Chapter 1A Real Numbers - Department of - math.tamu.eduscarboro/150TextChapter1.pdf · Exercises...

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Chapter 1A - Real Numbers

Properties of Real NumbersReal numbers are used in almost every human endeavor. Whenever we need to quantify objects, we usenumbers. Cooking recipes, prices, interest rates, blood pressure, height, age, voltage, and wind velocityare a few of the everyday objects that are quantified by real numbers. As we know, the two operationsof addition and multiplication are defined for real numbers. In other words, for any two realnumbers a and b, the sum a b and the product a b are uniquely defined real numbers. Two specialreal numbers are zero (0 ) and one (1 ). These operations satisfy:Properties of real numbers:

Commutative: a b b a ab baExample: 7 3 3 7 10 5 6 6 5 30

Associative: a b c a b c abc abcExample: 3 4 7 3 4 7 2 5 3 2 5 3

3 11 7 7 14 2 15 10 3 30

Identity: a 0 0 a a a 1 1 a aExample: 8 0 0 8 8 11 1 1 11 11

For each real number a, there is a real number, denoted by −a, called the negative of a, for whichInverse: a −a 0 −a aExample: 7 −7 0

Subtraction, denoted by a − b, is defined as follows: a − b a −b

For each a ≠ 0, there is a real number, denoted by 1a or 1/a or a−1, called the reciprocal of a, for which

Inverse: a 1a 1 1

a a

Example: 7 17 1

7 7 1

Division, denoted by a b or ab or a/b, where b ≠ 0, is defined as follows:

a b ab a/b a 1

b ab−1

Finally, there is a property which relates addition and multiplication:

Distributive: ab c ab ac a bc ac bcExample: −4 3 2 −4 3 −4 2

−4 5 −12 −8 −20

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Types of Real Numbers

IntegersThe real numbers are classified into several categories. The positive integers, also called countingnumbers, are

1,2,3,4,…The negative integers,

… ,−4,−3,−2,−1are the negatives (or additive inverses) of the positive integers. An integer is either a positive integer, anegative integer, or zero.

Rationals.The ratio a

b of any two integers a and b, where b ≠ 0, is called a rational number. Common fractionsare rationals. If b 1 then the number is also an integer, so integers are also rational numbers.Examples:

45 , 121

54 ,− 1631 ,−7

Irrationals.Real numbers that are not rational are called irrational. Examples of irrational numbers include , 2 ,and so on. One distinction between rational numbers and irrational numbers is that rational numbershave repeating or terminating decimal expansions, whereas irrational numbers do not.

SummaryThe diagram below shows the relationship among these types of numbers, all of which are a part of theset of real numbers: That is, whole numbers are a subset of the integers, which are a subset of therational numbers. The irrational numbers are disjoint from all of these and the real numbers arecomprised of the union of the set of irrational numbers and rational numbers.

Real NumbersIrrational Numbers

Rational NumbersIntegers

Whole Numbers

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Number Lines and Absolute Value

A number line is a method of picturing the set of real numbers. Each point on the number linecorresponds to exactly one real number, as in the picture below:

-5 -4 -3 -2 -1 0 1 2 3 4 5

P 3/2O π Q-3/2 5/2

The absolute value of a number x, denoted by |x|, refers to the distance from that number to the origin,or zero (point O in the picture above). Note that, since distance is always positive or zero, the absolutevalue of a number will always be positive or zero. A more precise definition is as follows: if x ispositive or zero, then |x| will be the same as x, but if x is negative, we must change the sign to positive,indicated by −x. Therefore,

|x| x if x ≥ 0−x if x 0

For instance, |−7| 7 |10| 10 |0| 0 |3 − 5| |−2| 2.

Example: Evaluate |6 − 9|.Solution: |6 − 9| |−3| 3.Notice that in this example, we did not simply change the subtraction to addition. The absolute valuebars act as parentheses, so what is inside must be evaluated first. Therefore, in general, it is incorrect tosay that |x − 3| |x| 3, or |x − 3| x 3!!

Also notice that |6 − 9| gave us 3, the distance between the numbers 6 and 9 on the number line. Ingeneral, |a − b| gives us the distance between the numbers a and b on the number line.

Example: What is the distance between the numbers 2 and −24?

Solution: The distance between any two numbers is the absolute value of their difference. Thus,distance between 2 and −24 equals |2 − −24| |2 24| |26| 26.

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Exercises for Chapter 1A - Real Numbers

For problems 1-6, determine which property of the real numbers is used.1. −6x 6x 02. 17 x x 173. 28 x − 7 28 x − 74. 14 x − 3 x − 3 145. 13x 1

x 136. x 3 − y −y x 3

For problems 7-10, simplify and name the properties used.7. x 7 − x8. 4x − 3 129. 3x 4 4x10. x 7x − 3

For problems 11-20, evaluate the given expression.11. |−18 − 3|12. |7 − 14|13. −|15 − 4|14. −2 − 4 − |−6|15. −334 − 25 − |10|16. −48 −37 48 38 99917. |−24 −24| 75 −49 − |74|18. 2 1

4 −512 1

319. 2 2

5 2 23 − 5 1

820. 1

12 1

312 2

321. Choose , , or : −|2| ? |−2|22. Describe the following in absolute-value notation: The distance between x and 3 is at most

2 .23. Describe in words: |y 2| a (Hint: y 2 y − −2 )24. Order the following list of numbers from smallest to largest:

|3 − 5|, |−3 − 5|, |−3 5|, − |3 − −5|, − |3 − 5|, |− |−3 − 5||

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Answers to Exercises for Chapter 1A - Real Numbers

1. Additive Inverse2. Commutative Property of Addition3. Associative Property of Addition4. Commutative Property of Multiplication5. Multiplicative Inverse and Multiplicative Identity (Since 13x 1

x 131 13 )6. Commutative Property of Addition (remember that x 3 − y x 3 −y )7.

x 7 − x Commutative Property 7 x − x Associative Property 7 x − x Additive Inverse Property 7 0 Identity for Addition 7

NOTE: While you can simply say that x 7 − x 7, the properties of real numbers helpexplain why this simplifying is allowed.

8.4x − 3 12 Distributive Property

4x − 12 12 Associative Property 4x −12 12 Additive Inverse* 4x 0 Identity for Addition 4x

*-remember that 4x − 12 4x −129.

3x 4 4x Commutative Property-last 2 terms 3x 4x 4 Distributive Property (in reverse) 3 4x 4 7x 4

Again you could simplify immediately: 3x 4 4x 7x 4. The properties allow you tosee why this works.

10. Later we will learn another method (FOIL) for doing this. It is based on the distributiveproperty as shown below:

x 7x − 3 x 7x −3 x 7x x 7−3 x2 7x x 7−3 x2 7x − 3x − 21 then simplify as #9 x2 4x − 21

11. 21

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12. 713. −1114. − 2 − 4 − |−6| (work parentheses from the inside out)

−2 − 4 − 6 −2 − −2 −2 2 −4

15. − 334 − 25 − |10| (do multiplication before subtraction) −334 − 10 − 10 −314 −42

16. − 48 −37 48 38 999 ( Here it may be best to rearrange the terms.) −48 48 38 −37 999

0 1 999 1000Note that the numbers may be added in any order (associative and commutative properties),but the above is the most expediant way.

17. |−24 −24| 75 −49 − |74| (Again rearranging terms may help.) |−48| 75 −49 − 74 48 −49 75 − 74

−1 1 018. 2 1

4 −512 1

3 Find a common denominator

2 312 −

512 4

12 Write the mixed number as an improper fraction

2712 −

912 18

12 32 or 1 1

219. 2 2

5 2 23 − 5 1

8 Start by writing improper fractions:

125

83 −

418 Common denominator inside absolute value

125

6424 −

12324

125

5924 cross cancel the 12 and the 24

5910 or 5 9

10 or 5.9

20. 112 1

312 2

3Work inside the parentheses first

136 2

636 4

6 1

56

76

13536

3635

21. −|2| |−2| Since −|2| −2 and |−2| 222. |x − 3| ≤ 223. y is more than a units from −224. − |3 − −5| −8

− |3 − 5| −2|3 − 5| |−3 5| 2

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Chapter 1B - Exponents and Radicals

Properties of ExponentsExponents can be used to represent repeated multiplication of a single factor. More formally, if nrepresents a positive integer, the expression xn means multiply x times itself n times.

If n 0, then we define xn x0 1, and for n 0 we define

xn x x x x

n times

Examples: 43 4 4 4 64 −152 −15 −15 225 Note that −152 −152 −225 since exponents are computed before multipliying (unless

there are parentheses as in the previous example) y3 y y y What does 54 equal? Answer: 54 5 5 5 5 625

We define x to a negative power as follows (we assume that x ≠ 0)x−1 1

xx−2 1

x2

x−n 1xn

Examples: 4−3 1

43 164

2−5 125 1

32

The following list of properties of exponentiation should be memorized, as you need to be able tomanipulate expressions involving exponents.

Property Example1 am an amn 23 24 234 27

2 am

an am−n x4

x2 x4−2 x2

3 a−n 1an 3−4 1

34 181

4 a0 1 for a ≠ 0 14 − x20 1 iff 14 − x2 ≠ 05 abn anbn 2y3 23y3 8y3

6 ab

n an

bn5x

2 52

x2 25x2

7 amn amn x3−4 x3−4 x−12 1x12

Although these properties work for any real exponents m and n, the properties are easy to see when m

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and n are integers."Proof" of Example 1(from previous page): 23 24

2 2 2 2 2 2 2 2 2 2 2 2 2 2 27

We can use these properties to simplify expressions with exponents in them.Example 1: Simplify the expression −z33z4.Solution:

−z33z4 −13z334z4

−181z34

−81z7

Example 2: Simplify the expression x−3y4

5y−1−3

.

Solution:x−3y4

5y−1−3

x9y−12

5−3y3

x953

y3y12

125x9

y15

Example 3: Simplify the expression 1545−6.Solution:

1545−6 3 545−6

34545−6

34545−6 3454−6

345−2

34

52

One could have stopped at 345−2 and not bothered about rewriting 5−2 as 1/52.

Example 4: Simplify the expression 534−210273.Solution: Before starting, scan the expression to find common terms. Here we note that 4 2 2, and10 5 2. This means that we should think to combine the common integer factors of the terms 53,102, and 4−2 .

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534−210273 5322−25 2273

532−4522273

53222−473

552−273

5573

22 .

We could have stopped at the line 552−273, and not rewritten this as 5573

22 .

If we have a complicated expression involving integers raised to a power, then the easiest first stepwould be to write the integers as a product of their prime factors. Remember that a positive integer(greater than 1) is said to be prime if its only integer divisors are itself and 1. Thus, 2, 3, 7, 23, and 31are prime, while 4, 9, 28, and 32 are not.

Example 5: The following example is fairly complicated. If you can follow all of the steps andunderstand the reasoning, you are well on your way to understanding and being able to use theexponential properties.

Simplify the expression 1983700−4

546−3

Solution: As we noted at the end of Example 4, when simplifying a complicated expression involvingintegers, factor each integer into its prime factors.

1983700−4

546−32 3211322527−4

542 3−3

23361132−85−87−4

542−33−3

23361132−85−87−45−42333

23−833635−8−47−4113

2−2395−127−4113

39113

2251274

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Radicals and Properties of RadicalsRadicals (or roots) are, in effect, the opposite of exponents. In other words, the n th root of a number ais a number b such that

b n a a1/n bn a

The number b is called an n th root of a. The number n is referred to as the index of the radical (if noindex appears, n is understood to be 2). The principal n th root of a number is the n th root of a whichhas the same sign as a. For example both 2 and −2 satisfy x2 4, but 2 is the (principal) square root of4.Examples: 3 27 3 since 33 27 4 16 2 since 24 16 (Note −24 16 also, but 2 is the principal 4th root 3 −64 −4 since −43 −64 4 −81 is not a real number and we will say that it does not exist. (In this course we won’t learn

how to take an eventh power of a negative number.)

Radicals are used to define rational exponents:

a1n n a

amn n am

The notation a1/n is extremely useful, and we encourage you to use it whenever you have to simplifyexpressions involving radicals.Examples: 125

13 3 125 5

−6423 3 −642 3 −64 2

−42 16

32−35 1

3235 1

5 32 3 123 1

8

Since radicals are nothing more than rational exponents, many of the properties of exponents also applyto radicals.

Property Example

1 n am n a m 5 323 5 32 3 23 8

2 n a n b n ab 27 3 81 9

3n an b

nab , b ≠ 0 3

278

3 273 8

32

4 m n a mn a 4 2x 8 2x

5a If n is odd n an a 3 −1273 −127

5b If n is even n an |a| 4 −1274 |−127| 127

The following list is a restatement of these properties, but in exponential notation. You need to be

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familiar with both radical and exponential notation, and be able to convert between the two.Property Example

1 am1/n am/n 3231/5 321/53 23 8

2 a1/nb1/n ab1/n 271/231/2 27 31/2 811/2 9

3 a1/n

b1/n ab

1/n, b ≠ 0 27

81/3

271/3

81/3 32

4 a1/m1/n a1/mn 2x1/4 1/2 2x1/8

5a If n is odd an1/n a −1273 1/3 −127

5b If n is even an1/n |a| −1274 1/4 |−127| 127

Examples: −32 3 (refer to Property 5b)

163/2 163/2 161/2 3 43 64 (refer to property 1-given the right hand side)

−163/2 −1632 −16

12

3

(refer to property 1) −16 3

There is no answer as we cannot take the square root of −16. 321/5271/3 321/5271/3 2 3 6 8 x8 |x|

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Simplifying Radicals

Properties 2 and 3 in the previous section can be used to simplify radical expressions.A radical expression is simplified when the following conditions hold:

1. All possible factors (”perfect roots”) have been removed from the radical.2. The index of the radical is as small as possible. Remember, the index of 3 10 is 3.3. No radicals appear in the denominator.

For example, to simplify the radical 40 , we factor the number into prime factors. Since the index is2, any square factors are ”pulled out” of the radical:

40 23 5

22 2 5

22 2 5 (Using Property 2)

2 10 , or 2 10 .

Notes for the above example:

Recall that Property 2 states that n ab n a n b or, in exponential notation ab1n a

1n b

1n

Since 23 22 2 , the perfect square is pulled out of the radical, while the remaining 2 staysinside the radical.

The previous example could also be done directly by finding a factor of 40 which is a perfectsquare:

40 4 10

4 10 (by Property 2 again-see previous note)

2 10This method works quite well for square roots.

Example1: Simplify 72x3

Solution:72x3 32 22 2 x2 x

32 22 x2 2x 3 2 |x| 2 x 6|x| 2x

Example 2: Simplify 3 54xy4

Solution: 3 54xy4 3 2 33 x y4

3 y 3 2 x y 3y 3 2xy

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Example 3: Simplify 480x2

y4

Solution: 480x2

y4 424 5 x2

y4

2|y|

4 5x2 (since 4 y4 |y| )

Finally, we can use the fact that radicals can be written as fractional exponents to be sure the radical isin ”lowest terms”.Example 4: Simplify 4 9Solution: 4 9 4 32

3 24

3 12

3

Example 5: Simplify 6 81x4

Solution: 6 81x4 6 34x4

3 46 x 4

6

3 23 x 2

3

3 32x2

3 9x2

These last two examples demonstrated the utility of exponential notation.

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Rationalizing DenominatorsRationalizing the Denominator is a technique used whenever a radical appears in the denominator of anexpression. To begin with, make sure the radical is simplified; that is, perfect roots are pulled out. Thenext step depends on how the radical appears in the denominator:

I. The Radical is a single termMultiply both the numerator and the denominator of the expression by something which will produce aperfect root in the denominator.Example 1: Rationalize the denominator: 8

3 54Solution: First simplify the radical in the denominator:

83 54

83 2 33

83 3 2

(Recall that 3 33 3)

Note that if the 2 was a cube, we would no longer need a radical in the denominator.To rationalize this denominator, we multiply both the numerator and denominator by 3 22 as follows:

83 3 2

3 22

3 22 8 3 22

3 3 23 8 3 4

3 2 4 3 43

Question: Why do we pick 3 22 ?Answer: We want to multiply the denominator of 8

3 3 2by a term which will enable us to remove

any radicals. Thus, we need to figure out what to multiply 3 2 by. Well, if we write this in exponentialnotation, we have

3 2 21/3 .Clearly if we multiply by 22/3 we get something nice.

21/322/3 21 2 .

Moreover 22/3 3 22 .

Example 2: Rationalize the denominator: 510

Solution: 510

1010

5 1010

102

Example 3: Rationalize the denominator: 13 5x2

Solution: 13 5x2

3 5x3 5x

3 5x

3 5x3(note that you only need one more 5x )

3 5x5x

II. The Denominator is a Sum of TermsIn this example the denominator has the form a b m (here we only concern ourselves with squareroots). The conjugate of this expression is a − b m . Similarly, the conjugate of the expression

a b is a − b . In the example on the next page, notice what happens when we multiply the

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numerator and denominator by the conjugate:

55 7

5 − 75 − 7

5 5 − 7

5 7 5 − 7

25 − 5 752 − 7 2

25 − 5 718 .

Here the multiplication in the denominator is done using a distributive property technique called FOIL(shown below). You can also use the special product a ba − b a2 − b2 which we will discussin detail in another section.

5 7 5 − 7

5 5 5 − 7 5 7 7 − 7

First Outer Inner Last

52 − 5 7 5 7 − 72

52 − 72

Note how the inner terms cancel.

25 − 7 18

Question: What should you multiply 5 − 3 by to rationalize it?Answer: Multiply 5 − 3 by 5 3 .

5 − 3 5 3 25 − 9 16.

Remembera − ba b a2 − b2.

Question: What should you multiply a b c by to rationalize it?Answer: Multiply a b c by a − b c .

a b c a − b c a2 − b2 c 2

a2 − b2|c| .

Example 4: Rationalize the denominator: 2x5 − 3

Solution: 2x5 − 3

5 35 3

2x 5 3

52 − 3 2

2x 5 3

22

x 5 3

11

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Example 5: Rationalize the denominator: 45 6

Solution:

45 6

5 − 65 − 6

4 5 − 6

5 2 − 6 2 4 5 − 6

5 − 6 −4 5 − 6 4 6 − 4 5

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Combining Radical Expressions

Radical expressions can only be combined by addition or subtraction if there are like terms-terms withthe same index and same radicand (the expression inside the radical).

2 2 3 4 2 − 4 3 ,In the above expression, the terms with 2 are alike and the terms with 3 are alike. Therefore, theabove expressions simplifies as follows:

2 2 3 4 2 − 4 3 2 4 2 2 3 − 4 3

1 4 2 2 − 4 3

5 2 − 2 3

Notice that the two remaining terms are not alike and, hence, cannot be simplified.

Some terms which do not look alike at first glance may be alike after simplifying. Therefore, it isimportant that you simplify all radicals before combining like terms.

Examples: 50 − 32 2 52 2 − 25 2

5 2 − 4 2 2 5 − 4 1 2 2 2

7 3 80x − 2 3 270x 4 3 10 7 3 24 5x − 2 3 33 2 5x 4 3 10 7 2 3 2 5x − 2 3 3 2 5x 4 3 10 14 3 10x − 6 3 10x 4 3 10 8 3 10x 4 3 10

(note that these terms are NOT alike even though they both have a 10)

In simplifying 12 − 1

8 , we begin by using Property 3 ( nab

n an b

), then simplifying and

rationalizing the denominator before combining:12 − 1

8 12− 1

8(Note that 8 4 2)

12− 1

2 2(Multiply both by 2

2)

1 22 2

− 1 22 2 2

22 − 2

4 (now get a common denominator)

2 24 − 2

4

2 2 − 24 (now combine radicals)

24

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Exercises for Chapter 1B - Exponents and Radicals

For problems 1-3, evaluate the expression.1. −34

2. 5−3

3. 25

−2

For problems 4-6, simplify the expression.4. −3ab44ab−35. 5a3a2−116a100

6. 15x−2y5x3y−4

−2

For problems 7-13, compute the following, if possible:7. 3 −343

8. 48116

9. −4910. Write 3 y4 using exponents

11. Write x10

− 23 using radicals

12. −6413

13. 12527

− 23

For problems 14-18, simplify the expression.14. 4815. 3 54x4

16. 3 5617. 4 162x3y6

18. 10 32x5

For problems 19-26, rationalize the denominator.19. 3

220. 2

1821. 6

3 1622. 9

4 8x23. 2

1 524. 3

5 3

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25. 47 − 3

26. 54 − 11

For problems 27-31, simplify the expression.27. 27 − 5 48 2 4528. 4 48 5 4 2 − 3 4 329. 75x3 − 27x 3x5

30. 26

3 32− 2

331. 45 − 6 12532. 16x3 3 16x4y 3 54x4y x 25x

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Answers to Exercises for Chapter 1B - Exponents andRadicals

1. −3 −3 −3 −3 812. 1

53 1125

3. 52

2 52

22 254

4. −3ab44ab−3 −12a11b4−3 −12a2b

5. 5a3a2−116a100 125a3 1a2 1 125a

6. 15x−2y5x3y−4

−2

3y y4

x3 x2

−2

3y5

x5

−2

x5

3y5

2

x10

9y10

(Other methods are possible)7. −7 , since −73 −343

8.4 814 16

32

9. Not possible; there is no real number whose square is −4910. y 4

3

11. 3x

10−2

310x

2 3

100x2

12. −4

13. 27125

23

3 273 125

2

35

2 9

25

14. 48 24 3 22 3 4 315. 3 54x4 3 2 33 x4 3x 3 2x16. 3 56 3 23 7 2 3 717. 4 162x3y6 4 2 34 x3 y6 3y 4 2x3y2 (Since y6 y4 y2 )18. 10 32x5 10 25x5 2 5

10 x 510 2 1

2 x 12 2x

19. 32 2

2 3 2

2

20. 218

22 32

(simplify radical first)

23 2

22

2 23 2

23 (Note the cancellation of the 2’s)

21. 63 16

63 24

(simplify radical first)

62 3 2

(OK to cancel first)

33 2

3 22

3 22

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3 3 22

3 23

3 3 42

22. 94 8x

94 23 x

(factor first to see what you need)

94 23 x

4 2x3

4 2x3(to get 4th powers you need one 2 and 3 x’s)

9 4 2x3

4 24x4

9 4 2x3

2x

23. 21 5

1 − 51 − 5

2 1 − 5

12 − 5 2

21 − 5 −4

− 1 − 52

24. 35 3

5 − 35 − 3

3 5 − 3

5 2 − 32

3 5 − 9−4 or −3 5 − 9

4

25. 47 − 3

7 37 3

4 7 3

7 2 − 3 2

4 7 3

4 7 3 (since the 4’s cancel)

26. 54 − 11

4 114 11

5 4 11

42 − 11 2 4 5 555

27. 27 − 5 48 2 45 32 3 − 5 24 3 2 32 5 3 3 − 5 4 3 6 5 3 4 3 −1 6 5 7 3 5 5

28. 4 48 5 4 2 − 3 4 3 4 24 3 5 4 2 − 3 4 3 2 4 3 5 4 2 − 3 4 3 5 4 2 2 − 3 4 3 5 4 2 − 4 3

29. 75x3 − 27x 3x5 52 3 x3 − 33 x 3x5

5x 3x − 3 3x x2 3x x2 5x − 3 3x

30. (rationalize the denominators first)26

66

3 32

22− 2

3

33

2 66 3 6

2 − 2 33

2 66 9 6

6 − 4 36

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2 6 9 6 − 4 36

11 6 − 4 36

31. 45 − 6 125 32 5 − 6 53 3 5 − 5 36 3 5 − 5 2 5

32. 16x3 3 16x4y 3 54x4y x 25x 4x x 2x 3 2xy 3x 3 2xy 5x x 9x x 5x3 2xy

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© : Pre-Calculus - Chapter 1C

Chapter 1C - Polynomials

Definition of a Polynomial

A polynomial is the most common algebraic expression. Polynomials are expressions which containterms of the form axk , where a is a real constant and k is a nonnegative integer. More formally, apolynomial expression is an expression of the form

anxn an−1xn−1 a1x a0

In the above expression, an is a nonzero real number, a0, a1, ... , an−1 are real numbers, and n is apositive integer. n is called the degree of the polynomial, an is called the leading coefficient, and a0 iscalled the constant term.

If n 2 the polynomial is called a binomial or quadratic.If n 3 the polynomial is called a trinomial or cubic.

Example The height, s (in feet), of a ball thrown in the air at time t 0 seconds is given by thepolynomial equation s −16t2 80t 5 . Here the degree of the polynomial is 2 (the highestexponent), the leading coefficient is −16, and the constant is 5.

Example The volume, V, of a cube with edge length x is given by V x3 . Here the degree of thepolynomial is 3, the leading coefficient is 1 (remember that x3 1x3), and the constant is 0.

Polynomial expressions do not contain negative exponents nor do they contain radicals. So anexpression such as 9 − x2 is not a polynomial.

Question: Is x3 1x a polynomial?

Answer: No, the 1x term is the same as x−1 and the exponent is not a positive integer.

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Sums and Differences of PolynomialsAddition and Subtraction of Polynomials is done by combining like terms, that is, terms which havethe same variable and exponent.

Example 1: Simplify the expression: 15x2 − 6 8x3 − 14x2 17Solution: 15x2 − 6 8x3 −14x2 17

8x3 15 − 14x2 −6 17 8x3 x2 11

To subtract polynomials, recall that subtraction is defined as adding the opposite (inverse); that is,a − b a −b . It is important that you distribute the negative sign over the entire polynomial.

Example 2: Simplify the expression: 15x2 − 6 − 8x3 − 14x2 17Solution: 15x2 − 6 − 8x3 − 14x2 17

15x2 − 6 −8x3 14x2 − 17 (NOT −8x3 − 14x2 17 !!)15x2 −6 − 8x3 14x2 − 17 −8x3 29x2 − 23

Example 3: Simplify the expression: 3x3 − 2x2 8 3x4 2x2 − 5Solution: 3x3 − 2x2 8 3x4 2x2 − 5

3x3 − 2x2 8 3x4 2x2 − 5 3x4 3x3 −2 2x2 8 − 5 3x4 3x3 3

Example 4: Simplify the expression: −5x2 − 1 − −3x2 5Solution: − 5x2 − 1 − −3x2 5 (distribute negative signs)

−5x2 1 3x2 − 5 −2x2 − 4

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Products of PolynomialsMulitplication of Polynomials is based on the Distributive Property of Real Numbers as illustrated inthe example below:Example 1: Expand the following product x − 68x 7.Solution:

x − 68x 7 x8x 7 − 68x 7 (treat 8x 7 as a single number first)

x8x x7 − 68x − 67

8x2 7x − 48x − 42 8x2 − 41x − 42

Notice, in the underlined step, that the distributive properties follow the familiar pattern FOIL, which isexplained below.

x8x x7 − 68x − 67First Outer Inner Lastterms terms terms terms

Although FOIL only works with binomials, the distributive property method works on anypolynomials.

Example 2: Expand 5x − 6x3 − 4x2 2.Solution 1

5x − 6x3 − 4x2 2 5xx3 − 4x2 2 − 6x3 − 4x2 2 5x4 − 20x3 10x − 6x3 24x2 − 12 5x4 − 26x3 24x2 10x − 12

(Don’t forget to distribute the negative sign as well!!!)

Solution 2: The distributive method is easy to organize and picture using a rectangle, or box, whoselengths are the two factors as illustrated below:

x3 −4x2 25x 5x4 −20x3 10x−6 −6x3 24x2 −12

5x4 − 20x3 − 6x3 24x2 10x − 12 5x4 − 26x3 24x2 10x − 12

Example 3: Expand x2 9x2 − 6x 9.Solution: x2 9x2 − 6x 9 x2x2 − 6x 9 9x2 − 6x 9

x4 − 6x3 9x2 9x2 − 54x 81 x4 − 6x3 18x2 − 54x 81

Example 4: Expand a ba − b.Solution: a ba − b aa − b ba − b

a2 − ab ba − b2

a2 − ab ab − b2 (By the commutative property)

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a2 − b2

The previous example a ba − b a2 − b2 is an example of a Special Product. Special productscan not only make multiplication of certain polynomials easier, they are also useful when you want tofactor a polynomial; that is, given a polynomial, find two or more polynomials whose product is thegiven polynomial. The most common special products are listed below (here a and b represent anynumbers, variables, or algebraic expressions).

Difference of Squares: a ba − b a2 − b2

Square of a Binomial: a b2 a2 2ab b2

a − b2 a2 − 2ab b2

Sum of Cubes: a ba2 − ab b2 a3 b3

Difference of Cubes: a − ba2 ab b2 a3 − b3

Cube of a Binomial: a b3 a3 3a2b 3ab2 b3

a − b3 a3 − 3a2b 3ab2 − b3

NOTE that there is no special product for the sum of squares that involves only real numbers..

Historical Note: The French mathematician Blaise Pascal (1623-1662) discovered an easy way toremember the coefficients of a binomial power using what is now called Pascal’s Triangle. After thefirst two rows, each succesive row is obtained by adding the two numbers immediately above it asshown below:

1 a b0 11 1 a b1 1a 1b

1 2 1 a b2 1a2 2ab 1b2

1 3 3 1 etc. a b3 1a3 3a2b 3ab2 1b3

Using this triangle, we can see thata b4 1a4 4a3b 6a2b2 4ab3 1b4

The numbers in Pascal’s Triangle are also useful in Probability.

Example 5: Expand x − 23.Solution: x − 23 x3 − 3x22 3x22 − 23

x3 − 6x2 12x − 8

Example 6: Expand x 2yx − 2y2.Solution: x 2yx − 2y2 x2 − 2y2 2 (work inside the parenthesis first)

x2 − 4y2 2

x22 − 2x24y2 4y22

x4 − 8x2y2 16y4

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Quotients of PolynomialsDivision of Polynomials is done using standard long division techniques.Example 1: Review basic long division for numbers. Use long division to find 1350 18.Solution: The algorithm for long division is: Divide

Multiply Subtract Bring Down. Repeat these steps as long as it is possible to divide.

Note that 18 7 126, so 18 divides 135 seven times:

7 (divide)18 1 3 5 0

1 2 6 (multiply)9 (subtract)

Now bring down the 0 and repeat, noting that 18 5 90:

7 518 1 3 5 0

1 2 69 09 0

0

So 1350 18 75.

Example 2: Now use long division to divide polynomials: 8x2 13x − 6 x 2Solution: First, divide the highest terms. 8x2

x 8x. Then follow the steps discussed in the aboveexample.

8x −3x 2 8x2 13x −6

8x2 16x−3x −6−3x −6

0

so 8x2 13x − 6 x 2 8x − 3Another way of writing this is to say that 8x2 13x − 6 x 28x − 3.

Example 3: Perform the following operation: x4 − 3x2 − 4 x2 xSolution: We begin by writing out the long division, using 0’s for any missing powers of x:

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x2 x x4 0x3 −3x2 0x −4

We need to determine how many times x2 will divide into x4. Since x4

x2 x2, the first term of ourquotient is x2:

Divide x2

x2 x x4 0x3 −3x2 0x −4

Now multiply and subtract:

x2

x2 x x4 0x3 −3x2 0x −4Multiply x4 x3

Subtract −x3

Bring down the next term:

x2

x2 x x4 0x3 −3x2 0x −4x4 x3

−x3 −3x2 Bring down

The next term of the quotient will be −x3

x2 −x, so we repeat the process:

x2 −x Dividex2 x x4 0x3 −3x2 0x −4

x4 x3

−x3 −3x2

x2 −xx2 x x4 0x3 −3x2 0x −4

x4 x3

−x3 −3x2

Multiply −x3 −x2

x2 −xx2 x x4 0x3 −3x2 0x −4

x4 x3

−x3 −3x2

−x3 −x2

Subtract −2x2 0x Bring down

The next term of our quotient will be −2x2

x2 −2, so bring down the next term and repeat:

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x2 −x −2 Dividex2 x x4 0x3 −3x2 0x −4

x4 x3

−x3 −3x2

−x3 −x2

−2x2 0x

x2 −x −2x2 x x4 0x3 −3x2 0x −4

x4 x3

−x3 −3x2

−x3 −x2

−2x2 0xMultiply −2x2 −2x

x2 −x −2x2 x x4 0x3 −3x2 0x −4

x4 x3

−x3 −3x2

−x3 −x2

−2x2 0x−2x2 −2x

Subtract 2x −4 Bring down

Since x2 does not evenly divide 2x, the long division is complete.We can say that x4 − 3x2 − 4 x2 x x2 − x − 2 with a remainder of 2x − 4. We could also sayeither of the following:

x4 − 3x2 − 4x2 x

x2 − x − 2 2x − 4x2 x

or:x4 − 3x2 − 4 x2 xx2 − x − 2 2x − 4

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FactoringThe process of factoring is a type of ”reverse-multiplication”, where you are given a polynomial andhave to write it as a product of factors. A polynomial is completely factored when each factor is prime,or cannot be factored again. The idea is very similar to factoring numbers:

60 5 12 is one factoring of 60 5 6 2 5 2 3 2 5 22 3 is the prime factorization of 60

There are several strategies to determine how to factor polynomials.

I. Common Factors

A common factor is a factor of every term of an expression. Common factors can be pulled out of anexpression using the distributive property in reverse:

ab ac ab cExamples: x3 4x xx2 4 −2x2 6x −2xx − 3 (Note the negative in the second term) 6mn2 15m2n − 30m3n3 3mn2n 5m − 10m2n2

Finding common factors should always be the first step in factoring an expression.

II. Factoring by Grouping

Factoring by Grouping is especially useful when you have more than three terms in the polynomial.The technique of factoring by grouping is really using common factors creatively, as shown in thefollowing example:

Example 1: Factor 2x3 − x2 6x − 3Solution: Although there is no factor common to all terms (except 1, which we ignore), we can”group” the polynomial by twos, each of which have a common factor:

2x3 − x2 6x − 3

x22x − 1 32x − 1Now notice that there are two terms, each of which has the factor 2x − 1.

2x − 1x2 3

It is very easy to check your answer by mulitplying it out:2x − 1x2 3

2xx2 3 − 1x2 3 2x3 6x − x2 − 3 2x3 − x2 6x − 3✓

III. Factoring Using Special Products

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The special products we learned in the section ”Products of Polynomials” can be used to help factorexpressions which have an appropriate form:

Example 2: Factor completely x3 − 9xSolution: As mentioned in the previous section, the first thing we do is look for a common factor:

x3 − 9x xx2 − 9

Notice how the second factor can now be written as x2 − 32 , a difference of two squares. Recall thata2 − b2 a ba − b.

x3 − 9x xx2 − 9

xx 3x − 3The expression is now completely factored.

Example 3: Factor completely x3 8y3

Solution: Notice how the expression can be rewritten as x3 2y3, a sum of two cubes.Recall that a ba2 − ab b2 a3 b3

x3 8y3

x 2y x2 x2y 2y2

x 2yx2 2xy 4y2

Example 4: Factor completely x2 25Solution: Recall that there is no special product for a sum of squares a2 b2 . In fact, thisexpression cannot be factored; it is prime over the real numbers.

IV. Factoring Binomials (x2 bx c )

If possible, a binomial of this form must factor as x mx n , where m and n are integers. Notethat if this is true:

x mx n x2 bx cx2 nx mx mn x2 bx c

x2 m nx mn x2 bx c

We can see in the last statement that the numbers m and n that we seek must add up to b and multiply toc.

Example 5: Factor completely x2 7x 12Solution: We are looking for two numbers which will add up to 7 and multiply to 12. Thenumbers (obtained by trial and error) are 3 and 4. Therefore,

x2 7x 12 x 3x 4

Example 6: Factor completely x2 − 4x − 21

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Solution: We are looking for two numbers which will add up to −3 and multiply to −21. Notethat since the product is negative, the numbers must be different signs (one negative and one positive).The numbers are −7 and 3 (Note that if we had chosen 7 and −3, the numbers add up to 4. In general,if the signs are different, the larger number will have the same sign as b, the middle term). Therefore,

x2 − 3x − 21 x − 7x 3

Example 7: Factor completely x2 − 6x 9Solution: We are looking for two numbers which will add up to −6 and multiply to 9. Note thatsince the product is positive, the numbers must be the same sign (both positive or both negative). Sincethe middle term is negative, the numbers must both be negative. The numbers are −3 and −3.Therefore

x2 − 6x 9 x − 3x − 3 x − 32

Note that we could have factored this using the special product a − b2 a2 − 2ab b2

V. Factoring Trinomials (ax2 bx c )

If possible, a trinomial of this form (where the leading coefficient is not 1) must factor in the formmx snx t , where m,n, s, and t are all integers. It is possible to find these numbers by trial anderror; however, such a method may at times be haphazard or tedious. The following is a morestraightforward approach utilizing the technique of factoring by grouping.

First, we need to split the middle term bx into two terms which will allow us to factor by grouping. Todo this, we use a similar technique as before, only now we look for two numbers whose sum is b andwhose product is ac .

If you are curious as to why, see below:As before, multiply the desired form out:mx snx t ax2 bx cmnx2 mtx nsx st ax2 bx cmnx2 mt nsx st ax2 bx c

Now note that b mt ns and ac mnst mtns . The numbers we arelooking for are mt and ns. How they split up into m,n, s, and t are determined when wefactor by grouping.

Example 8: Factor completely 9x2 − 3x − 2Solution: We first must find two numbers whose sum is −3 and whose product is9−2 −18. The numbers are −6 and 3. Now we rewrite −3x as −6x 3x and factor by grouping:

9x2 − 3x − 2

9x2 − 6x 3x − 2

3x3x − 2 13x − 2

3x 13x − 2(remember that you can check your answer by multiplying)

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Example 9: Factor completely 6x2 − x − 15Solution: We must first find two numbers whose sum is −1 and whose product is6−15 −90. The numbers are 9 and −10. Now we rewrite −x as 9x − 10x and factor by grouping:

6x2 − x − 15

6x2 9x −10x − 15

3x2x 3 − 52x 3(Notice that both signs change in the last term)

3x − 52x 3

If you are afraid of getting the middle terms in the wrong order, not to worry...

6x2 − x − 15

6x2 − 10x 9x − 15

2x3x − 5 33x − 5

2x 33x − 5the same as above.

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Exercises for Chapter 1C - Polynomials

1. Determine which of the following are polynomials:4x3 − 8x2 17x − 12 , −6x 6x−1 , −24 , 3 − x2 , 16 − x2

2. For the polynomials in #1, give the degree, leading coefficient, and constant termFor problems 3-6, simplify the expression.

3. −5x7 2x4 3x2 − 2 −9x2 44. 2x3 − 5x2 1 −3x3 7x − 6 − 3x3 6x2 − 7x5. y3 1 − y3 − 1 y2 − 2y 16. y3 1 − y3 − 1 − y2 − 2y 17. Find a polynomial which, when added to x3 3x2 − 3x 2, yields 2x2 4x 7

For problems 8-14, simplify the expression.8. 3x − 25x 79. −9x2 4x − 510. 3x − 2x2 − 2x 211. 4x 52

12. 3x 723x − 713. 2x 1x − 3 5x − 314. 2x − 13

15. An open box made from a 9-inch square piece of material by cutting equal squares from allcorners and folding the sides. The volume of the box is given by x9 − 2x2. Give thedegree, leading coefficient, and constant term of this polynomial.

For problems 16-23, divide by long division.16. x2 − 6x 10 2x 317. x3 − 1 x − 118. x4 − 1

x − 119. x5 − 1

x − 120. If n is a positive integer, what do you think xn − 1

x − 1 equals ?

21. x2 1x − 1

22. x3 − 6x2 12x − 8x − 2

23. x4 − 2x3 4x2 − 2x − 7x2 − 2x 3

For problems 24-38, factor each expression as completely as possible.24. 6x3 − 4x25. x − 22x − 3x − 226. y3 − 27z3

27. 2x3 1628. x4 − 2x3 − 8x 1629. a3 − 27a

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30. 4x3 4x2 − 9x − 931. x2 11x 1032. y2 − 5y 633. x2 x − 234. x2 8x − 2035. y3 − 2y2 − 3y36. 2x2 x − 1537. 3x2 10x 838. 15x2 − 11x 2

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Answers to Exercises for Chapter 1C - Polynomials

1. The polynomials are 4x3 − 8x2 17x − 12 , −24, and 3 − x2

− 6x 6x−1 is not a polynomial because of the negative exponent16 − x2 is not a polynomial because of the radical.

2. 4x3 − 8x2 17x − 12 : degree 3, leading coefficient 4, and constant term −12−24 : degree 0, leading coefficient −24, and constant term −243 − x2 −x2 3 : degree 2, leading coefficient −1, and constant term 3

3. −5x7 2x4 3x2 − 2 −9x2 4 −5x7 2x43x2−2−9x24 −5x7 2x4 − 6x2 2

4. 2x3 − 5x2 1 −3x3 7x − 6 − 3x3 6x2 − 7x 2x3 − 5x2 1 − 3x3 7x − 6 − 3x3 − 6x2 7x 2 − 3 − 3x3 −5 − 6x2 7 7x 1 − 6 −4x3 − 11x2 14x − 5

5. y3 1 − y3 − 1 y2 − 2y 1 y3 1 − y3 − 1 − y2 − 2y 1 y3 1 − y3 1 − y2 2y − 1 1 − 1y3 − y2 2y 1 1 − 1 −y2 2y 1

6. y3 1 − y3 − 1 − y2 − 2y 1 y3 1 − y3 − 1 y2 − 2y 1 y3 1 − y3 1 y2 2y 1 1 − 1y3 y2 2y 1 1 1 y2 2y 3

7. Recall that if a b c, then a c − b, so the polynomial we seek is2x2 4x 7 − x3 3x2 − 3x 2 2x2 4x 7 − x3 − 3x2 3x − 2

−x3 2 − 3x2 4 3x 7 − 2 −x3 − x2 7x 5

8. 3x − 25x 7

3x −25x 15x2 −10x7 21x −14

15x2 21x − 10x − 14 15x2 11x − 14

9. −9x2 4x − 5

x −5−9x2 −9x3 −45x2

4 4x −20

2

−9x3 − 45x2 4x − 20

10. 3x − 2x2 − 2x 2

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x2 −2x 23x 3x3 −6x2 6x−2 −2x2 4x −4

3x3 − 6x2 − 2x2 6x 4x − 4 3x3 − 8x2 10x − 4

11. 4x 52 4x 54x 5

4x 54x 16x2 20x5 20x 25

16x2 20x 20x 25 16x2 40x 25

12. We could square the first term using a b2 a2 2ab b2, but notice the following:3x 723x − 7

3x 73x 73x − 7Now use difference of squares on the term in brackets:

3x 79x2 − 49

9x2 −493x 27x3 −147x7 63x2 −343

27x3 63x2 − 147x − 343

13. We will use FOIL here because of the multiple operations:2x 1x − 3 5x − 3

2xx 2x−3 1x 1−3 5x − 15 2x2 − 6x x − 3 5x − 15 2x2 − 18

14.2x − 13

2x − 122x − 1 4x2 − 4x 12x − 1

2x −14x2 8x3 −4x2

−4x −8x2 4x1 2x −1

8x3 − 12x2 6x − 1

15. Multiply out the expression:x9 − 2x2 x18 − 36x 4x2 4x3 − 36x2 18x

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Therefore, the degree of the polynomial is 3, the leading coefficient is 4, and the constantterm is 0.

16. x2 − 6x 102x 3 1

2 x − 154 85

42x 317. x3 − 1

x − 1 x2 x 1

18. x4 − 1x − 1 x3 x2 x 1

19. x5 − 1x − 1 x4 x3 x2 x 1

20. xn − 1x − 1 xn−1 xn−2 x 1

21. x2 1x − 1 1 x 2

x − 122. x3 − 6x2 12x − 8

x − 2 x2 − 4x 4

23. x4 − 2x3 4x2 − 2x − 7x2 − 2x 3

x2 1 − 10x2 − 2x 3

24. 6x3 − 4x 2x3x2 − 225. x − 22x − 3x − 2 x − 22x − 3

NOTE: you could multiply out the expression first and then factor the resulting trinomial.26. Use difference of cubes:

y3 − 27z3 y3 − 3z3 y − 3zy2 3yz 9z2

27. Factor the common factor, then use sum of cubes:2x3 16 2x3 8 2x 2x2 − 2x 4

28. Factor by grouping, then use difference of cubes:x4 − 2x3 − 8x 16

x3x − 2 − 8x − 2 x3 − 8x − 2 x − 2x2 2x 4x − 2 x − 22x2 2x 4

29. a3 − 27a aa2 − 27 We cannot factor this any further (using only integers).30. Factor by grouping, then use difference of squares:

4x3 4x2 − 9x − 9 4x2x 1 − 9x 1 4x2 − 9x 1 2x 32x − 3x 1

31. We need two numbers whose product is 10 and whose sum is 11; the numbers are 10 and 1:x 10x 1

32. We need two numbers whose product is 6 and whose sum is −5; the numbers are −2 and −3y − 2y − 3

33. We need two numbers whose product is −2 and whose sum is 1; the numbers are −1 and 2:x 2x − 1

34. We need two numbers whose product is −20 and whose sum is 8; the numbers are 10 and−2:

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x 10x − 235. Common factor first:

y3 − 2y2 − 3y yy2 − 2y − 3

We need two numbers whose product is −3 and whose sum is −2; the numbers are −3 and 1: yy − 3y 1

36. We need two numbers whose product is 2−15 −30 and whose sum is 1; the numbers are6 and -5:

2x2 6x − 5 − 15 2xx 3 − 5x 3 2x − 5x 3

37. We need two numbers whose product is 38 24 and whose sum is 10; the numbers are 6and 4:

3x2 6x 4x 8 3xx 2 4x 2 3x 4x 2

38. We need two numbers whose product is 152 30 and whose sum is −11; the numbers are−5 and −6:

15x2 − 5x − 6x 2 5x3x − 1 − 23x − 1 5x − 23x − 1

© : Pre-Calculus

© : Pre-Calculus - Chapter 1D

Chapter 1D - Rational Expressions

Definition of a Rational Expression

A rational expression is the quotient of two polynomials. (Recall: A function px is a polynomial in xof degree n, if there are constants a0, a1, , an, with an ≠ 0 such that px a0 a1x anxn.)

More formally, a rational expression is an expression of the form pq , where p and q are polynomials,

and qx cannot be the zero polynomial. The denominator of a rational expression can be a constantpolynomial though. For example, rx x is a polynomial of degree 1, and it is also a rationalexpression, for rx x x

1 .

Example 1: 4x − 76x − 16 , 2

x − 9 , x3 − 282 , and x 5

x2 25are examples of rational expressions.

Example 2: x2 12x 7 is not a rational expression.

A rational expression involves a quotient, and since division by 0 is not defined for real numbers, thereare sometimes restrictions on the variable x. In particular, the restrictions occur wherever thedenominator is zero.Example 3: For the rational expression 4x − 7

6x − 16 , the values of x which make the denominator zerocan be found by solving the equation

6x − 16 06x 16

x 166 8

3Therefore, the variable x cannot equal 8

3 in the rational expression 4x − 76x − 16 . We also say that the

domain of this expression is all x not equal to 83 .

Example 4: For the rational expression 2x − 9 , the values of x which make the denominator zero

can be found by solving the equationx − 9 0

x 9Therefore, the variable x cannot equal 9 in the rational expression 2

x − 9 . The domain of thisexpression is all x not equal to 9.

Example 5: For the rational expression x 5x2 25

, there are no values of x which make thedenominator zero since x2 25 is always positive. Therefore, there are no restrictions on the variable x.The domain of this expression is all x.

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Example 6: For the rational expression x2 − 9x2 − 1

, the values of x which make the denominator zerocan be found by solving the equation

x2 − 1 0x2 1x 1

Therefore, the variable x cannot equal 1 in the rational expression x2 − 9x2 − 1

. The domain of thisexpression is all x not equal to 1.

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Simplifying Rational ExpressionsTo simplify a rational expression means to reduce it to lowest terms. From working with fractions, youmay recall that simplifying is done by cancelling common factors. Therefore, the key to simplifyingrational expressions (and to most problems involving rational expressions) is to factor the polynomialswhenever possible.

Example 1: Write 8x4

12x6 in reduced form.Solution: Factor the numbers and cancel common factors. Use properties of exponents to helpcancel the x’s:

8x4

12x6 4 2 x4

4 3 x4 x2

23x2

Example 2: Write y3 5y2 6yy2 − 4

in reduced form.

Solution:y3 5y2 6y

y2 − 4

yy2 5y 6y2 − 4

yy 2y 3y 2y − 2

yy 3

y − 2 y2 3yy − 2 ; y ≠ 2

Question: Why is y ≠ 2?Answer: In order for our answer to be equivalent to the original fraction, the variable must havethe same restrictions. Since y cannot equal 2 in the original expression (the denominator would thenbe zero), we must restrict the domain of our answer in order for these fractions to be equivalent.

Example 3: Write 4x − x3

x2 − x − 2in reduced form.

Solution:4x − x3

x2 − x − 2

x4 − x2x2 − x − 2

x2 x2 − xx − 2x 1

At this point, it doesn’t look like anything will cancel. However, if we factor −1 from the last term inthe numerator, we obtain the following:

−x2 xx − 2x − 2x 1

−xx 2

x 1 −x2 − 2x

x 1 ; x ≠ 2 or − 1

Question: What property of real numbers tells us that −x2 x −xx 2?Answer: The commutative property of addition.

x 2 2 x .

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Operations with Rational ExpressionsThe operations of addition, subtraction, multiplication, and division with rational expressions followthe same rules that are used with common fractions. The key, as with simplifying rational expressions,is to factor the polynomials.

I. Multiplication of Rational ExpressionsRecall with fractions a

b and cd that a

b cd ac

bd . Also recall when mulitplying fractions that youmay cross-cancel; that is, reduce common factors of any numerator with any denominator.Example 1:

29 3

10 2

3 3 32 5

13 1

5 1

15

The same is true of multiplying rational expressions. As before, the key is to factor the polynomialsfirst.Example 2: Simplify 5

x − 1 x2 − 125x − 50

Solution: First factor the polynomials. Don’t forget to cancel common factors!5

x − 1 x2 − 125x − 50 5

x − 1 x 1x − 1

25x − 2

11 x 1

5x − 2

x 15x − 10 ; x ≠ 1 or 2

Before looking at the next example, recall our strategies for factoring:1. Factor out any common factors2. If more than 3 terms, try factoring by grouping3. Recognize special products4. Factor trinomials using product/sum strategies:

a) x2 bx c: factors into x mx n wheremn c and m n b

b) ax2 bx c: split bx into mx nx, wheremn ac and m n b, then factor by grouping.

Example 3: Simplify 2x2 x − 6x2 4x − 5

x3 − 3x2 2x4x2 − 6x

x 5x3 − 8

Solution:2x2 x − 6x2 4x − 5

x3 − 3x2 2x4x2 − 6x

x 5x3 − 8

2x − 3x 2x 5x − 1

xx − 2x − 12x2x − 3 x 5

x − 2x2 2x 4

x 22x2 2x 4

x 22x2 4x 8

; x ≠ −5,1,0, 32 ,2

Example 4: Simplify 2x − 4x2 − 4

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Solution: 2x − 4x2 − 4

2x − 2

x − 2x 2 2x 2 , x ≠ 2.

II. Division of Rational ExpressionsRecall with fractions a

b and cd that a

b cd a/b

c/d ab d

c . In other words, dividing by a fraction

is the same as multiplying by its reciprocal. The same is true for rational expressions. As before, thekey to making the work easier is to factor the polynomials first.Example 5: x2 2x − 15

x 2 x2 7x 10

Solution: Begin by writing the second term as x2 7x 101 :

x2 2x − 15x 2 x2 7x 10

1 x2 2x − 15x 2 1

x2 7x 10

x 5x − 3x 2 1

x 5x 2

x − 3x 22 x − 3

x2 4x 4; x ≠ −5, − 2

Notice that the x 2 terms cannot be cancelled since both are in the denominator.

Multiplication and Division can also appear together. Only take the reciprocal of the fractions you aredividing.

Example 6: y2 − 7y 12y2 3y − 18

y2 3y − 28y2 12y 36

3y 214y 24

Solution: We only take the reciprocal of the second fraction:y2 − 7y 12y2 3y − 18

y2 3y − 28y2 12y 36

3y 214y 24

y2 − 7y 12y2 3y − 18

y2 12y 36y2 3y − 28

3y 214y 24

y − 3y − 4y 6y − 3

y 62

y 7y − 4 3y 74y 6

3; y ≠ −7, − 6, 3, 4

Question: Explain why the values −7, −6, 3, and 4 are excluded in the previous example.Answer: The easiest way to understand why these values have been excluded is to write

y2 − 7y 12y2 3y − 18

y2 3y − 28y2 12y 36

3y 214y 24 as a fraction.

y2 − 7y 12y2 3y − 18y2 3y − 28y2 12y 36

3y 214y 24

y ≠ −6 follows since 4y 24 cannot equal 0.We also get y ≠ 3 because the term

y2 − 7y 12y2 3y − 18

y2 − 7y 12y − 6y 3

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The next observation is that the denominator ofy2 − 7y 12y2 3y − 18y2 3y − 28y2 12y 36

that is, y2 3y − 28y2 12y 36

cannot equal zero. Factoring the numerator and denominator of this rational

expression we havey2 3y − 28 y 7y − 4

The numerator is zero if y −7 or if y 4. This accounts for excluding the numbers −7 and 4. Thedenominator of

y2 3y − 28y2 12y 36

factors into y 62 which means we have to exclude −6, but we have already done that.

Question: What values of x are not allowed in the rational expression x − 1x − 6 x − 3

x − 4 ?

Answer: 3, 4, 6 (The value 3 is not allowed because x − 3x − 4 is zero at x 3 and we cannot

divide by 0. x 4 is not allowed because we have the term x − 4 in the denominator. Similarly x 6 isnot allowed because the term x − 6 is the denominator of x − 1

x − 6 .)

III. Addition and Subtraction of Rational ExpressionsRecall that addition and subtraction is done by first finding a common denominator. The least commondenominator (LCD) of several fractions is the product of all prime factors of the denominators. Afactor only occurs more than once in the LCD if it occurs more than once in any one fraction. Onceyou have the LCD, convert each fraction to an equivalent fraction with the LCD, then add or subtractthe numerators.Example 7: 1

x −1

x − 1Solution: The LCD here is xx − 1. We convert each fraction to an equivalent one with thisdenominator: That is, 1

x x − 1xx − 1 and 1

x − 1 xxx − 1

1x −

1x − 1

1x − 1xx − 1 −

1xx − 1x

x − 1xx − 1 −

xxx − 1

x − 1 − xxx − 1 −1

xx − 1

Example 8: y − 4y 6 −

y2 3y − 28y2 12y 36

Solution: We must factor the denominators first to find the LCD:

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y − 4y 6 −

y2 3y − 28y 62

The LCD is y 62. We now convert each fraction to an equivalent fraction using this denominator:y − 4y 6 y − 4y 6

y 62

y2 3y − 28y 62 y2 3y − 28

y 62

Next subtract the two expressionsy − 4y 6 −

y2 3y − 28y2 12y 36

y − 4y 6y 62 − y2 3y − 28

y 62

y2 2y − 24y 62 − y2 3y − 28

y 62

y2 2y − 24 − y2 3y − 28

y 62

y2 2y − 24 − y2 − 3y 28y 62

−y 4y 62

Note that the minus sign in the numerator was distributed across the expression−y2 3y − 28 −y2 − 3y 28.

Question: What is the common denominator in x − 5x − 7x 1 −

1x2 − 4

?

Answer: x − 7x 1x2 − 4

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Compound FractionsA compound fraction (also called a complex fraction) is an expression which contains nested fractions.In general, there is one ”main fraction” which will have one or more fractions in the numerator and/ordenominator. There are two strategies which may be used to simplify compound fractions. First,simplify both the numerator and the denominator individually, then divide the numerator by thedenominator by multiplying with the reciprocal of the denominator.

Example 1:6y −

52y 1

6y 4

Solution: First simplify the numerator using the common denominator y2y 1:Example

6y −

52y 1

6y 4

62y 1y2y 1 −

5y2y 1y

6y 4

12y 6 − 5yy2y 1

6y 4

7y 6y2y 1

6y 4

Now simplify the denominator using the common denominator y (remember that 4 41 :

6y −

52y 1

6y 4

7y 6y2y 16y

4y1y

7y 6y2y 1

6 4yy

We now divide the two fractions 7y 6y2y 1 and 6 4y

y :

6y −

52y 1

6y 4

7y 6y2y 1

6 4yy

7y 6y2y 1

y6 4y

7y 62y 16 4y ; y ≠ −32 , −12 , 0 .

As you can see, this is often a long, tedious process. A second technique is to multiply thenumerator and the denominator of the ”main fraction” by the LCD of all the smaller fractions andsimplify the result.

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Example 2:6y −

52y 1

6y 4

Solution: The LCD of all the smaller fractions is y2y 1. We multiply the numerator anddenominator by this LCD:

6y −

52y 1

6y 4

6y −

52y 1 y2y 1

6y 4 y2y 1

6y y2y 1 − 5

2y 1 y2y 1

6y y2y 1 4y2y 1

62y 1 − 5y

62y 1 4y2y 1

12y 6 − 5y2y 16 4y

7y 62y 16 4y

The following example is a common one in many calculus classes:

Example 3: Simplify the expression 2x h 1−1 − 2x 1−1

h .

Solution: First rewrite the expression without negative exponents. (Recall x−1 1x )

2x h 1−1 − 2x 1−1

h 2

x h 1 −2

x 1h

Method I Solution: Now simplify the numerator using the common denominator x h 1x 1:

2x h 1−1 − 2x 1−1

h

2x 1x h 1x 1 −

2x h 1x 1x h 1

h

2x 2 − 2x 2h 2x h 1x 1

h

2x 2 − 2x − 2h − 2x h 1x 1

h

−2hx h 1x 1

hThe denominator is h h

1 which is already simplified. We now divide the two fractions−2h

x h 1x 1 and h1 :

2x h 1−1 − 2x 1−1

h −2hx h 1x 1

1h

−2x h 1x 1 ; h ≠ 0

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Method II Solution: The LCD of all the smaller fractions is x h 1x 1. We mulitply thenumerator and denominator by this LCD:

2x h 1 −

2x 1 x h 1x 1

hx h 1x 1

2

x h 1 x h 1x 1 − 2x 1 x h 1x 1

hx h 1x 1

2x 1 − 2x h 1

hx h 1x 1

2x 2 − 2x − 2h − 2hx h 1x 1

−2hhx h 1x 1 −2

x h 1x 1 ; h ≠ 0

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Exercises for Chapter 1D - Rational Expressions

For problems 1-5, determine if the given expression is a rational expression. If it is arational expression, state any restrictions on the variable.

1. 3x2 − 6x 72. 5x − 10

5x 10

3. x 8x2 4x 6

4. 5x 74

5. 2x − 4

For problems 6-8, reduce the rational expression if possible.

6. x3yx2y2

7. x2 4x − 125x − 10

8. x3 − 4xx2 x − 2

For problems 9-20, perform the indicated operations and simplify.9. x3 − 8

x2 − 4 x3 8

x2 2x 410. z2 − 5z − 24

z2 − 9 9z2 − 12z 4

z2 − 10z 16 z2 − z − 6

9z − 1211. 3

x − 1 6x 9x2 − 9x 8

12. y2 8y − 20y2 11y 10

y3 1y3 − 8

y2 − y 1y2 2y 4

13. y3 − 272y3 4y

y2 − 5y 6 2y − 6

y2 − 2y14. x

x − 3 23x 4

15. 1x − 2 −

3x 3

16. 1x 3 1

x − 3 −10

x2 − 917. 7

2x 1 −8x

2x − 1 4

18. 3x − 1 −

2x x 3

x2 − 119. y2 − 7y 12

y2 3y − 18 y2 − 25

y2 − 8y 15(HINT: Simplify first)

20. y − 7y2 2y 1

− 5y 1

For problems 21-25, simplify the expression completely.

21.3x y

4x y

2

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Answers to Exercises for Chapter 1D - RationalExpressions

1. This is a rational function and there are no restrictions on the variable x.2. Is a rational function

Restriction:5x 10 ≠ 0

5x ≠ −10x ≠ −2

3. Not a rational function because of the radical in the numerator.4. This is a rational function and there are no restrictions on the variable x.5. Is a rational function.

Restriction:x − 4 ≠ 0

x ≠ 4

6.x3yx2y2 x2 x y

x2 y y x

y ; x ≠ 0 and y ≠ 0

7.x2 4x − 12

5x − 10 x 6x − 25x − 2 x 6

5 ; x ≠ 2

8.x3 − 4x

x2 x − 2

xx 2x − 2x 2x − 1

xx − 2x − 1 ; x ≠ −2 and x ≠ 1

9.x3 − 8x2 − 4

x3 8x2 2x 4

x − 2x2 2x 4x 2x − 2

x 2x2 − 2x 4x2 2x 4

x2 − 2x 4; x ≠ 2,−2

10.z2 − 5z − 24

z2 − 9 9z2 − 12z 4

z2 − 10z 16 z2 − z − 6

9z − 12

z − 8z 3z 3z − 3

3z − 22

z − 2z − 8 z − 3z 2

33z − 4

13 3z − 22 z 2

z − 23z − 4 z ≠ −3,3,8,−2, 43

11.3

x − 1 6x 9x2 − 9x 8

3x − 1 x2 − 9x 8

6x 9 3x − 1

x − 8x − 132x 3 x − 8

2x 3 ; x ≠ 1,8

12.

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y2 8y − 20y2 11y 10

y3 1y3 − 8

y2 − y 1y2 2y 4

y 10y − 2y 10y 1

y 1y2 − y 1y − 2y2 2y 4

y2 2y 4y2 − y 1

1; y ≠ −10,−1,2

13.y3 − 27

2y3 4yy2 − 5y 6

2y − 6y2 − 2y

y − 3y2 3y 9

2y3 4yy − 2y − 3

yy − 22y − 3

y2 3y 9yy − 3 ; y ≠ 2

14. LCD: x − 33x 4x

x − 3 23x 4

x3x 4x − 33x 4

2x − 33x 4x − 3

3x2 4x 2x − 6x − 33x 4 3x2 6x − 6

x − 33x 415. LCD: x − 2x 3

1x − 2 −

3x 3

1x 3x − 2x 3 −

3x − 2x 3x − 2 x 3 − 3x − 6

x − 2x 3 −2x 9x − 2x 3

16. LCD: x 3x − 3

1x − 3

x 3x − 3 1x 3

x − 3x 3 −10

x 3x − 3

x − 3 x 3 − 10x 3x − 3

2x − 10x 3x − 3

17.7

2x 1 −8x

2x − 1 41

LCD: 2x 12x − 1

72x − 1

2x 12x − 1 −8x2x 1

2x − 12x 1 42x − 12x 12x − 12x 1

14x − 7 − 16x2 8x 16x2 − 4

2x 12x − 1

6x − 112x 12x − 1

18.3

x − 1 −2x x 3

x2 − 1 3

x − 1 −2x x 3

x 1x − 1LCD: xx 1x − 1

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3xx 1

x − 1xx 1 −2x 1x − 1xx 1x − 1 x 3x

x 1x − 1x

3x2 3x − 2x2 − 2 x2 3x

xx − 1x 1

2x2 6x 2xx − 1x 1

19.y2 − 7y 12y2 3y − 18

y2 − 25y2 − 8y 15

y − 4y − 3y 6y − 3

y 5y − 5y − 5y − 3

y − 4y 6 y 5

y − 3 ; y ≠ 5

So LCD is y 6y − 3

y − 4y − 3y 6y − 3

y 5y 6y − 3y 6

y2 − 7y 12 y2 11y 30

y 6y − 3

2y2 4y 42y 6y − 3 ; y ≠ 5

20.y − 7

y2 2y 1− 5

y 1

y − 7y 12 −

5y 1

So LCD y 12

y − 7y 12 −

5y 1y 12

y − 7 − 5y 5y 12

−4y − 12y 12

21.3x y

4 4x y

2 4

3x y2x y 3

2

22.

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5x 1 x

x − 12

x2 − 1 1

5

x 1 xx − 1 x 1x − 1

2x 1x − 1 1 x 1x − 1

5x − 1 xx 12 x 1x − 1

5x − 5 x2 x2 x2 − 1

x2 6x − 5x2 1

; x ≠ 1,−1

23.1ab

1a 1

b ab ab

b a ; a ≠ 0;b ≠ 0

24.1x −

12

x − 2

1x −

12 2x

x − 21 2x

2 − x2xx − 2

−x − 22xx − 2

−12x25.

x h−2 − x−2h

1x h2 −

1x2 x h2x2

hx h2x2

x2 − x h2

hx2x h2

x2 − x2 2xh h2hx2x h2

x2 − x2 − 2xh − h2

hx2x h2

−2xh − h2

hx2x h2

h−2x − hhx2x h2

−2x − hx2x h2 ; h ≠ 0

© : Pre-Calculus

© : Pre-Calculus - Chapter 1E

Chapter 1E - Complex Numbers

Definition of a Complex Number

Not everything can be done using real numbers. For example, there is no real number x such thatx2 −1. To handle this, mathematicians (such as William R. Hamilton 1805-1865) expanded the realnumber system by introducing the imaginary unit i −1 . Putting real and imaginary numberstogether creates the Complex Number System.

Definition A complex number is a number that can be written in the form a bi,where a and b are real numbers. a is referred to as the real part and b is referred to asthe imaginary part. This is called the standard form of the complex number.

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts areequal.

Example 1: Write the number 24 − −64 in its standard form a bi.Solution: We begin by simplifying the radicals, noting that −64 64 −1 :

24 − −64

4 6 − 64 −1

2 6 − 8 −1

2 6 − 8iThe real part is 2 6 and the imaginary part is −8.

Example 2: Find real numbers a and b such that 2a − 1 − 4i −7 b3 i

Solution: Find a by setting the real parts equal:2a − 1 −7

2a −6a −3

Then find b by setting the imaginary parts equal:− 4 b

3− 12 b

We now define the absolute value of a complex number.If z a bi is a complex number in standard form, the absolute value of z, |z|, is defined as

|z| |a bi| a2 b2 .If z1 and z2 are two complex numbers, then the distance between these two numbers isdefined to be (We’ll see how to subtract complex numbers soon.)

|z1 − z2 |If z is a real number, then the absolute value of z is the same as its real number absolute value.

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Example 3: Compute the absolute values of the following complex numbers. |2 − 3i| 22 −32 4 9 13 . |5| |5 0i| 52 02 52 5. |−7| |−7 0i| −72 02 49 7. |7 11i| 72 112 49 121 170 . |16 − 5i| 162 52 281 . |−5| −52 02 5.

Some properties of the absolute value of a complex number:1. For z a bi a complex number, the absolute value of z represents the distance from the

origin 0,0 to the point with coordinates a,b. If you have not learned about the Cartesiancoordinate system for the Euclidean plane yet, then ignore this for now.

2. Let z1 and z2 be any two complex numbers, then|z1z2 | |z1 ||z2 | .

3. Let z1 and z2 be any two complex numbers, thenz1z2

|z1 ||z2 | .

4. Let z1 and z2 be any two complex numbers, then|z1 z2 | ≤ |z1 | |z2 | .

This inequality is called the triangle inequality.

Example 4: Compute the absolute value of the following compex numbers.2 − 5i, 7 3i, −11 − 4i, and 3i.

Solution:|2 − 5i| 4 25 29 ≈ 5. 3852

|7 3i| 49 9 58 ≈ 7. 6158

|−11 − 4i| 121 16 137 ≈ 11. 705

|3i| 9 3

Example 5: Show that the absolute value of the product of 3 − 2i and −4 i equals the product oftheir absolute values.Solution:

|3 − 2i−4 i| |−10 11i| 100 121 221

|3 − 2i||−4 i| 9 4 16 1 13 17 13 ∘ 17 221 .

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Example 6: Show that the absolute value of the sum of −8 i and 4 3i is less than or equal to thesum of their absolute values.Solution:

|−8 i 4 3i| |−4 4i| 16 16 32

|−8 i| |4 3i| 64 1 16 9 65 25 65 5 .

We now need to show that 32 ≤ 65 5, but this is obvious since 32 65 and therefore32 65 .

We now define the conjugate of a complex number.

The conjugate of the complex number z a bi is defined to be a − bi, where we assumethat z is in standard form.

The conjugate of a complex number z is commonly denoted by writing a line above z. That is, z̄ is usedto denote the conjugate of z.

Example 7: Compute the conjugate of the following complex numbers:1 − i, 3i, −8, and −3 − 6i.

Solution: The conjugate of 1 − i 1 − i 1 i. The conjugate of 3i 3i −3i. The conjugate of −8 −8 −8. The conjugate of −3 − 6i −3 − 6i −3 6i.

Some Properties of the Conjugate of a Complex NumberIn the following z, z1, and z2 represent arbitrary complex numbers.1. z z |z|2, a bia − bi a2 b2 0i a2 b2 |a bi|2.2. If x is a real number, then x x.3. z̄̄ z (The conjugate of the conjugate of z equals z.)4. z1 z2 z1 z2 (The conjugate of a sum is the sum of the conjugates.)5. z1 z2 z1 z2 (The conjugate of a product is the product of the conjugates.)6. z1

z2 z1

z2(The conjugate of a quotient is the quotient of the conjugates.)

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Exercises for Chapter 1E - Complex Numbers

1. Write each of the following in standard form:−2 −16 , 5 − −50 , 2i2 − 4i, − 9

4 − 34

2. Find real numbers a and b such that a bi 14i − 33. Find real numbers a and b such that 2a − 5 − 6i 11 − 3bi4. What is the absolute value and conjugate of z 5?5. What is the absolute value and conjugate of z −7i?6. If the conjugate of the complex number z equals −2 7i, what does z equal?7. What is the absolute value and conjugate of z −3 − 5i?8. If the real part of z equals −5 and the absolute value of z 5, what must z equal?9. If the imaginary part of z equals −13, what does the imaginary part of z̄ equal?10. If the real part of z equals −23, what does the real part of z̄ equal?11. If the real and imaginary parts of a complex number are equal, and the real part equals −5,

what is the complex number?12. What are the real and imaginary parts of the complex number 2 − 5i?

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Answers to Exercises for Chapter 1E - Complex Numbers

1. −2 −16 −2 4i5 − −50 5 − 5 2 i2i2 − 4i −2 − 4i

− 94 − 3

4 − 34 3

2 i

2. a −3 b 143. 2a − 5 − 6i 11 − 3bi

This equation of complex numbers implies the following two real equations.2a − 5 11− 6 −3b

Thus, 2a 16 or a 8, and b −6−3 2.

4. |5| 5 5̄ 55. |−7i| 7 −7i 7i6. Since z̄ −2 7i, and z̄̄ z, we have z z̄̄ −2 7i −2 − 7i.7. |−3 − 5i| 9 25 34 −3 − 5i −3 5i8. Suppose z a bi. We know that a −5. Thus, 5 |z| −52 b2 .Squaring both

sides of this equality, we have 25 25 b2, which implies that b 0. Thus, z −5.9. 1310. −2311. −5 − 5i12. Real part of 2 − 5i is 2, and the imaginary part of 2 − 5i is −5.

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