So f(3) = r = 4405.

2. f(x) = 2x5 − 30x3 + 6x2 + 12x+ 3 at x = −4.



2 0 − 30 6 12 3− 4 − 8 32 − 8 8 − 80

2 − 8 2 − 2 20 − 77

So f(−4) = r = −77.


Math 103 Rational Equations and Inequalities

7.6 Rational Equations and Inequalities

Goals:1. Solve rational equa-

tions.

2. Solve problemsinvolving ratio-nal functions thatmodel appliedsituations.

3. Solve rational in-equalities.

The purpose of the last five sections was to build up your tool box and your understandingof rational expressions. Now that we understand rational expressions and their function, aswell as the restrictions made on the domain, we can solve rational equations and inequalities.

7.6.1 Rational Equations

Rational equations are equations that have a rational expression in them.

f(x)g(x) = c

Where f(x) and g(x) are polynomials.

There are a different ways of solving rational equations. We will discuss two. Onething to note throughout this book, there are different ways to solve all these problems.The questions is, when you do solve them a different way, are you certain that they aremathematical correct? Can you justify that what you did is okay? If you can do thesethings, then you may perform the math anyway you like.

Instructions: 7.6.1:Solving Rational Equations option 1:

1. Find the LCD of all the rational expressions and list the inputs that result in adivision by zero.

2. Multiply both sides by the LCD so there are no fractions left.

3. Solve the equation.

4. Reject any solution which gave us a division by zero.

Instructions: 7.6.2:Solving Rational Equations option 2:

1. Bring all rational expressions to one side, and add/subtract them 2 at a time untilyou have one fraction.

f(x)g(x) + p(x)

q(x) = 0

f(x)q(x) + p(x)g(x)g(x)q(x) = 0

2. The denominator is the LCD. Since it is equal to zero, we can multiply both sidesby the denominator to eliminate it and get.

f(x)q(x) + p(x)g(x) = 0

3. Solve the equation.

4. Reject any solution which gave us a division by zero.

The second version may seem more difficult, but this is the same way that you solvenon-rational equations with fractional coefficients. The reason that I stay away from option1, is because it only works on rational expressions, and given more than 2 fractions, findingthe LCD of multiple expressions can become tedious and difficult. I will show both ways sothere will be more references.



Example: 7.6.1:

Solve the rational equations.

1. 16k2 = 1

3k3 − 1k

LCD=6k3, and k 6= 0.

16k2 = 1

3k3 −1k

16k2 · 6k

3 =(

13k3 −

1k

)6k3

k = 2− 6k2

6k2 + k − 2 = 0

6k2 + k − 2 = 0

(3k + 2)(2k − 1) = 0

So k = − 23 , or 1

2 . Both are okay sincewe won’t have division by zero whenwe plug them in.

16k2 = 1

3k3 −1k

16k2 −

13k3 + 1

k= 0

16k2 ·

k

k− 1

3k3 ·22 + 1

k= 0

k

6k3 −2

6k3 + 1k

= 0

k − 26k3 + 1

k· 6k2

6k2 = 0

k − 26k3 + 6k2

6k3 = 0

k − 2 + 6k2

6k3 = 0

6k2 + k − 2 = 0

Solving the same way we will get thesame answer.

2. b+64b2 + 3

2b2 = b+42b2

LCD=4b2, and b 6= 0.

b+ 64b2 + 3

2b2 = b+ 42b2(

b+ 64b2 + 3

2b2

)· 4b2 = b+ 4

2b2 · 4b2

b+ 6 + 6 = 2b+ 8

b = 4

There is no restriction on 4. So this isa solution.

b+ 64b2 + 3

2b2 = b+ 42b2

b+ 64b2 + 3− b− 4

2b2 = 0

b+ 64b2 + 3− b− 4

2b2 · 22 = 0

b+ 64b2 + −2b− 2

4b2 = 0

b+ 6− 2b− 24b2 = 0

−b+ 4 = 0

b = 4



3. 1v + 3v+12

v2−5v = 7v−56v2−5v

(a) In this problem I will use a common denominator instead of the LCD. Thiswill illustrate multiple ways to solve these problems. CD=v(v2− 5v), andv 6= 0 or 5.

(1v

+ 3v + 12v2 − 5v

)v(v2 − 5v) = 7v − 56

v2 − 5v · v(v2 − 5v)

v2 − 5v + 3v2 + 12v = 7v2 − 56v

3v2 − 63v = 0

3v(v − 21) = 0

So v = 0, or 21. We have a restriction at v = 0 This will give us divisionby zero in 1

v . So the solution is v = 21.(b)

1v

+ 3v + 12v2 − 5v = 7v − 56

v2 − 5v

1v

+ 3v + 12v2 − 5v −

7v − 56v2 − 5v = 0

1v

+ 3v + 12− 7v + 56v2 − 5v = 0

1v· v − 5v − 5 + −4v + 68

v2 − 5v = 0

v − 5v(v − 5) + −4v + 68

v2 − 5v = 0

v − 5− 4v + 68v(v − 5) = 0

−3v + 63 = 0

v = 21

Notice depending on how you solve these problems you get a different end resultfor the second to the last step. However, when you have the restrictions thesolution set will be the same.

4. 1r−2 + 1

r2−7r+10 = 6r−2

(a) LCD=(r − 2)(r − 5), and r 6= 2 or 5.



1r − 2 + 1

r2 − 7r + 10 = 6r − 2(

1r − 2 + 1

(r − 2)(r − 5)

)· (r − 2)(r − 5) = 6

r − 2 · (r − 2)(r − 5)

(r − 5) + 1 = 6(r − 5)

r − 4 = 6r − 30

−5r = −26

r = 265

So r = 265 . This solution is okay. No restriction.

(b)

1r − 2 + 1

r2 − 7r + 10 = 6r − 2

1r − 2 −

6r − 2 + 1

r2 − 7r + 10 = 0

−5r − 2 ·

r − 5r − 5 + 1

(r − 2)(r − 5) = 0

−5r + 25(r − 2)(r − 5) + 1

(r − 2)(r − 5) = 0

−5r + 25 + 1(r − 2)(r − 5) = 0

−5r + 26 = 0

r = 265

5. p+5p2+p = 1

p2+p −p−6p+1

(a) LCD=p(p+ 1), and p 6= 0 or −1.

p+ 5p2 + p

= 1p2 + p

− p− 6p+ 1

p+ 5p2 + p

· p(p+ 1) =(

1p2 + p

− p− 6p+ 1

)· p(p+ 1)p+ 5 = 1− p(p− 6)

p+ 4 = −p2 + 6p

p2 − 5p+ 4 = 0

(p− 4)(p− 1) = 0

So p = 4, or 1. Both are okay since we won’t have division by zero whenwe plug them in.



(b)

p+ 5p2 + p

= 1p2 + p

− p− 6p+ 1

p+ 5p2 + p

− 1p2 + p

+ p− 6p+ 1 = 0

p+ 5− 1p2 + p

+ p− 6p+ 1 ·

p

p= 0

p+ 4p2 + p

+ p2 − 6pp(p+ 1) = 0

p2 − 5p+ 4p2 + p

= 0

p2 − 5p+ 4 = 0

(p− 4)(p− 1) = 0

Solving the same way we will get the same answer.

6. x+3x2−6x−16 + x−14

x2−4x−32 = x+1x2+6x+8

(a) First factor all the denominators.

x+ 3x2 − 6x− 16 + x− 14

x2 − 4x− 32 = x+ 1x2 + 6x+ 8

x+ 3(x− 8)(x+ 2) + x− 14

(x− 8)(x+ 4) = x+ 1(x+ 4)(x+ 2)

LCD=(x− 8)(x+ 4)(x+ 2), and x 6= 8, −4, or −2.Notice the cancelation when we multiply both sides by the LCD.

i. (x+ 3

(x− 8)(x+ 2)

)(x− 8)(x+ 4)(x+ 2)

=(x+ 3)(x+ 4)

=x2 + 7x+ 12

ii. (x− 14

(x− 8)(x+ 4)

)(x− 8)(x+ 4)(x+ 2)

=(x− 14)(x+ 2)

=x2 − 12x− 28



iii.x+ 1

(x+ 4)(x+ 2) · (x− 8)(x+ 4)(x+ 2)

=(x+ 1)(x− 8)

=x2 − 7x− 8

Plug back in to get

x2 + 7x+ 12 + x2 − 12x− 28 = x2 − 7x− 8

2x2 − 5x− 16− x2 + 7x+ 8 = 0

x2 + 2x− 8 = 0

(x+ 4)(x− 2) = 0

So x = −4, or 2. We have a restriction for x = −4, so x = 2 is the onlysolution.

(b) Using the factored form already solved for above,

x+ 3(x− 8)(x+ 2) + x− 14

(x− 8)(x+ 4) −x+ 1

(x+ 4)(x+ 2) = 0

x+ 3(x− 8)(x+ 2) ·

x+ 4x+ 4 + x− 14

(x− 8)(x+ 4) ·x+ 2x+ 2 −

x+ 1(x+ 4)(x+ 2) = 0

(x+ 3)(x+ 4)(x− 8)(x+ 2)(x+ 4) + (x− 14)(x+ 2)

(x− 8)(x+ 2)(x+ 4) −x+ 1

(x+ 4)(x+ 2) = 0

(x+ 3)(x+ 4) + (x− 14)(x+ 2)(x− 8)(x+ 2)(x+ 4) − x+ 1

(x+ 4)(x+ 2) ·(x− 8)(x− 8) = 0

(x+ 3)(x+ 4) + (x− 14)(x+ 2)(x− 8)(x+ 2)(x+ 4) − (x+ 1)(x− 8)

(x− 8)(x+ 4)(x+ 2) = 0

(x+ 3)(x+ 4) + (x− 14)(x+ 2)− (x+ 1)(x− 8)(x− 8)(x+ 2)(x+ 4) = 0

(x+ 3)(x+ 4) + (x− 14)(x+ 2)− (x+ 1)(x− 8) = 0

Doing the distribution and combining like terms will lead use to factorand get the same solution

7.6.2 Rational Inequalities

We will now define what a rational inequality is. Remember an inequality requires, well aninequality. Hence the only difference between what we defined as a rational function and arational inequality is the inequality and the constant that the function is above or below.

f(x)g(x) = anx

n + an−1xn−1 + . . .+ a2x

2 + a1x+ a0

bmxm + bm−1xm−1 + . . .+ b2x2 + b1x+ b0> 0

where the ai and bj are real numbers, and g(x) 6= 0. Also we may substitute > with , <, ≥,or ≤.

Note: When comput-ing inequalities, rememberthat you may not divide ormultiply to both sides byanything that has a vari-able in it. This is becausewe do not know if what wemultiplied or divided by isnegative.



Instructions: 7.6.3:Solving Rational Inequalities:

For each of the statements we may replace > with , <, ≥, or ≤. Let f(x) and g(x) bepolynomials.

f(x)g(x) = anx

n + an−1xn−1 + . . .+ a2x

2 + a1x+ a0

bmxm + bm−1xm−1 + . . .+ b2x2 + b1x+ b0> 0

1. Write the inequality in the form f(x)g(x) > 0.

2. Determine where f(x)g(x) = 0. This happens when f(x) = 0.

3. Determine where f(x)g(x) is undefined. This happens when g(x) = 0.

4. Draw a number line broken into parts, labeling the roots and the number wheref(x)g(x) is undefined. For the sake of ease, we will only plot c1, c2 and c3.

c1 c2 c3

x

5. Test one number in each of the intervals by picking a number in each interval andplugging it into f(x)

g(x) . We only care about if it is positive or negative, so it maybe easier to have f(x)

g(x) factored.

(a) If f(x)g(x) is negative, then f(x)

g(x) < 0 on that interval.

(b) If f(x)g(x) is positive, then f(x)

g(x) > 0 on that interval.

6. Write the solution set using interval notation.

When using ≥ or ≤, you must include the roots in your solution set.

Similarly you may use the multiplicity rule to determine if the sign changes or stays thesame.

Example: 7.6.2:

Solve the rational inequalities.

1. x−7x−1 < 0

1) Done2) x− 7 = 0 =⇒ x = 73) x− 1 = 0 =⇒ x = 14) Label

−9 −8 −7 −6 −5 −4 −3 −2 −1 0 2 3 4 5 6 8 971x

5) Using zero for ease,0− 70− 1 = positive

Using the multiplicity we get



−9 −8 −7 −6 −5 −4 −3 −2 −1 0 2 3 4 5 6 8 971

+ − +

x

6) For every number in (1, 7), x−7x−1 is less than zero.

−9 −6 −3 3 6 9

−15

−10

−5

5

10

15

• n

y

2. n+15n+6 ≤ 3

1)

n+ 15n+ 6 − 3 ≤ 0

n+ 15n+ 6 −

3(n+ 6)n+ 6 ≤ 0

n+ 15− 3n− 18n+ 6 ≤ 0

−2n− 3n+ 6 ≤ 0

2) −2n− 3 = 0 =⇒ n = − 32

3) n+ 6 = 0 =⇒ n = −64) Label

−9 −8 −7 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9−1.5−6n

5) Using zero for ease,−2(0)− 3

0 + 6 = negative


−9 −8 −7 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9−1.5−6

−+−

n

6) For every number in (−∞,−6]∪ [−1.5,∞), −2n−3n+6 is less than or equal to

zero.



−9 −6 −3 3 6 9

−15

−10

−5

5

10

15

• n

y

3. p2−3p+2p2−2p−3 > 0

1) Done2)

p2 − 3p+ 2 = 0

(p− 2)(p− 1) = 0

p− 2 = 0 or p− 1 = 0

p = 2 or p = 1

3)

p2 − 2p− 3 = 0

(p− 3)(p+ 1) = 0

p− 3 = 0 or p+ 1 = 0

p = 3 or p = −1

4) Label

−9 −8 −7 −6 −5 −4 −3 −2 0 4 5 6 7 8 9−1 1 2 3

p

5) Using zero for ease,02 − 3(0) + 202 − 2(0)− 3 = negative


−9 −8 −7 −6 −5 −4 −3 −2 0 4 5 6 7 8 9−1 1 2 3

−+ + − +p



6) For every number in (−∞,−1) ∪ (1, 2) ∪ (3,∞), p2−3p+2p2−2p−3 is less than or

equal to zero.

−9 −6 −3 3 6 9

−15

−10

−5

5

10

15

n

y

It is hard to see but the part in green is ever so slightly above the y = 0line. We know this because the zeros are at 1 and 2.

4. (m+7)(m−3)(m−5)2 > 0

1) Done2)

(m+ 7)(m− 3) = 0

m+ 7 = 0 or m− 3 = 0

m = −7 or m = 3

3)

(m− 5)2 = 0

m− 5 = 0

m = 5

With multiplicity 2.

4) Label

−9 −8 −6 −5 −4 −3 −2 −1 0 1 2 4 6 7 8 953−7m

5) Using zero for ease,

(0 + 7)(0− 3)(0− 5)2 = negative




−9 −8 −6 −5 −4 −3 −2 −1 0 1 2 4 6 7 8 953−7

−+ + +

m

6) For every number in (−∞,−7)∪(3, 5)∪(5,∞), (m+7)(m−3)(m−5)2 is greater than

zero.

−9 −6 −3 3 6 9

−15

−10

−5

5

10

15

• • n

y


Math 103 Formulas and Applications of Rational Equations

7.7 Formulas and Applications of Rational Equations

Goals:1. Solve a formula with

a rational expressionfor a variable.

2. Solve business prob-lems involving aver-age cost.

3. Solve problems in-volving time in mo-tion.

4. Solve problems in-volving work.

When solving rational equation we need to bring the variable that we want to solve for tothe top and to one side. For example; solve for x in

4 = 3r + x

x

Get rid of the x in the denominator by multiplying both sides by x. Then solve

4 = 3r + 2xx

4x = 3r + 2x

4x− 2x = 3r

2x = 3r

x = 32r

Some examples of these types of problems are, average cost, time in motions, and work.We may calculate these in the same manner as above. What I would like to do is give youthe general forms of all these types of problems. Your goal is to look at each of the problemsto determine which formula is needed.

Theorem 7.7.1:

The average cost of an object is determined by

C̄(x) = fixed cost + cx

x

where x represents the amount of units produced, and c represents the cost of theproduct. This formula is derived from the cost function.

C(x) = fixed cost + cx

Without the division we do not have an average.

Theorem 7.7.2:

Time in motion is given byt = d

r

t is time, d is distance, and r is rate. You can get this from knowing that the distancetraveled is tr = d.

Theorem 7.7.3:

Time for multiple people to work and finish a job is given by

t

a+ t

b= 1.

t is the time it takes to finish together, a is the time it takes for one person to finishsolo, and b is the time it takes for the other person to finish alone.

Example: 7.7.1:

1. It cost Okuhara kamaboko (fish cake) $10,000 to begin production of kamaboko,plus $0.50 for every unit of the good produced. Let x be the number of units



produced by the company.

(a) What is the formula for the cost?

Clearly this uses theorem 7.7.1. The cost is given by

$10, 000 + $0.5(x)

(b) What is the average cost formula?

$10, 000 + $0.5(x)x

(c) What is the average cost when producing 10,000 units?

C(x) = $10, 000 + $0.5(x)x

C(10, 000) = $10, 000 + $0.5(10, 000)10, 000

= $1.5

The average cost of each kamaboko when producing 10,000 units is $1.50.Notice if we produce 50,000 units we would have

$10, 000 + $0.5(50, 000)50, 000 = $0.7

This should make sense. The more product produced, the lower the aver-age cost because the initial cost does not change.

2. There are two farmer. These farmers work in the loi kalo (taro farm). Farmerone can harvest one plot in 2 hours, while the other farmer takes 3.5 hours.How long will it take them to harvest a plot together?

This is a work problem. We will use theorem 7.7.3. We know that a is 2 and bis 3.5. Plugging in to the formula we get

t

2 + t

3.5 = 1

t

2 + t3510

= 1

t

2 + 10t35 = 1

t

2 + 2t7 = 1

7t14 + 4t

14 = 1

11t14 = 1

t = 1411



t = 1.27

This says that it will take them approximately 1411 or 1.27 hours to harvest one

plot in the loi if they work together.

3. In still water, Keala can travel on his paddle board 2 times as fast as thecurrent in Ala Moana beach. He takes a 3-mile round trip down Ala Moanabeach and returns in 2 hours. Find the rate of the current.

This is clearly a motion problem theorem 7.7.2.Since rate of the current is what we want to find, let x=rate of the current. Ifyou move with the current, then the current supports (+) your speed. If yougo against the current, then the current decreases (−) your speed. Hence wehave

rate of paddle board distance time

with current 2x+ x 3 Time down= 32x+x

against current 2x− x 3 Time back= 32x−x

We also knowTime down + Time back = 4

Hence3

2x+ x+ 3

2x− x = 4

33x + 3

x= 4

33x + 9

3x = 4

3 + 93x = 4

12 = 12x

x = 1

This says that the rate of the current was 1 mile per hour.


Math 103 Modeling Using Variation

7.8 Modeling Using Variation

Goals:1. Solve direct varia-

tion problems.

2. Solve inverse varia-tion problems.

3. Solve combined vari-ation problems.

4. Solve problems in-volving joint varia-tion.

Variation problems are 100% depending on your ability to read the problem. You need tofind the key words that tell you what formula you need to use.

Definition 7.8.1: Some translations.

• y varies as x means y = kx.

• y varies jointly as x and z means y = kxz.

• y varies inversely (is inversely proportional) as x means y = kx = k 1

x .

• y varies directly as x2 means y = kx2.

• y varies as the nth power of x means y = kxn. This is another way to say y = kx2.This can be written as, y varies as the 2nd power of x.

Instructions: 7.8.1:Variation problems:

1. Determine what type of variation problem. Write the formula.

2. Use the information given to plug in so that you can solve for k.

3. Plug back into the equation.

4. Now use the remaining information to answer the question.

Lets look at a multitude of problems so that we can determine what type of variationproblem they are.

Example: 7.8.1:

1. The variable y- varies directly with x. Find an equation to represent thisrelationship if x=9, when y=30. Find y when x = 15.

Varies directly says that this looks like

y = kx

We know x=9, when y=30, so

30 = k(9) =⇒ k = 309 = 10

3

We now have y = 103 x. Lastly we want to know what y is when x = 15. Hence

y = 103 (15) = 10(5) = 50.

So y is 50 when x = 15.

2. The variable y- varies directly with x. Find an equation to represent thisrelationship if x=20, when y = 27. Find y when x = 12.


y = kx



We know x=20, when y=27, so

27 = k(20) =⇒ k = 2720

We now have y = 2720x. Lastly we want to know what y is when x = 12. Hence

y = 2720(12) = 81

5 .

So y is 815 when x = 12.

3. The variable d- varies inversely with x. Find an equation to represent thisrelationship if x=4, when d=3. Find d when x = 7.

Varies inversely says that this looks like

d = k1x

We know x=4, when d=3, so

3 = k14 =⇒ k = 12

We now have d = 12 1x Lastly we want to know what d is when x = 7. Hence

d = 1217 = 12

7 .

So d is 127 when x = 12.

4. The variable P - varies inversely with n. Find an equation to represent thisrelationship if n = 1

2 , when P = 8. Find P when n = 12.

Varies inversely says that this looks like

P = k1n

We know n = 12 , when P=8, so

8 = k112

=⇒ k = 4

We now have P = 4 1n Lastly we want to know what P is when n = 12. Hence

P = 4 112 = 1

3 .

So P is 13 when n = 12.

Lets look at a word problem to see that the entire problem can be seen if we read itproperly.

a) For a given distance, the speed s in miles per hour of a car being driven is inverselyproportional to the time t it has been drive. Find an equation to express this relationshipif the speed is 45 miles per hour for 2 hours of driving. Find the speed for 1.5 hours of driving.



Lets rewrite the important parts of the first sentence of the problem

The speed sis inversely proportional to the time t.

s=k 1t

Remember that “is” represents an = sign, and the inverse of t represents 1t . See how the

math equation can be read exactly like the words. Knowing this, we can now solveWe know s = 45, when t=2, so

45 = k12 =⇒ k = 90

We now have s = 90 · 1t Lastly we want to know what s is when t = 1.5. Hence

s = 90 11.5

= 90 11510

= 90 · 23

= 60

So the speed s is 60 miles per hour when the time t is 1.5.Let us continue with more problems so that we may have many examples of how different

variation problems are worded.Example: 7.8.2:

1. Hooke’s Law in physics says that the force to stretch a spring past its restingstate varies directly with the distance the spring is stretched. Find an equationto express this relationship if a force of 9 pounds stretches a spring 7 inchespast resting. How many inches will a force of 15 pounds stretch the spring?


F = kd

We know F = 9, when d=7, so

9 = k(7) =⇒ k = 97

We now have F = 97d Be careful. We are asked something slightly different

here. We want to know what d is when F = 15. Hence

15 = 97d =⇒ d = 35

3 .

So we will stretch the spring 353 inches when F = 15.

2. The number of calories burned c while playing full-court basketball variesdirectly with the number of hours h spent playing. Find an equation to expressthis relationship if the number of calories burned for a 175lb man during 1 hourof playing full court basketball is 871. Use this equation to find how many calo-ries are burned by a 175lb man during 2.5 hours of playing full court basketball.




c = kh

We know c = 871, when h=1, so

871 = k(1) =⇒ k = 871

We now have c = 871h We want to know what c is when h = 2.5. Hence

c = 871(2.5) =⇒ c = 2177.5

So a 175lb man will burn 2177.5 calories when he plays full court basketballfor 2.5 hours.

3. The distance of a body falling from rest varies directly as the square of thetime it falls (ignoring air resistance). If a ball falls 144 ft in 3 seconds, how farwill the ball fall in seven seconds?

Varies directly as a square says that this looks like

d = kt2

We know d = 144, when t=3, so

144 = k(3)2 =⇒ k = 16

We now have d = 16t2 We want to know what d is when t = 7. Hence

d = 16(72) =⇒ d = 784

So a ball will fall 784 feet from rest in 7 seconds.

4. The weight of a body varies inversely as the square of its distance from thecenter of the earth. If the radius of the earth is 4000 miles, how much would a200-pound man weight 1000 miles above the surface of the earth?

This is an interesting problem that is easier to see with a picture.

•

F

r=4000

man1000 m above earth

Based on the wording we have

w = k1d2 .



The information that is given is that the man is 200lb. It is implied that he iscurrently at ground level of the earth. So when d = 4000, w = 200. Hence

200 = k1

40002

k = 200 · 40002

k = 3, 200, 000, 000.

So w = 3, 200, 000, 000 1d2 . We want to know the mans weight if he was 1000m

above the earth. This means that he is 5000m from the center, so d = 5000.

w = 3, 200, 000, 000 · 150002 = 128

Hence the man who weight 200lb on earth will weigh 128lb, 1000 miles abovethe surface of the planet.

Now that we have the fundamentals of variation problems, we need to solve problems withmultiple variations. These problems will involve writing a formula that contains more thanone of the variations that we have been working on. These are not tremendously differentthan what we have worked on already. Again the key is to read the problem carefully. Theproblem will tell you what to do word for word.

Example: 7.8.3:

1. The stopping distance of a car varies inversely with the friction of the roadbut directly with the speed of the car. At the speed of 20 mph, car takes 50feet to stop on the road with friction value of 2. Find the stopping distance ata speed of 40 mph with road friction value to be 4.

Lets again look at this word for word.

The stopping distance of a car varies inversely with the friction of the roadbut directly with the speed of the car.

This translates asd= k

1f·s

We know s = 20, d = 50, and f = 2. Hence

50 = k12 · 20

50 = k(10)

k = 5

The equation is not d = 5 1f · s. We want to know the stopping distance d at a

speed of 40 mph with friction value 4.

d = 514 · 40

d = 5(10)

k = 50

So it will take 50 feet to stop with a car traveling at a speed of 40 mph with



road friction value of 4.

2. The number of hours h that it takes m men to assemble x machines variesdirectly as the number of machines and inversely as the number of men. Iffour men can assemble 12 machines in four hours, how many men are neededto assemble 36 machines in eight hours?

Translating this word for word we have

The number of hours h that it takes m men to assemble x machines variesdirectly as the number of machines and inversely as the number of men.

h= kx· 1m

We know that 4 men can assemble 12 machines in 4 hours. So

4 = k(12)14

4 = k(3)

k = 43

The formula is h = 43x ·

1m . We want to find how many men it will take to

assemble 36 machines in 8 hours.

8 = 43(36) · 1

m

8 = 4(12) · 1m

m = 4(12)8

m = 6

It will take 6 men to assemble 36 machines in 8 hours.

Radicals, Radi-cal Functions, andRational Expo-nents

Math 103 Radical Expressions and Functions

8 Radicals, Radical Functions, and Rational Ex-ponents

Radical functions are defined asf(x) = n

√x

Here are some examples of the graphs of radical functions.

−5 −4 −3 −2 −1 1 2 3 4 5

−5−4−3−2−1

12345

x

y

Figure 10: f(x) =√x = x

12

−5 −4 −3 −2 −1 1 2 3 4 5

−5−4−3−2−1

12345

x

y

Figure 11: f(x) = 3√x = x

13

8.1 Radical Expressions and FunctionsGoals:

1. Evaluate squareroots.

2. Evaluate squareroot functions.

3. Find the domain ofsquare root func-tions.

4. Use models that aresquare root func-tions.

5. Simplify expressionsof the form 2

√a .

6. Evaluate cube rootfunctions.

7. Simplify expressionsof the form 3

√a .

8. Find even and oddnth roots.

9. Simplify expressionsof the form n

√a .

The domain of this function is dependent on whether or not n is even or odd. If n is even,then the domain is [0,∞). If n is odd, then the domain is (−∞,∞). We can see this intheir respective graphs. This is possible because of how we evaluate radicals. Remember,we want to find repeated multiplication n time when trying to simplify nth roots. Becauseof the odd n, this will allow us to have for example 3

√−27 = 3

√−3 · −3 · −3 = −3. So we can

plug in a number that make the interior of the radical negative and get a negative numberout. So these numbers are in the domain of odd nth roots. This is not true for even roots.There is no way to multiply a negative number by itself an even number of times and havethe end result be negative.

Example: 8.1.1:

Determine if the given input is in the domain of the radical function. If it is what isthe output? What is the domain of the function?

1) f(x) =√x− 2 + 5, f(1)

The restriction on even roots is thatthe interior of the root is greaterthan or equal to 0. Want

x− 2 ≥ 0

x ≥ 2

So the domain of f is [2,∞). Sincewe are using an input of 1, it is notin the domain.

2) g(m) = 3√x+ 1− 4

Since the domain of any odd root isall real numbers (−∞,∞), −2 is inthe domain of g.

g(−2) = 3√−2 + 1− 4

= −1− 4

= −5



3) p(n) =√

2n− 5, p(7)

We have an even root, so we needthe interior to be greater than orequal to zero.

2n− 5 ≥ 0

2n ≥ 5

n ≥ 52

So the domain is [ 52 ,∞). We want

to find the output of input 7. Since7 is in [ 5

2 ,∞), we have

p(7) =√

2(7)− 5

=√

14− 5

=√

9

= 4

4) h(t) = 4√−t+ 5− 1, g(-11)

Since we have an even root, wewant −t+ 5 ≥ 0

−t+ 5 ≥ 0

−t ≥ −5

t ≤ 5

The domain of h is (−∞, 5]. Since−11 is in this interval we may findthe output.

h(−11) = 4√−(−11) + 5− 1

= 4√

11 + 5− 1

= 4√

16− 1

= 4√

2 · 2 · 2 · 2− 1

= 2− 1

= 1

When simplifying radical expressions we want to take out as much as possible from theinterior of the radical. We have done this previously with simple numbers. Now lets do thisusing algebraic expressions. We do this the exact same way that we have with numbers.Write out the factors and pull out one for every group of size n. In case you forgot

Theorem 8.1.1:

1. If n is even, then n√xn = |x|.

2. If n is odd, then n√xn = x.

Example: 8.1.2:

1)√

75x7y5

√75x7y5 =

√5 · 5 · 5x7y5

= |5|√

5 · x2 · x2 · x2 · xy5

= |5x · x · x|√

5xy2 · y2 · y

= |5x3y · y|√

5xy

= |5x3y2|√

5xy

2) 6√

218(x+ 1)18

Since we have an even root, the endresult will be in absolute values.

6√

218(x+ 1)18 = 6√

[2(x+ 1)]18

= 6√

([2(x+ 1)]3)6

= |[2(x+ 1)]3|

= 8|(x+ 1)3|



3) 3√

40a7b4

3√40a7b4 = 3√5 · 2 · 2 · 2a7b4

= 2 3√5a3 · a3 · ab4

= 2a · a 3√5ab3 · b

= 2a2b3√

5ab

4) 4√

48m21n100

Grouping by four is dividing. So wecan divide the exponents to get theexponents of the variables fast.

4√48m21n100 = 4√48m20mn100

= |m5n25| 4√

48m

= |m5n25| 4√24 · 3m

= |2m5n25| 4√

3m

5) 9√

(m2 + 1)9

There is an odd root so the endresult is just the interior of the root.

9√

(m2 + 1)9 = m2 + 1

6)√

100x5y

√100x5y =

√10 · 10x5y

= 10√x4 · xy

= 10|x2|√xy

7)√x2 + 2x+ 1

√x2 + 2x+ 1 =

√(x+ 1)(x+ 1)

=√

(x+ 1)2

= |x+ 1|

8)√

16(x+ 1)3y6

√16(x+ 1)3y6

= 4√

(x+ 1)3y6

= 4|(x+ 1)2|√

(x+ 1)y6

= 4(x+ 1)2|y3|√

(x+ 1)


Math 103 Rational Exponents

8.2 Rational Exponents

Goals:1. Use the definition ofa

1n

2. Use the definition ofa

mn

3. Use the definition ofa

−mn

4. Simplify expressionswith rational expo-nents.

5. Simplify radical ex-pressions using ra-tional exponents.

This is in my opinion, the most important section for radical expressions. Understandingthese rules allows us to solve all of the previous problems. It also looks cleaner. All of theserules follow the exponent rules from theorem 1.2.3, but using fractions instead.

We define the radical function as

f(x) = n√x = x

1n .

Using this new definition, lets understand the rules of exponents, and how they play a majorrole in radical functions. Lets look at some examples using these rules

1. Product rule: Same base, add the exponents.√a+ 3√a = a

12 · a 1

3

= a12 + 1

3

= a36 + 2

6

= a56

2. Quotient rule: Same base, subtract the denominator’s exponent from the exponent ofthe numerator.

√a

4√a

= a12

a14

= a12−

14

= a24−

14

= a14

3. Zero-exponent rule:

5√a

5√a

= a15

a15

= a15−

15

= a0

= 1 a 6= 0

This should make sense because we are dividing a number by itself.5√a5√a = 1.

4. Negative exponent rule: Remember a negative exponent is done by erasing the negativesign and writing 1

the rest .



5√a

3√a

= a15

a13

= a15−

13

= a3

15−5

15

= a−2

15

= 1a

215

Note: Notice that noneof these rules have a ad-dition attached to them.One of the most commonmistakes is applying theserules to problems like

(a+ b) 12 .

Many will distribute thepower because powers dis-tribute. This is the rea-son why it is importantto say the entire rule, notjust part of it. Powersdistribute when the inte-rior is strictly multiplica-tion or division. Peoplewill write

(a+ b) 12 = a

12 + b

12 .

This is incorrect, andis called the freshmensdream.

5. Fraction and exponent rule: Exponents distribute in fractions and in multiplication.(√a

b

)=(ab

) 12

= a12

b12

(√a · bc · d

)=(a · bc · d

) 12

= (a · b) 12

(c · d) 12

= a12 · b 1

2

c12 · d 1

2

This is the reason why we can separate square roots as √xy =√x · √y.

6. Exponents to exponents rule: Exponents raised to exponents multiply together.

3√√

a = 3√a

12

=(a

12

) 13

= a12 ·

13

= a16



Similarly in 5, we can distribute the exponent and then multiply in an example like

3√√

a4√b = 3

√a

12 · b 1

4

=(a

12 · b 1

4

) 13

= a12 ·

13 b

14 ·

13

= a16 b

112

These exponential rules allows us to have a fairly simple theorem which is useful insituations where finding the root or finding the exponent first is easier.

Theorem 8.2.1:

Let n√a be a real number and m

n be positive with n 6= 0. Then

amn = n

√am = ( n

√a)m

Example: 8.2.1:

1) 9 32

You will find that this is easier bytaking the square root first.

9 32 = (

√9)3

= 33

= 27

2) 4 23

This one on the other hand is easierperforming the square first.

4 23 = 3√42

= 3√

4 · 4

= 3√

2 · 2 · 2 · 2

= 2 3√

2

Lets apply all these rules in the following examples. Similarly to many of the otherproblems, there are many ways to solve them. As long as you are following the rules ofexponents you will be fine.

Example: 8.2.2:

Convert into rational exponents, then simplify.

1. 2m2 · 4√m3 · 4m−2

2m2 · 4√m3 · 4m−2 = 2m2 · 4(m3) 1

2 · 4m−2 Def of rational exp.

= 2m2 · 4m3· 12 · 4m−2 Exp to exp

= 2m2 · 4m 32 · 4m−2

= 2(4)(4)m2 ·m 32 ·m−2 Commutative prop.

= 32m2+ 32 +(−2) Mult prop.

= 32m 32 Or 32

√m3



2.(

2√p3)−2

(2√p3)−2

=(p

32

)−2Rational exp

= p32 ·(−2) exp to exp

= p−3

= 1p3 Neg exp

3. 3(√x)−1·3

√xy− 1

3

3( 4√y)−7

3(√x)−1 · 3

√xy−

13

3( 4√y)−7 = 3x− 1

2 · 3x 12 y−

13

3(y−7) 14

Rational exp

= 3x− 12 · 3x 1

2 y−13

3y− 74

exp to exp

= 9x− 12 + 1

2 y−13

3y− 74

Mult prop and Comm prop

= 93x

0y−13−(− 7

4 ) Div prop

= 3(1)y− 412 + 21

12 0 exp rule

= 3y 1712

4.(a−1 3√b · 3√a−4b2

)2

(a−1 3√b · 3√

a−4b2)2

=(a−1b

13 · (a−4) 1

3 b2)2

Rational exp

=(a−1b

13 · a− 4

3 b2)2

exp to exp

=(a−1+(− 4

3 )b13 +2)2

Mult prop

=(a−

33−

43 )b

13 + 6

3

)2

=(a−

73 b

73

)2

= a−73 ·2b

73 ·2 Exp to exp



= a−143 b

143

= b143

a143

Neg exp

This can also be written as3

√b14

a14 ,

3

√(b

a

)14,

or (b

a

) 143

.

5.4√

(√x−1y2)−5

x2√y

4

√(√x−1y2

)−5

x2√y=

[((x−1) 1

2 y2)−5

]− 14

x2y12

Rational exp

=

(x−1· 12 y2

)−5· 14

x2y12

Exp to exp

=

(x−

12 y2)− 5

4

x2y12

=

(x−

12 ·(− 5

4 )y2·(− 54 ))

x2y12

Exp to exp

= x58 y−

52

x2y12

= x58−2y−

52−

12 Div prop

= x58−

168 y−

52−

12

= x−118 y−

62

= x−118 y−3

= 1x

118 y3

Neg exp

6.4√√

x−1y43√x2 2√y3·

√x−3√y



This last problem is quite difficult. However this will really determine if youunderstand how to use the exponent rules with fractions properly.

4√√

x−1y4

3√x2 2√y3 ·√x−3√

y=

((x−1) 1

2 y4) 1

4

(x2) 13 (y3) 1

2 · (x−3) 12 y

12

Rational exp

=

(x−1· 12 y4

) 14

x2· 13 y3· 12 · x−3· 12 y12

Exp to exp

=

(x−

12 y4) 1

4

x23 y

32 · x− 3

2 y12

= x−12 ·

14 y4· 14

x23 y

32 · x− 3

2 y12

Exp to exp

= x−18 y1

x23 y

32 · x− 3

2 y12

= x−18 y1

x23 +(− 3

2 )y 32 + 1

2

Mult prop

= x−18 y1

x46−

96 y

42

= x−18 y1

x−56 y

42

= x−18−(− 5

6 )y1− 42 Div prop

= x−3

24 + 2024 y

22−

42

= x1724 y−1

= x1724

yNeg exp


Math 103 Multiplying and Simplifying Radical Expressions

8.3 Multiplying and Simplifying Radical Expressions

Goals:1. Use the product rule

to multiply radicals.

2. Use factoring andthe product rule tosimplify radicals.

3. Multiply radicalsand then simplify.

In the next two sections we will do more examples specific to different rules of exponentsbut in radical form. I personally belief that if you can perform these operations usingfractional exponents, then you should convert all radicals into rational exponents and applythe corresponding rule. I never use radicals. I always convert into exponents.

Lets write Some common rules that show up with radicals. We will prove these usingthe exponent rules.

Theorem 8.3.1:

For a ≥ 0, n√a · n√b = n√ab

Proof.

n√a · n√b = a

1n · b 1

n

= (ab) 1n

= n√ab

For the following problems in all sections, assume that the variables represent positivenumbers. If this assumption is not made, we need to have the absolute values.

Example: 8.3.1:

Simplify the following expressions

1.√

2x5 ·√

52

(a) Radicals:√2x5 ·

√52 =

√2x5 ·

52

=√x

(b) Exponents:√2x5 ·

√52 =

(2x5

) 12

·(

52

) 12

=(

2x5 ·

52

) 12

= x12

2. 9√

12x2y3 · 9√

3x7y6

(a) Radicals:

9√

12x2y3 · 9√

3x7y6 = 9√

12x2y3(3)x7y6

= 9√

36x9y9

=9

√36xxxxxxxxx︸︷︷︸

9 repeats

yyyyyyyyy︸︷︷︸9 repeats

= xy9√

36



(b) Exponents:

9√

12x2y3 · 9√

3x7y6 = (12x2y3) 19 (3x7y6) 1

9

= (12x2y3(3)x7y6) 19

= (36x9y9) 19

= 36 19x9·

19 y9· 19

= 36 19x1y1

3.√

81(x− 2)4

(a) Radicals: √81(x− 2)4 =

√9 · 9︸︷︷︸

2 times

(x− 2)(x− 2)︸︷︷︸2 times

(x− 2)(x− 2)︸︷︷︸2 times

= 9(x− 2)(x− 2)

= 9(x− 2)2

(b) Exponents: √81(x− 2)4 = (92(x− 2)4) 1

2

= 92· 12 (x− 2)4· 12

= 91(x− 2)2

= 9(x− 2)2

4. 3√

6x7y · 3√

9x4y12

(a) Radicals:

3√

6x7y · 3√

9x4y12 = 3√

6x7y(9)x4y12

=3

√3(3)(3)︸︷︷︸3 times

(2) xxx︸︷︷︸3 times

xxx︸︷︷︸3 times

xxx︸︷︷︸3 times

xxy13

= 3x33

√2x2 yyy︸︷︷︸

3 times

yyy︸︷︷︸3 times



y

= 3x3y4 3√

2x2y



(b) Exponents:

3√

6x7y · 3√

9x4y12 = (6x7y) 13 (9x4y12) 1

3

= (6x7y(9)x4y12) 13

= (33(2)x11y13) 13

= 33· 13 (2) 13x11· 13 y13· 13

= 33· 13 (2) 13x11· 13 y13· 13

= 3x3y4(2 13x

23 y

13 ) Exp of x, 11

3 = 3 r2

5. 3√x− 6 · 3

√(x− 6)7

(a) Radicals:

3√x− 6 · 3

√(x− 6)7

= 3√

(x− 6)(x− 6)7

=3

√(x− 6)(x− 6)(x− 6)︸︷︷︸

3 times

(x− 6)(x− 6)(x− 6)︸︷︷︸3 times

(x− 6)(x− 6)

= (x− 6)2 3√

(x− 6)2

(b) Exponents:

3√x− 6 · 3

√(x− 6)7 = (x− 6) 1

3 ·((x− 6)7) 1

3

=((x− 6)(x− 6)7) 1

3

=((x− 6)8) 1

3

= (x− 6)8· 13

= (x− 6) 83

= (x− 6)2(x− 6) 23

6. (−5x2y3z√

2xyz)(−x4z√

10x√z)

(a) Radicals:

(−5x2y3z√

2xyz)(−x4z√

10x√z) = (−5x2y3z)(−x4x)

√(2xyz)(10xz)

= 5x6y3z2√

20x2yz2



= 5x6y3z2(2xz)√

5y D = [0,∞)

= 10x7y3z3√5y

(b) Exponents:

(−5x2y3z√

2xyz)(−x4z√

10xz) = (−5x2y3z(2xyz) 12 )(−x4z(10xz) 1

2 )

= (−5x2y3z)(−x4z)(2xyz) 12 (10xz) 1

2 )

= (5x6y3z2)((2xyz)(10xz)) 12

= (5x6y3z2)((5)22x2yz2) 12

= (5x6y3z2)(2xz)(5y) 12

= (10x7y3z3)(5y) 12


Math 103 Adding, Subtracting, and Dividing Radical Expressions

8.4 Adding, Subtracting, and Dividing Radical Expressions

Goals:1. Add and subtract

radical expressions.

2. Use the quotientrule to simplifyradical expressions.

3. Use the quotientrule to divide radicalexpressions.

Division using radicals also follow the same rules of exponents that we have studied earlier.So we may either use the exponent rules or the following theorem. Similarly in this section,assume that the variables represent positive numbers.

Theorem 8.4.1:

For a ≥ 0, b > 0,n√a

n√b

= n√

ab

Proof.

n√a

n√b

= a1n

b1n

=(ab

) 1n

= n

√a

b

What is nice about the exponent rules instead of theorem 8.4.1 is the ability to do divisionwith the interior of the radicals being the same.

Example: 8.4.1:

Simplify the quotient.

1.√

54a7b11√3a−4b−2

(a) Radicals:

√54a7b11

√3a−4b−2

=√

54a7b11

3a−4b−2

=√

543 a

7−(−4)b11−(−2)

=√

18a11b13

= 3a5b6√

2ab

(b) Exponents:

√54a7b11

√3a−4b−2

= (54a7b11) 12

(3a−4b−2) 12

=(

54a7b11

3a−4b−2

) 12

=(

543 a

7−(−4)b11−(−2)) 1

2

=(18a11b13) 1

2



= (32(2)a10ab12b) 12

= 3a5b6(2ab) 12

2.3√

250q5p3

3√

2q3

(a) Radicals:

3√

250q5p3

3√

2q3= 3

√250q5p3

2q3

= 3√

125q5−3p3

= 3√

53q2p3

= 5p 3√q2

(b) Exponents:

3√

250q5p3

3√

2q3=(

250q5p3

2q3

) 13

= (125q5−3p3) 13

= (53q2p3) 13

= 5p(q2) 13

= 5pq 23

3.3√m2+7m+12

3√m+3

(a) Radicals:

3√m2 + 7m+ 12

3√m+ 3

= 3

√m2 + 7m+ 12

m+ 3

= 3

√(m+ 4)(m+ 3)

m+ 3

= 3√m+ 4

(b) Exponents:



3√m2 + 7m+ 12

3√m+ 3

=(m2 + 7m+ 12

m+ 3

) 13

=(

(m+ 4)(m+ 3)m+ 3

) 13

= (m+ 4) 13

4.3√a3−b3

3√a−b

(a) Radicals:

3√a3 − b3

3√a− b

= 3

√a3 − b3

a− b

= 3

√(a− b)(a2 + ab+ b2)

a− bDiff of cubes

= 3√a2 + ab+ b2

(b) Exponents:

3√a3 − b3

3√a− b

=(a3 − b3

a− b

) 13

=(

(a− b)(a2 + ab+ b2)a− b

) 13

= (a2 + ab+ b2) 13

The most common mistake with exponents in general is the addition and subtraction ofterms that have different variables. Remember when adding we need to have like terms. Wecan consider addition and subtraction like counting. Another way to think about this is weneed to be able to factor out something in common with all the terms before we can add.

Example: 8.4.2:

Perform the indicated operations.

1. 3√

12ab2 + 3b√

75a− 3√

27ab2

3√

12ab2 + 3b√

75a− 3√

27ab2 = 3b(2)√

3a+ 3b(5)√

3a− 3(3)b√

3a



= 6b√

3a+ 15b√

3a− 9b√

3a

= (6b+ 15b− 9b)√

3a

= (6 + 15− 9)b√

3a

= 12b√

3a

2. 2n2√

2m3 − 2m√

50n4m− 3√

18n4m3

2n2√

2m3 − 2m√

50n4m− 3√

18n4m3 = 2n2m√

2m− 10n2m√

2m− 9n2m√

2m

= (2n2m− 10n2m− 9n2m)√

2m

= (2− 10− 9)n2m√

2m

= −17n2m√

2m

3.√

272 +

√757

√272 +

√757 = 7

√27

14 + 2√

7514

= 7√

3 · 3 · 314 + 2

√5 · 5 · 314

= 7(3)√

314 + 2(5)

√3

14

= 21√

3 + 10√

314

= (21 + 10)√

314

= 31√

314

4. 7√

2x3 + 40x3√150x2

5x2√

3x

7√

2x3 + 40x3√

150x2

5x2√

3x= 7√

2x3 · 5x2√3x5x2√

3x+ 40x3

√150x2

5x2√

3x

= 35x2√

2x3√

3x5x2√

3x+ 40x3

√150x2

5x2√

3x



= 35x2√

6x4

5x2√

3x+ 40x3

√150x2

5x2√

3x

= 35x2√

6x4 + 40x3√

5 · 5 · 3 · 2x2

5x2√

3x

= 35x2x2√6 + 40x3(5x)√

3 · 25x2√

3x

= 35x4√6 + 200x4√65x2√

3x

= (35 + 200)x4√65x2√

3x

= 235x4√65x2√

3x

= 47x2√6√3x

= 47x2√

63x

= 47x2√

2x−1

= 47x2x−12√

2

= 47x2− 12√

2

= 47x 32√

2

= 47x√x√

2

= 47x√

2x

Note: I want to showthat sometimes the firstway that you do the prob-lem may work, but can bemuch more difficult thananother way. Either wayis correct, but notice howmuch easier the secondway is.

For problem 4 the solution was the first way that I thought of. However there is an easierway. We should have simplified the right hand side first.

7√

2x3 + 40x3√

150x2

5x2√

3x= 7√

2x3 + 8x√

150x2√

3x

= 7√

2x3 + 8x√

150x2

3x

= 7√

2x3 + 8x√

50x

= 7x√

2x+ 8x(5)√

2x

= 7x√

2x+ 40x√

2x

= (7 + 40)x√

2x

= 47x√

2x


Math 103 Multiplying with More Than One Term and Rationalizing Denominators

8.5 Multiplying with More Than One Term and Rationalizing Denomi-nators

Goals:1. Multiply radical ex-

pressions with morethan one term.

2. Use polynomial spe-cial products to mul-tiply radicals.

3. Rationalize denom-inators containingone term.

4. Rationalize denom-inators containingtwo terms.

5. Rationalize numera-tors.

Throughout the entirety of this book we have been writing out many of the steps thatsometimes may have seemed trivial. However, when there is a time when you are learningsomething new or difficult and you have been skipping the trivial step because it was toeasy, then you won’t know what to do when the problems get more complicated. What dowe do when we multiply a expression to a sum or difference of expressions? We distribute.This can be written generally as

f(x)(g(x) + h(x)) = f(x)g(x) + f(x)h(x)

where f , g, and h are mathematical algebraic. This idea applies when radicals are involved.The distributive property is what we will be applying with the radical expressions.

Example: 8.5.1:


1. 3√x( 3√

24x2 − 3√x)

3√x( 3√24x2 − 3

√x) = 3

√x

3√24x2 − 3√x 3√x

= 3√

24x2(x)− 3√x(x)

= 3√24x3 − 3√x2

= 3√

23(3)x3 − 3√x2

= 2x 3√

3− 3√x2

2. (4√

3 + 3√

2)(4√

3− 3√

2)

(4√

3 + 3√

2)(4√

3− 3√

2)11

22= 4√

3 · 4√

3− 4√

3 · 3√

2 + 3√

2 · 4√

3− 3√

2 · 3√

2

= 16√

3 · 3− 12√

3 · 2 + 12√

2 · 3− 9√

2 · 2

= 16(3)−12√

6 + 12√

6− 9(2)

= 48− 18

= 30

3. ( 3√m− 3)( 3

√m+ 7)

( 3√m+ (−3))( 3

√m+ 7)

11

22



3√m · 3√m+ 3

√m · 7 + (−3) 3

√m+ (−3)7 = 3

√m ·m+ 7 3

√m− 3 3

√m− 21

= 3√m2 + 7 3

√m− 3 3

√m− 21

= 3√m2 + 4 3

√m− 21

4. (3 4√p2 + 1− 3

√2p+ 1)(

√5p− 4 4

√2p3 + 3p2)

(3 4√p2 + 1− 3

√2p+ 1)(

√5p− 4 4

√2p3 + 3p2)

11

22

3 4√p2 + 1 ·

√5p− 3 4

√p2 + 1 · 4 4

√2p3 + 3p2

+(−3)√

2p+ 1 ·√

5p− (−3)√

2p+ 1 · 4 4√

2p3 + 3p2

We can write the colored parts under one radical, but other than that we aredone.

We will now learn how to rationalize. This is exactly as it sounds. We want to rationalize.What is rational in math? We need to make a section capable of being written in the formpq such that p and q are integers and q 6= 0.

This is because radicals of number are irrational unless they are of the form n√xn. Our

goal is to remove the radical to make either the numerator or denominator rational.

Instructions: 8.5.1:Rationalizing:

1. What are we rationalizing?

2. Simplify what you want to rationalize, then determine what to multiply to theitem we want to rationalize so that the end result has no radical. Use exponentrules.

• To rationalize u 1n , you may also multiply by un−1

n . This however will not besimplified, and you must simplify after.

3. Multiply this to the top and bottom.

Example: 8.5.2:

Rationalize the part of the fraction asked.

1. Rationalize the denominator of√

3−9√24 .

What do I need to multiply to√

24 so the square root disappears? We couldmultiply by

√24, since

√24 ·√

24 = 24, however we would need to simplifymore than necessary at the end if we did it this way. We should simplify firstas√

24 =√

4 · 6 = 2√

6. Now to get rid of the square root we only need to getrid of

√6 by multiplying it by

√6. Simplifying at all possible times will make

future calculations simpler.



√3− 9√

24=√

3− 92√

6

=√

3− 92√

6·√

6√6

√6 ·√

6 = 6

= (√

3− 9)√

62√

62

=√

3 ·√

6− 9√

62(6)

= 3√

2− 9√

612

=√

2− 3√

64

2. Rationalize the denominator of 2√

20x5−√

12x2√18x .

Similar to example one, we can simplify√

18x =√

9 · 2x = 3√

2x. So we onlyneed to multiply the top and bottom by

√2x.

2√

20x5 −√

12x2√

18x= 2√

5 · 4x4x−√

3 · 4x2√

2 · 9x

= 4x2√5x− 2x√

33√

2x

= 4x2√5x− 2x√

33√

2x·√

2x√2x

√2x ·√

2x = 2x

= (4x2√5x− 2x√

3) ·√

2x3√

2x ·√

2x

= 4x2√5x ·√

2x− 2x√

3 ·√

2x3(2x)

= 4x2√

5x(2x)− 2x√

3(2x)6x

= 4x2√

10x2 − 2x√

6x6x

= 4x3√10− 2x√

6x6x

= 2x(2x2√10−√

6x)2x(3)



= 2x2√10−√

6x3

3. Rationalize the numerator of3√54n2m7

3√m+2+ 3√16n2 .

If you have been getting in the habit of thinking that when we rationalize we justmultiply by the same thing to get rid of the square root, then you are sorelymistaken. First lets simplify the numerator as 3

√54n2m7 = 3

√27 · 2n2m7 =

3m2 3√

2n2m. We want to rationalize 3√

2n2m. If we multiply by itself we get3√2n2m · 3√2n2m = 3√2n2m · 2n2m

= 3√4n4m2

= n3√4nm2

Which cannot be simplified anymore. There are two ways to do this. We cando it algebraically as

3√2n2m(something) = 2n2m

something = 2n2m3√

2n2m

something = 2n2m

2 13n

23m

13

something = 21− 13n2− 2

3m1− 13

something = 2 23n

43m

23

something = 3√22n4m2

Notice that this give us the same answer as if we used un−1

n . In this case

(2n2m)3−1

3 = (2n2m) 23 = 3√22n4m2.

This is not the simplest thing to multiply. It’s like finding the LCD. The LCDis the easiest thing to multiply to the numerator and denominator so that wecan add the two fractions. However, it is not requires. All we need is a commondenominator. This is similar. To get the simplest, lets consider how the 3

√works. We need to have a multiplication of an item 3 times in order to pull itout of the radical. So how many 2’s, n’s, and m’s are we missing? We need 2more twos, one n, and two m’s. So we need 3

√22nm2. Notice that

3√2n2m · 3√22nm2 = 3√

2(22)n2nmm2

= 3√23n3m3

= 2nm



We were able to rationalize. Now to solve the problem we need to multiply tothe top and the bottom.

3√

54n2m7

3√m+ 2 + 3

√16n2

=3√

54n2m7

3√m+ 2 + 3

√16n2

·3√

22nm2

3√

22nm2

= 3m2 3√

2n2m · 3√

22nm2

( 3√m+ 2 + 3

√16n2) 3

√22nm2

= 3m2(2nm)3√m+ 2 · 3

√22nm2 + 3

√16n2 · 3

√22nm2

= 6m3n3√

4nm2(m+ 2) + 3√

22nm2(16n2)

= 6m3n3√

4nm2(m+ 2) + 3√

22(24)n3m2

= 6m3n3√

4nm2(m+ 2) + 22n3√m2

There is no reason to try and simplify 4nm2(m+2). Since there is an addition,we will not be able to pull anything out.

There are more cases than this that may need to be rationalized. What if we have aproblem that requires a rationalization of a sum of square roots.

√n+√m

We already have a technique that can solve these problems. What can we multiply to thisso that the two terms end up being squared? The difference of square that we learned insection 5.5.1.

A2 −B2 = (A+B)(A−B)

Definition 8.5.1: The conjugate of A+B is A−B and the conjugate of A−B is A+B. Theconjugate of a two termed expression is the expression where the middle sign is changed.

Notice that if we multiply the problem by the conjugate we will be left with the firstterm squared minus the second term squared. So we are able to remove the square roots.This holds whether or not there is a plus or a minus in the middle.

Example: 8.5.3:

Rationalize the given part of the fraction.

1. Rationalize the denominator of 15√6+1 .

15√6 + 1

= 15√6 + 1

·√

6− 1√6− 1

= 15(√

6− 1)(√

6 + 1)(√

6− 1)

= 15(√

6− 1)6− 12 Diff of squares



= 15(√

6− 1)5

= 3(√

6− 1)

2. Rationalize the numerator of√

11−√

5√11+√

5 .

√11−

√5√

11 +√

5=√

11−√

5√11 +

√5·√

11 +√

5√11 +

√5

Rat the num

= (√

11−√

5)(√

11 +√

5)(√

11 +√

5)(√

11 +√

5)

=√

112 −√

52

(√

11 +√

5)2Diff of squares

= 11− 5(√

11 +√

5)2

3. Rationalize the numerator of√x+5−

√x

5 .

√x+ 5−

√x

5 =√x+ 5−

√x

5 ·√x+ 5 +

√x√

x+ 5 +√x

= (√x+ 5−

√x)(√x+ 5 +

√x)

5(√x+ 5 +

√x)

= x+ 5− x5(√x+ 5 +

√x)

Diff of squares

= 1√x+ 5 +

√x

4. Rationalize the denominator of 3√x+√y

√y−3√x

.

3√x+√y

√y − 3

√x

=3√x+√y

√y − 3

√x·√y + 3

√x

√y + 3

√x

=(3√x+√y)(√y + 3

√x)

(√y − 3√x)(√y + 3

√x)

=(3√x+√y)(√y + 3

√x)

y − 32xDiff of squares



=(3√x+√y)(√y + 3

√x)

y − 9x

5. Rationalize the denominator of 2x+1√f(x)+

√g(x)

.

2x+ 1√f(x) +

√g(x)

= 2x+ 1√f(x) +

√g(x)

·√f(x)−

√g(x)√

f(x)−√g(x)

= (2x+ 1)(√f(x)−

√g(x))

(√f(x) +

√g(x))(

√f(x)−

√g(x))

= (2x+ 1)(√f(x)−

√g(x))

f(x) + g(x)

Rationalizing is something that will be used in future classes and you should learn howto do this efficiently.


Math 103 Radical Equations

8.6 Radical Equations

Goals:1. Solve radical equa-

tions.

2. Use models that areradical functions tosolve problems.

Radical equations are as you suspect, equations with radicals in them. So far all the problemsthat we have been working on have only been using polynomials. But many situations arisewhere radicals are necessary. There will be two situations that we will discuss.

Instructions: 8.6.1:Solving radical equations:

Determine if the equation has one or two radicals.

1. If there is one radical, then

(a) Isolate it and raise both sides to the nth power.(b) Solve.(c) When raising both sides by an even power we must check that the solution

works.

2. If there are two radicals, then

(a) Isolate one radical, then raise each side to the nth power, or leave them onone side and raise both sides to the nth power.

(b) You will more than likely have another radical to deal with. Isolate thisradical and raise each side to the nth power.

(c) Solve.(d) When raising both sides by an even power we must check that the solution

works.

Example: 8.6.1:

Solve to given equations.

1.√x− 7 = 3

√x− 7 = 3

(√x− 7)2 = 32

x− 7 = 9

x = 16

Now plug back in to check that 16 is a valid solution.√

16− 7 = 3√

9 = 3

3 = 3

So when we plug 16 into√x− 7 we get 3.

2.√

2m− 2 = m− 1



√2m− 2 = m− 1

(√

2m− 2)2 = (m− 1)2

2m− 2 = m2 − 2m+ 1

0 = m2 − 2m+ 1− (2m− 2)

m2 − 4m+ 3 = 0

(m− 3)(m− 1) = 0

So m = 3 or 1. Check your solution.

√2(3)− 2 = (3)− 1

√4 = 2

2 = 2

√2(1)− 2 = 1− 1

√0 = 0

0 = 0

So when we use 3 and 0 as an input, the statement√

2m− 2 = m− 1 is true.

3.√

5p+ 11− (1 + p) = 0

If you try to square both sides right away you will get

[√

5p+ 11− (1 + p)]2 = 02

(√

5p+ 11)2 − 2√

5p+ 11(1− p) + (1 + p)2 = 0

5p+ 11− 2√

5p+ 11(1− p) + (1 + p)2 = 0

Notice that we still have a square root in the middle. This did not help usprogress in our goal of eliminating the radical. This is why we need to have theradical by itself first.√

5p+ 11− (1 + p) = 0√5p+ 11 = 1 + p

(√

5p+ 11)2 = (1 + p)2

5p+ 11 = p2 + 2p+ 1

0 = p2 + 2p+ 1− (5p+ 11)

p2 − 3p− 10 = 0

(p− 5)(p+ 2) = 0

Check is p = 5 and p = −2 are valid solutions.



√5(5) + 11− (1 + 5) = 0

√36− (6) = 0

6− 6 = 0

√5(−2) + 11− (1 + (−2)) = 0

√−10 + 11− (−1) = 0

√1 + 1 = 02 6= 0

Since −2 give us a false statement, the only solution is 5. When we plug in 5into

√5p+ 11− (1 + p) we get 0.

4. 7√x2 + 4x+ 2 = −1

7√x2 + 4x+ 2 = −1

( 7√x2 + 4x+ 2)7 = (−1)7

x2 + 4x+ 2 = −1

x2 + 4x+ 3 = 0

(x+ 3)(x+ 1) = 0

So when we plug −3 and −1 into 7√x2 + 4x+ 2 we get -1.

Let us now consider the types of problems with more than one radical.Example: 8.6.2:

Solve to following equations.

1.√p− 4−√p = −2

We will do it in two ways to show that you can do either in the instructions.First, leave both the radicals on one side.

√p− 4−√p = −2

(√p− 4−√p)2 = (−2)2

(√p− 4)2 − 2

√p− 4 · √p+ (√p)2 = 4

p− 4− 2√p2 − 4p+ p = 4

−2√p2 − 4p = −2p+ 8√p2 − 4p = p− 4

(√p2 − 4p)2 = (p− 4)2

p2 − 4p = p2 − 8p+ 16

4p = 16

p = 4



Now lets have one radical on each side. We will see that we get the sameanswer. √

p− 4−√p = −2√p− 4 = √p− 2

(√p− 4)2 = (√p− 2)2

p− 4 = (√p)2 − 4√p+ (−2)2

p− 4 = p− 4√p+ 4

−8 = −4√p

2 = √p

p = 4

Now that we can see that both of these methods works, lets check the solutionp = 4.

√4− 4−

√4 = −2

√0− 2 = −2

−2 = −2

So when we plug in 4 into√p− 4−√p we get −2.

2.√m− 2 +

√m− 1 = 1

√m− 2 +

√m− 1 = 1√m− 2 = 1−

√m− 1

(√m− 2)2 = (1−

√m− 1)2

m− 2 = (1)2 − 2(1)(√m− 1) + (

√m− 1)2

m− 2 = 1− 2√m− 1 +m− 1

−2 = −2√m− 1

1 =√m− 1

12 = (√m− 1)2

1 = m− 1

m = 2

So m = 2. Check your solution.



√2− 2 +

√2− 1 = 1

√0 +√

1 = 1

1 = 1

So when we use 2 as an input, the statement√m− 2 +

√m− 1 = 1 is true.

3.√

2q + 5−√q − 2 =

√4q + 1

Notice in this problem we have no choice but to square both sides.√2q + 5−

√q − 2 =

√4q + 1

(√

2q + 5−√q − 2)2 = (

√4q + 1)2

(√

2q + 5)2 − 2√

2q + 5 ·√q − 2 + (

√q − 2)2 = 4q + 1

2q + 5− 2√

(2q + 5)(q − 2) + q − 2 = 4q + 1

3q + 3− 2√

(2q + 5)(q − 2) = 4q + 1

−2√

(2q + 5)(q − 2) = q − 2

(−2√

(2q + 5)(q − 2))2 = (q − 2)2

4(2q + 5)(q − 2) = (q − 2)2

4(2q2 − 4q + 5q − 10) = q2 − 4q + 4

8q2 + 4q − 40 = q2 − 4q + 4

7q2 + 8q − 44 = 0

(7q + 22)(q − 2) = 0

If you cannot factor this, then use the quadratic equation. Check your solutionq = − 22

7 or 2.With solving can we see that − 22

7 is not a valid solution? This is because√q − 2 will have an imaginary part. The interior will be negative. Lets check

q = 2.

√2(2) + 5−

√2− 2 =

√4(2) + 1

√9−√

0 =√

9√

9 =√

9

So 2 is the only input that will make√

2q + 5−√q − 2 =

√4q + 1 true.



4.√

2x+ 3 ·√x− 2 = 3

First notice that we may combine the radicals.√

2x+ 3 ·√x− 2 = 3√

(2x+ 3)(x− 2) = 3

(√

(2x+ 3)(x− 2))2 = 32

(2x+ 3)(x− 2) = 9

2x2 − 4x+ 3x− 6 = 9

2x2 − x− 6 = 9

2x2 − x− 15 = 0

(2x+ 5)(x− 3) = 0

So x = − 53 or 3. Check your solution.

Again without doing the math, can wesee that

√x− 2 will have a negative

interior when we plug in − 53 . This

solution will be complex. Now checkx = 3.

√2(3) + 3 ·

√3− 2 = 3

√9 ·√

1 = 3

3 · 1 = 3

So when we use 3 as an input, to make the statement√

2x+ 3 ·√x− 2 = 3

true.


Math 103 Complex Numbers

8.7 Complex Numbers

Goals:1. Express square roots

of negative numbersin terms of i.

2. Add and subtractcomplex numbers.

3. Multiply complexnumbers.

4. Divide complexnumbers.

5. Simplify powers of i.

The complex numbers are the set of elements that are of the form

a+ bi, and i =√−1

where a and b are real numbers. This can be written formally asDefinition 8.7.1: The complex numbers is the set

C = {a+ bi | a, b ∈ R}.

The set of complex numbers is isomorphic to R2 = R× R = {(a, b) | a, b ∈ R}.Definition 8.7.2: The real part of a complex number is a. The number that is not attachedto i.Definition 8.7.3: The complex part of a complex number is b. The number that is attachedto i.

We will see that the algebra that we used for the real numbers is the same for complexnumbers. First lets consider how i works when we have more than one of them.

Addition/subtraction with i works in the same manner as adding/subtracting with x.

1) 3i+ 4i = 7i 2) 5i− 7i = −2i

Multiplication requires understanding that i was originally√−1. Using this we can

evaluate products of i.

1) 3i · 2i = 6i2 = 6(√−1)2 = −6 2) 4i · 2i · i = 8i3 = 8i2i = 8(

√−1)2i = −8i

We could continue this patter to see that we would have

in 3 Degree

i i 1

i2 (√−1)2 = −1 2

i3 (√−1)2i = −i 3

i4 (√−1)2(

√−1)2 = 1 4

i5 (√−1)2(

√−1)2i 5

i6 (√−1)2(

√−1)2(

√−1)2(

√−1)2 = −1 6

Notice that we are using the basic idea that (√−1)2 = −1 to evaluate all the powers of

i.

Instructions: 8.7.1:Evaluate powers of i.

1. Even degree.

(a) If the exponent is even, then write it in the form (i2)n. For example i4 =(i2)2, exponents raised to an exponent multiply together.

(b) The value of n will tell you how many times we multiply −1 to itself. Fromthis we know that if n is even, then the end result will be 1, and if n is odd,then the end result will be −1.

2. Odd degree



(a) If the exponent is odd, then write it in the form (i2)ni. For example i5 =i2 · i2i = (i2)2i, multiplication of same bases raised to exponents, add theexponents.

(b) Similarly with even degree, we will use n to determine if we have a negativei or a positive one.

Example: 8.7.1:

1) i36

This is an even degree so it can bewritten as (i2)18 = (−1)18. Sincen = 18 is even, we will have 1 as theend result. So

i36 = 1

2) i59

Since this is an odd degree, we canwrite it as i59 = (i2)29i = (−1)29i.This has a n value of 29, so we willend up with a −1. Hence

i59 = −i

You may be wondering why i is a complex number if all complex numbers are of theform a + bi. Well, it’s because it is of that form. We may write i as 0 + (1)i. Let’s nowconsider problems that have more than just i in them. There are a couple of ways to dobasic algebra using complex numbers. We may use the basic algebra properties used whenusing real numbers or the second definition of complex number. First lets add and subtracttwo complex numbers.

(3 + 5i) + (6− 7i)

We can write this problem without parenthesis and use the commutative property tomove the real parts together and the complex parts together.

(3 + 5i) + (6− 7i) = 3 + 6 + 5i− 7i = 9− 2i.

We may also write the complex numbers 3 + 5i and 6−7i as (3, 5) and (6,−7). Now addthese values straight down.

(3,5)

+ (6,-7)

(9,-2)

Notice that when we convert back we get 9 − 2i which is the same as what we hadbefore.You can also do this using subtraction. Lets consider (3 + 5i)− (6− 7i) instead.

(3,5)

- (6,-7)

(-3,12)

Example: 8.7.2:


1. (3 + 5i) + (6− 2i)



(3 + 5i) + (6− 2i) = 3 + 6 + 5i− 2i

= 9 + 3i

2. (24− 6i) + (−24 + 5i)

(24− 6i) + (−24 + 5i) = 24− 24− 6i+ 5i

= 0− i

= −i

3. 12− (13 + 15i)

12− (13 + 15i) = 12− 13− 15i

= −1− 15i

4. (−12− 3i)− (−2− 8i)

(−12− 3i)− (−2− 8i) = −12− 3i+ 2 + 8i

= −12 + 2− 3i+ 8i

= −10 + 5i

5. (2 + 6πi) + (4− 4πi)

(2 + 6πi) + (4− 4πi) = 2 + 4 + 6πi− 4πi

= 6 + 2πi

6. (3√

5− 2i√

3)− (5√

5− 7i√

3)



(3√

5− 2i√

3)− (5√

5− 7i√

3) = 3√

5− 5√

5− 2i√

3 + 7i√

3

= −2√

5 + 5i√

3

Of course, the algebraic way is more useful because it requires you to have knowledge ofalgebra. However the second way is useful when understanding graphs of complex numbers.We will not discuss this now, but understanding things in multiple ways will help you learn.

Multiplication however is much easier using the standard algebra used for real numbers.There is however one case that you need to be careful of. You cannot write

√−4 ·

√−4 as√

(−4)(−4) =√

42 = 4. We must, when using complex numbers, always convert into itsstandard notation before applying algebraic properties. Hence

√−4 ·√−4 = i

√4 · i√

4 = i24 = −4.

When finding products of complex numbers we must use the distributive property.Example: 8.7.3:

Perform the given operation.

1. 3i(−2 + 6i)

3i(−2 + 6i) = 3i(−2) + (3i)(6i)

= −6i+ 18i2

= −6i− 18

Written is standard form we have −18− 6i.

2. (−3 + 5i)(−5 + 7i)

(−3 + 5i)(−5 + 7i) = (−3 + 5i)(−5 + 7i)11

22= (−3)(−5) + (−3)(7i) + (5i)(−5) + (5i)(7i)

= 15− 21i− 25i+ 35i2

= 15− 35− 46i

= −20− 46i

3. (i)(6i)(4 + 6i)

(i)(6i)(4 + 6i) = (6i2)(4 + 6i)

= −6(4 + 6i)

= −24− 36i



4. (1 + 4i)(3 + 6i)

(1 + 4i)(3 + 6i) = (1 + 4i)(3 + 6i)11

22= (1)(3) + 1(6i) + (4i)(3) + (4i)(6i)

= 3 + 6i+ 12i+ 24i2

= 3− 24 + 18i

= −21 + 18i

5. (7 + 3i)(−7 + 5i)(−8− 4i)

(7 + 3i)(−7 + 5i)(−8− 4i) = (7 + 3i)(−7 + 5i)(−8− 4i)11

22= [(7)(−7) + (7)(5i) + (3i)(−7) + (3i)(5i)](−8− 4i)

= [−49 + 35i− 21i+ 15i2](−8− 4i)

= [−49− 15 + 14i](−8− 4i)

= (−64 + 14i)(−8− 4i)11

22= (−64)(−8) + (−64)(−4i) + (14i)(−8) + (14i)(−4i)

= 512 + 256i− 112i− 56i2

= 512 + 56 + 144i

= 568 + 144i

Finally the last operation, division. In order to solve a division of complex numberswe need to multiply the numerator and the denominator by the complex conjugate. Theconjugate is the same as when we rationalized. However, it is called the complex conjugatebecause it is a complex number. For example; the complex conjugate of 5 + 3i is 5 − 3i.Nothing changes except for the fact that the number is now complex. To make it easier,don’t forget that when multiplying by the conjugate we can use the difference of squaresproperty to get (A−B)(A+B) = A2 −B2.

Example: 8.7.4:

Perform the indicated operation.

1. 2i3−9i

2i3− 9i = 2i

3− 9i ·3 + 9i3 + 9i



= 2i(3 + 9i)(3− 9i)(3 + 9i)

= 6i+ 18i232 − (9i)2

= −18 + 6i9− 81i2

= −18 + 6i9 + 81

= −18 + 6i90

= −1890 + 6

90 i

= −15 + 1

15 i

2. 5i6+8i

5i6 + 8i = 5i

6 + 8i ·6− 8i6− 8i

= 5i(6− 8i)(6 + 8i)(6− 8i)

= 30i− 40i262 − (8i)2

= 40 + 30i36 + 64

= 40100 + 30

100 i

= 25 + 3

10 i

3. −1+5i−8−7i

−1 + 5i−8− 7i = −1 + 5i

−8− 7i ·−8 + 7i−8 + 7i

= (−1 + 5i)(−8 + 7i)(−8− 7i)(−8 + 7i)

11

22

= (−1)(−8) + (−1)(7i) + (5i)(−8) + (5i)(7i)(−8− 7i)(−8 + 7i)



= 8− 7i− 40i+ 35i2(−8)2 − (7i)2

= 8− 35− 47i64 + 49

= −27− 47i113

= − 27113 −

47113 i

4. 4+i2−5i

4 + i

2− 5i = 4 + i

2− 5i ·2 + 5i2 + 5i

= (4 + i)(2 + 5i)(2− 5i)(2 + 5i)

11

22

= (4)(2) + 4(5i) + (i)(2) + (i)(5i)(2− 5i)(2 + 5i)

= 8 + 20i+ 2i+ 5i2(2)2 − (5i)2

= 3 + 22i4 + 25

= 329 + 22

29 i

5. −3−9i5−8i

−3− 9i5− 8i = −3− 9i

5− 8i ·5 + 8i5 + 8i

= (−3− 9i)(5 + 8i)(5− 8i)(5 + 8i)

11

22

= (−3)(5) + (−3)(8i) + (−9i)(5) + (−9i)(8i)(5− 8i)(5 + 8i)

= −15− 24i− 45i− 72i2(5)2 − (8i)2

= 57− 69i25 + 64



= 5789 −

6989 i

6. −3−2i−10−3i

−3− 2i−10− 3i = −3− 2i

−10− 3i ·−10 + 3i−10 + 3i

= (−3− 2i)(−10 + 3i)(−10− 3i)(−10 + 3i)

11

22

= (−3)(−10) + (−3)(3i) + (−2i)(−10) + (−2i)3i)(−10− 3i)(−10 + 3i)

= 30− 9i+ 20i− 6i2(−10)2 − (3i)2

= 36 + 11i100 + 9

= 36109 + 11

109 i

This is probably the most difficult operation that we have done. It requires conjugates(which you may need later), multiplication of complex numbers, distributive property, dif-ference of squares, and fractions.

Name: Math 103

8.7.1 Homework 8.7 Complex Numbers

Problem 1. Simplify the following expressions.

1. i+ 6i 2. 3 + 4 + 6i 3. 3i+ i

4. −8i− 7i 5. −1− 8i− 4− i 6. 7 + i+ 4 + 4

7. −3 + 6i− (−5− 3i)− 8i 8. 3 + 3i+ 8− 2i− 7 9. 4i(−2− 8i)

Name: Math 103

10. 5i · i 11. 5i · i · (−2i) 12. −4i · 5i

13. (−2− i)(4 + i) 14. (7− 6i)(−8 + 3i) 15. 7i · 3i(−8− 6i)

16. (4− 5i)(4 + i) 17. (2− 4i)(−6 + 4i) 18. (−3 + 2i)(−6− 8i)

Name: Math 103

19. (8− 6i)(−4− 4i) 20. (1− 7i)2 21. 6(−7 + 6i)(−4 + 2i)

22. (−2− 2i)(−4− 3i)(7 + 8i) 23. 5i+ 7i · i 24. (6i)3

25. 6i · (−4i) + 8 26. −6(4− 6i) 27. (8− 3i)2

Name: Math 103

28. 3 + 7i− 3i− 4 29. −3i · 6i− 3(−7 + 6i) 30. −6i(8− 6i)(−8− 8i)

Problem 2. Critical thinking questions: What steps are different when expanding the following expressions.

Expand: (2 + x)(3− 2x)

Expand: (2 + i)(3− 2i)

Name: Math 103

Congratulations,You are DONE!


Math 103

A Systems of Linear Equations

1) 0 = 4 + x− 2y7x = 4y + 12x = 2y − 4 using the first7(2y − 4) = 4y + 1214y − 28 = 4y + 1210y = 40y = 40 = 4 + x− 2(4)x = 4(4, 4)x− 2y = −47x− 4y = 12−2x+ 4y = 8 multiply the top blue by-25x = 20x = 4

2) 0 = −84 + 24x+ 21y27 = −3y + 15

7 xy = 5

7x− 9 using the second0 = −84 + 24x+ 21( 5

7x− 9)0 = −84 + 24x+ 15x− 1890 = 39x− 27339x = 273x = 727 = −3y + 15

7 (7)27 = −3y + 15−3y = 12y = −4(7,−4)0 = −27− 3y + 15

7 x0 = −84 + 21y + 24x0 = −189− 21y + 15x multiply the topblue by 7.0 = −273 + 39x39x = 273x = 7

3) 0 = −x− 12 + 3y−24− 6y = 14xx = −12 + 3y using the first−24− 6y = 14(−12 + 3y)−24− 6y = −168 + 42y48y = 144y = 30 = −x− 12 + 3(3)x = −12 + 9x = −3(−3, 3)0 = −x− 12 + 3y0 = −14x− 24− 6y0 = −2x− 24 + 6y multiplying the topblue by 2. 0 = −16x− 4816x = −48x = −3

4) 0 = x+ 83 + 2

3y−y = −3− 2xy = 2x+ 3 using the second0 = x+ 8

3 + 23 (2x+ 3)

0 = x+ 83 + 4

3x+ 273x = − 14

3x = −2−y = −3− 2(−2)−y = −3 + 4y = −1(−2,−1)0 = x+ 8

3 + 23y

0 = −2x− 3 + y0 = 2x+ 16

3 + 43y multiply the top blue

by 2.0 = 7

3 + 73y7

3y = − 73

y = −1


Math 103

5) −9 + 7x = −3y0 = 3 + x+ 3

2yx = −3− 3

2y using the second−9 + 7(−3− 3

2y) = −3y−9− 21− 21

2 y = −3y−30 = 15

2 yy = −4−9 + 7x = −3(−4)−9 + 7x = 127x = 21x = 3(3,−4)0 = 3 + x+ 3

2y0 = −9 + 7x+ 3y0 = −6− 2x− 3y multiplying the topblue by -2.0 = −15 + 5x5x = 15x = 3

6) 3y = −2x− 95x = 3y − 12y = − 2

3x− 3 using the first5x = 3(− 2

3x− 3)− 125x = −2x− 9− 127x = −21x = −33y = −2(−3)− 93y = 6− 9y = −1(−3,−1)0 = −3y − 2x− 90 = 3y − 5x− 120 = −7x− 217x = −21x = −3

7) 0 = x− 3−x = −12− 3yx = 3 using the first−3 = −12− 3y9 = −3yy = −3(3,−3)0 = x− 30 = x− 12− 3y0 = −x+ 3 multiplying the top blue by-1.0 = −9− 3y3y = −9y = −3

8) y − 12x = −3

5x− 6 = −2yy = −3 + 1

2x using the first5x− 6 = −2(−3 + 1

2x)5x− 6 = 6− x6x = 12x = 2y − 1

2 (2) = −3y − 1 = −3y = −2(2,−2)0 = y − 1

2x+ 30 = −2y − 5x+ 60 = 2y − x+ 6 multiplying the top blueby 2.0 = −6x+ 126x = 12x = 2


Math 103

9) 7x+ 4 = −y−1 + 1

2y + 12x = 0

y = −7x− 4 using the first−1 + 1

2 (−7x− 4) + 12x = 0

−1− 72x− 2 + 1

2x = 0−3x− 3 = 0−3x = 3x = −17(−1) + 4 = −y−7 + 4 = −yy = 3(−1, 3)0 = −1 + 1

2y + 12x

0 = 4 + y + 7x0 = 2− y − x multiplying the top blueby -2.0 = 6 + 6x6x = −6x = −1

10) −2y − 2 = −xx = −4y + 8x = 2y + 2 using the first2y + 2 = −4y + 86y = 6y = 1−2(1)− 2 = −x−2− 2 = −xx = 4(4, 1)0 = −x+ 2y + 20 = x+ 4y − 80 = 6y − 66y = 6y = 1

11) −2x− y = −12x = 3 + yy = −2x+ 1 using the first2x = 3 + (−2x+ 1)2x = 4− 2x4x = 4x = 12(1) = 3 + yy = −1(1,−1)0 = −2x− y + 10 = 2x− y − 30 = −2y − 22y = −2y = −1

12) −4− y = 00 = 3y − 6x− 12y = −4 using the first0 = 3(−4)− 6x− 120 = −12− 6x− 126x = −24x = −4(−4,−4)0 = −y − 40 = 3y − 12− 6x0 = −3y − 12 multiplying the top blueby 3.0 = −24− 6x6x = −24x = −4


Math 103

13) −3 + y = −xy = −3− 7x−3− 3− 7x = −x6x = −6x = −1y = −3− 7(−1)y = 4(−1, 4)0 = −y − x+ 30 = y + 7x+ 30 = 6x+ 66x = −6x = −1

14) x = 9− 3y0 = −x+ y + 10 = −(9− 3y) + y + 10 = −9 + 3y + y + 14y = 8y = 2x = 9− 3(2)x = 9− 6x = 3(3, 2)0 = x+ 3y − 90 = −x+ y + 10 = 4y − 84y = 8y = 2

15) −9y = −15x− 93x− 9y = 27y = 5

3x+ 1 using the first3x− 9( 5

3x+ 1) = 273x− 15x− 9 = 27−12x = 36x = −33(−3)− 9y = 27−9− 9y = 27−9y = 36y = −4(−3,−4)0 = 9y − 15x− 90 = −9y + 3x− 270 = −12x− 3612x = −36x = −3

16) 0 = 9− 3y − 4x4 + y = x0 = 9− 3y − 4(4 + y)0 = 9− 3y − 16− 4y0 = −7− 7y7y = −7y = −14− 1 = xx = 3(3,−1)0 = 4 + y − x0 = 9− 3y − 4x0 = 12 + 3y − 3x multiplying the topblue by 3.0 = 21− 7x7x = 21x = 3


Math 103

17) 0 = 18 + 9x− 6y−2− 2y = −xx = 2 + 2y using the second0 = 18 + 9(2 + 2y)− 6y0 = 18 + 18 + 18y − 6y0 = 36 + 12y12y = −36y = −3−2− 2(−3) = −x−2 + 6 = −xx = −4(−4, 3)0 = 2− x+ 2y0 = 18 + 9x− 6y0 = 6− 2x+ 6y multiplying the topblue by 3.0 = 24 + 7x7x = −24x = −4

18) 3y + 12 = 7x3y − 12 + x = 0x = −3y + 12 using the second3y + 12 = 7(−3y + 12)3y + 12 = −21y + 8424y = 72y = 33(3)− 12 + x = 09− 12 + x = 0x = 3(3, 3)0 = 3y + 12− 7x0 = 3y − 12 + x0 = −3y − 12 + 7x multiplying the topblue by -1.0 = −24 + 8x8x = 24x = 3

19) 2 = y + x−8y = 8 + 2xx = 2− y using the first−8y = 8 + 2(2− y)−8y = 8 + 4− 2y−6y = 12y = −22 = −2 + xx = 4(4,−2)0 = y + x− 20 = 8y + 2x+ 80 = −2y − 2x+ 4 multiplying the topblue by -2.0 = 6y + 126y = −12y = −2

20) 0 = −5x+ 2y + 40 = x+ 2y − 8x = −2y + 8 using the second0 = −5(−2y + 8) + 2y + 40 = 10y − 40 + 2y + 40 = 12y − 3612y = 36y = 30 = x+ 2(3)− 8x = 2(2, 3)0 = −5x+ 2y + 40 = x+ 2y − 80 = 5x− 2y − 4 multiplying the topblue by -1.0 = 6x− 126x = 12x = 2


Math 103

21) −1− 14y + 3

8x = 02y − 3x = −4y = 3

2x− 2 using the second−1− 1

4 ( 32x− 2) + 3

8x = 0−1− 3

8x+ 12 + 3

8x = 0−1 + 1

2 = 0− 1

2 = 0 which is false, so no solution.0 = −1− 1

4y + 38x

0 = 4 + 2y − 3x0 = −8− 2y + 3x multiplying the topblue by 8.0 = −4 again is false, so no solution.

22) 6 + y = 2x0 = −y − 2y = −2 using the second6− 2 = 2x2x = 4x = 2(2,−2)0 = y + 6− 2x0 = −y − 20 = 4− 2xx = 2

23) −4y = 12 + x−4y + 20 = 5xx = −4y − 12 using the first−4y + 20 = 5(−4y − 12)−4y + 20 = −20y − 6016y = −80y = −5−4(−5) + 20 = 5x20 + 20 = 5xx = 8(8,−5)0 = −4y − 12− x0 = −4y + 20− 5x0 = 4y + 12 + x multiplying the topblue by -1.0 = 32− 4xx = 8

24) 5y − 14x = 300 = −35 + x− 5yx = 5y + 35 using the second5y − 14(5y + 35) = 305y − 70y − 490 = 30−65y = 520y = −80 = −35 + x− 5(−8)0 = −35 + x+ 40x = −5(−5,−8)0 = −30− 14x+ 5y0 = −35 + x− 5y0 = −65− 13xx = 5


Math 103

Blue is always the first equation and red is always the second equation.

1)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(4,4)

x

y

2)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(7,-4)

x

y

3)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(-3,3)

x

y

4)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(-2,-1)

x

y


Math 103

5)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(3,-4)

x

y

6)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(-3,-1)

x

y

7)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(3,-3)

x

y

8)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(2,-2)

x

y


Math 103

9)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(-1,3)

x

y

10)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(4,1) x

y

11)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(1,-1)

x

y

12)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(-4,-4)

x

y


Math 103

13)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(-1,4)

x

y

14)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(3,2)x

y

15)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(-3,-4)

x

y

16)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(3,-1)

x

y


Math 103

17)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(-4,-3)

x

y

18)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(3,3)

x

y

19)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(4,-2)

x

y

20)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(2,3)

x

y

Some More No-tation

Math 103

21)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

x

y

22)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

12345678

(2,-2)

x

y

23)

−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8

−8−7−6−5−4−3−2−1

12345678

(8,-5)

x

y

24)

−7−6−5−4−3−2−1 1 2 3 4 5 6 7

−8−7−6−5−4−3−2−1

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(-5,-8)

x

y

B Some More Notation

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