Chapter 4 Polynomials TSWBAT determine if an expression is a monomial, binomial, or polynomial;...

ChapterChapter 4 4PolynomialsPolynomials

TSWBAT determine if an TSWBAT determine if an expression is a expression is a

monomial, binomial, or monomial, binomial, or polynomial; determine polynomial; determine

the degree of a monomial the degree of a monomial or polynomial; add and or polynomial; add and

subtract two polynomials. subtract two polynomials.

Section 1 – Polynomials Section 1 – Polynomials cont.cont.

• MonomialMonomial – a constant, a – a constant, a variable, or the product of a variable, or the product of a constant and one or more constant and one or more variables.variables.

Ex. -7, Ex. -7, uu, 1/3, 1/3mm 22, -, -s s 22t t 33, 6, 6xy xy 33

• BinomialBinomial – a polynomial that has – a polynomial that has two terms. Ex. 2x +3two terms. Ex. 2x +3

• TrinomialTrinomial – a polynomial that has – a polynomial that has three terms. Ex. Xthree terms. Ex. X22 + 4x – 5 + 4x – 5


• Degree of a variableDegree of a variable – the – the number of times the variable number of times the variable occurs as a factor in the occurs as a factor in the monomial or the exponent of monomial or the exponent of the variable.the variable.

Ex. 6Ex. 6xyxy 33 has variables x and y. has variables x and y.

X has degree 1 and y has X has degree 1 and y has degree 3.degree 3.


• Degree of a monomialDegree of a monomial – the sum – the sum of the degrees of the variables in of the degrees of the variables in the monomial. A nonzero the monomial. A nonzero constant has degree 0. The constant has degree 0. The constant 0 has no degree.constant 0 has no degree.

Ex. 6Ex. 6xy xy 33 has degree has degree 0+1+3=40+1+3=4

--s s 22t t 33 has degree 0+2+3=5 has degree 0+2+3=5

uu has degree 1 has degree 1

-7 has degree 0-7 has degree 0


• Similar (or like) MonomialsSimilar (or like) Monomials – – monomials that are identical or that monomials that are identical or that differ only in their coefficients.differ only in their coefficients.

Ex. -Ex. -s s 22t t 33 and 2 and 2s s 22t t 33 are similar are similar

66xy xy 33 and 6 and 6x x 33yy are not are not similarsimilar


• PolynomialPolynomial – a monomial or a sum – a monomial or a sum of monomials. The monomials in of monomials. The monomials in a polynomial are called the a polynomial are called the termsterms of the polynomial. of the polynomial.

Ex. xEx. x22 + (-4)x + 5 or x + (-4)x + 5 or x22 – 4x + 5 – 4x + 5

The terms are xThe terms are x22, -4x, and 5, -4x, and 5


• Simplified polynomialSimplified polynomial – a – a polynomial in which no two polynomial in which no two terms are similar. The terms are terms are similar. The terms are usually arranged in order of usually arranged in order of decreasing degree of one of the decreasing degree of one of the variables, usually x.variables, usually x.

Ex. 2xEx. 2x33 – 5 + 4x + x – 5 + 4x + x3 3

is not simplified butis not simplified but

3x3x33 + 4x – 5 is + 4x – 5 is simplified.simplified.


• Degree of a polynomialDegree of a polynomial – the – the greatest of the degrees of greatest of the degrees of its terms after it has been its terms after it has been simplified.simplified.

Ex. xEx. x44 – 2x – 2x22yy33 + 6y – 11 + 6y – 11 has terms with degree has terms with degree 4, 5, 4, 5, 1, and 0. Thus the 1, and 0. Thus the

polynomial has degree polynomial has degree 5 the 5 the largest degree of largest degree of its terms.its terms.


• Addition of PolynomialsAddition of Polynomials – to add two or – to add two or more polynomials, write their sum and more polynomials, write their sum and then simplify by combining similar then simplify by combining similar terms.terms.

Ex. Ex. Add 2xAdd 2x22 – 3x + 5 and x – 3x + 5 and x33 – 5x – 5x22 + 2x – 5 + 2x – 5

2x2x22 – 3x + 5 – 3x + 5 + x+ x33 – 5x – 5x22 + 2x – 5 + 2x – 5

xx33 – 3x – 3x22 - x - x


• Subtracting PolynomialsSubtracting Polynomials – to subtract one – to subtract one polynomial from another, add the opposite of polynomial from another, add the opposite of each term of the polynomial you are each term of the polynomial you are subtracting.subtracting.

Ex. Subtract Ex. Subtract

2x2x22 – 3x + 5 from x – 3x + 5 from x33 – 5x – 5x22 + 2x – 5 + 2x – 5

xx33 – 5x – 5x22 + 2x – 5 + 2x – 5 x x33 – 5x – 5x22 + 2x – 5 + 2x – 5

- (2x- (2x22 – 3x + 5) – 3x + 5) - 2x - 2x22 + 3x – 5 + 3x – 5

xx33 – 7x – 7x22 + 5x – 10 + 5x – 10

Multiplying PolynomialsMultiplying Polynomials

TSWBAT multiply Polynomials TSWBAT multiply Polynomials using the FOIL method, the using the FOIL method, the

Rules of Special Products, and Rules of Special Products, and the distributive Property.the distributive Property.

Section 3 – Multiplying Section 3 – Multiplying PolynomialsPolynomials

If you are multiplying two binomials If you are multiplying two binomials together it is best to use the together it is best to use the FOILFOIL method of multiplication. method of multiplication.

• FOIL MethodFOIL Method – –

FF - Multiply the two first terms - Multiply the two first terms

OO – Multiply the two outer terms – Multiply the two outer terms

II – Multiply the two inner terms – Multiply the two inner terms

LL – Multiply the two last terms – Multiply the two last terms

Section 3 – Multiplying Section 3 – Multiplying Polynomials cont.Polynomials cont.

Example of FOILExample of FOIL

(2a – b)(3a + 5b)(2a – b)(3a + 5b)

F – (2a)(3a) = 6aF – (2a)(3a) = 6a22

O – (2a)(5b) = 10abO – (2a)(5b) = 10ab

I – (-b)(3a) = -3abI – (-b)(3a) = -3ab

L – (-b)(5b) = -5bL – (-b)(5b) = -5b22

Combining the Combining the four terms into four terms into one polynomial one polynomial and combining and combining the like terms of the like terms of O and I we get O and I we get the following as the following as a result:a result:6a6a22 + 7ab – 5b + 7ab – 5b22


• Besides the FOIL Method Besides the FOIL Method there are also three special there are also three special products to know, that will products to know, that will help you multiply polynomials help you multiply polynomials and their terms faster.and their terms faster.

Special ProductsSpecial Products

(a+b)(a+b)22 = a = a22 + 2ab + b + 2ab + b22

(a-b)(a-b)22 = a = a22 -2ab + b -2ab + b22

(a+b)(a-b) = a(a+b)(a-b) = a22 + b + b22


• Besides these methods of FOIL and Besides these methods of FOIL and Special Products there is also the Special Products there is also the old fashioned Distribution Method.old fashioned Distribution Method.Ex. (2x+3)(xEx. (2x+3)(x22+4x-5)+4x-5)2x2x(x(x22+4x-5)+4x-5)++33(x(x22+4x-5)+4x-5)2x(x2x(x22)+2x(4x)-2x(5)+3(x)+2x(4x)-2x(5)+3(x22)+3(4x)-)+3(4x)-3(5)3(5)2x2x33+8x+8x22-10x+3x-10x+3x22+12x-15+12x-15

Combine Similar Terms Combine Similar Terms 2x2x33+11x+11x22+2x-15+2x-15

ExponentsExponents

TSWBAT Use the laws of TSWBAT Use the laws of exponents in polynomial exponents in polynomial

multiplication and multiplication and simplification.simplification.

Section 2 – Laws of Section 2 – Laws of ExponentsExponents

• There are There are five laws of five laws of exponentsexponents – –

Let a and b be real Let a and b be real numbers and m and n be numbers and m and n be positive integers. Then:positive integers. Then:

1. a1. amm * a * ann = a = am+nm+n

2. (ab)2. (ab)mm = a= ammbbmm

3. (a3. (amm))nn = a= am*nm*n

4. (a/b)4. (a/b)nn = a= ann/b/bnn

5. a5. amm/a/ann = a= am-nm-n

Section 2 – Laws of Section 2 – Laws of ExponentsExponents

• ExamplesExamples

FactoringFactoring

TSWBAT find the prime TSWBAT find the prime factorization of a number and factorization of a number and the greatest common factor or the greatest common factor or least common multiple of two least common multiple of two

or more numbers or or more numbers or monomials.monomials.

Section 4 – Prime Section 4 – Prime FactorizationFactorization

• Factor SetFactor Set – The set of all numbers – The set of all numbers that are a factor of a given number.that are a factor of a given number.

• FactorFactor – One of two numbers that – One of two numbers that multiplied together form a given multiplied together form a given number.number.

• Prime Number or PrimePrime Number or Prime – a number – a number whose only factors are 1 and itself.whose only factors are 1 and itself.

• Composite NumberComposite Number – any number – any number that has more than one set of that has more than one set of factors.factors.

Activity – Primes Chart

Section 4 – Prime Section 4 – Prime FactorizationFactorization• Greatest Common Greatest Common

Factor (GCF) – of two Factor (GCF) – of two or more integers is or more integers is the greatest integer the greatest integer that is a factor of that is a factor of each.each.

Ex. GCF of 15 Ex. GCF of 15 and 30and 30

1. Find all the 1. Find all the factors of each factors of each number.number.

2. Find the 2. Find the greatest number in greatest number in both sets. This is both sets. This is your GCF.your GCF.

FactorFactors of 15s of 15

Factors Factors of 30of 30

11 1515 11 3030

22 1515

33 55 33 1010

55 66

Section 4 – Prime Section 4 – Prime FactorizationFactorization• Least Common Least Common

Multiple (LCM) – Multiple (LCM) – of of two or more two or more integers is the least integers is the least positive integer positive integer having each as a having each as a factor.factor.

Ex. Find the Ex. Find the LCM of 12 and 30. LCM of 12 and 30.

1212 2424 3636 4848 6060

3030 6060 9090 121200

151500

1. Find the multiples of the two numbers.

2. Find the least common number among both numbers. That is your LCM.


• The GCF and LCM also hold true for The GCF and LCM also hold true for Monomials with a slight variation as Monomials with a slight variation as you now need to be aware of the you now need to be aware of the variables.variables.

• GCF of Monomials – the common GCF of Monomials – the common factor with the greatest degree and factor with the greatest degree and greatest numerical coefficient.greatest numerical coefficient.

• LCM of Monomials – the common LCM of Monomials – the common multiple that has the least degree and multiple that has the least degree and least positive numerical coefficient.least positive numerical coefficient.


Ex. of GCF of Monomials 48uEx. of GCF of Monomials 48u22vv22 and 60uv and 60uv33w.w.

1.1. Find GCF of CoefficientsFind GCF of Coefficients

48 – 1,48, 2,24, 3,16, 4,48 – 1,48, 2,24, 3,16, 4,1212, 6,8, 6,8

60 – 1,60, 2,30, 3,20, 4,15, 5,60 – 1,60, 2,30, 3,20, 4,15, 5,1212, 6,10, 6,10

2. Compare the variables. Only variables in 2. Compare the variables. Only variables in both can be used in GCF. Then use the both can be used in GCF. Then use the smaller exponent variable.smaller exponent variable.

uu22 or u – use u v or u – use u v22 or v or v33 – use v – use v22

3. Therefore our GCF is 12uv3. Therefore our GCF is 12uv22


Ex. of LCM of Monomials 48uEx. of LCM of Monomials 48u22vv22 and and 60uv60uv33w w

1.1. Find LCM of coefficients.Find LCM of coefficients.48,96,144,192,48,96,144,192,240240,288,28860,120,180,60,120,180,240240,300,360,300,3602. Compare the variables. All variables 2. Compare the variables. All variables

in both must be used in LCM. Then in both must be used in LCM. Then use the greater exponent variable.use the greater exponent variable.

uu22 or u – use u or u – use u22 v v22 or v or v33 – use v – use v33 use w use w3. Therefore the LCM is 240u3. Therefore the LCM is 240u22vv33ww


• Prime FactorizationPrime Factorization – is the – is the factorization of a factorization of a number down to number down to the product of a set the product of a set of prime numbers. of prime numbers. The most common The most common way to find the way to find the prime factorization prime factorization is to use a factor is to use a factor tree working down tree working down to all prime factors.to all prime factors.

• Ex. of Prime Ex. of Prime FactorizationFactorization936936=2*468=2*468

2*2*2342*2*234 2*2*2*1172*2*2*117 2*2*2*3*392*2*2*3*39

2*2*2*3*3*132*2*2*3*3*13Therefore Therefore

the Prime the Prime Factorization isFactorization is

2233*3*322*13*13

Factoring Quadratic Factoring Quadratic PolynomialsPolynomials

• Quadratic PolynomialsQuadratic Polynomials – or second-degree – or second-degree polynomials are polynomials of the form polynomials are polynomials of the form axax22+bx+c where (a≠0).+bx+c where (a≠0).

Ex. XEx. X22+0x+25+0x+25

• Quadratic Polynomials have three terms Quadratic Polynomials have three terms each with its own name.each with its own name.– Quadratic TermQuadratic Term – is the ax– is the ax22 term term – Linear TermLinear Term – is the bx term– is the bx term– Constant TermConstant Term – is the c term– is the c term

Factoring Quadratic Factoring Quadratic PolynomialsPolynomials

• Quadratic TrinomialQuadratic Trinomial – a quadratic – a quadratic polynomial in which a, b, and c are polynomial in which a, b, and c are all nonzero integers.all nonzero integers.

Ex. xEx. x22+2x-15+2x-15

• In this section we will factor a In this section we will factor a special type of quadratic polynomial special type of quadratic polynomial the perfect square trinomial. In the the perfect square trinomial. In the next section we will look at how to next section we will look at how to factor any Quadratic Polynomial.factor any Quadratic Polynomial.

Section 5 – Factoring Section 5 – Factoring PolynomialsPolynomials

• 5 Special Factorings5 Special Factorings – In section 3 we – In section 3 we had 3 special products. Those had 3 special products. Those products play a role here in factoring products play a role here in factoring as well. Besides those three there are as well. Besides those three there are also two other special cases in also two other special cases in factoring. On the following slide you factoring. On the following slide you will find these 5 cases.will find these 5 cases. (Hint – Take an (Hint – Take an index card and write these 5 cases on index card and write these 5 cases on it to help you when doing Problems!!!)it to help you when doing Problems!!!)


Special FactoringsSpecial Factorings

Perfect Square TrinomialsPerfect Square Trinomials

Positive aPositive a22 + 2ab + b + 2ab + b22 = (a + b) = (a + b)22

Negative aNegative a22 - 2ab + b - 2ab + b22 = (a - b) = (a - b)22

Difference of SquaresDifference of Squares

aa22 – b – b22 = (a + b)(a – b) = (a + b)(a – b)

Sum and Difference of CubesSum and Difference of Cubes

Sum aSum a33 + b + b33 = (a + b)(a = (a + b)(a22 – ab + b – ab + b22))

Difference aDifference a33 - b - b33 = (a - b)(a = (a - b)(a22 + ab + b + ab + b22))

Section 5 – Factoring Section 5 – Factoring PolynomialsPolynomials• Example of a Perfect Square TrinomialExample of a Perfect Square Trinomial

PositivePositive –> z–> z22+6z+9= (z+3)+6z+9= (z+3)22

NegativeNegative -> 4s-> 4s22-4st+t-4st+t22 = (2s-t) = (2s-t)22

• Example of a Difference of SquaresExample of a Difference of Squares

25x25x22-16a-16a22 = (5x+4a)(5x-4a) = (5x+4a)(5x-4a)

• Example of CubesExample of Cubes

SumSum ->8u->8u33+v+v33 = (2u+v)(4u = (2u+v)(4u22--2uv+v2uv+v22))

DifferenceDifference -> y-> y33-1 = (y-1)(y-1 = (y-1)(y22+y+1)+y+1)


• You can also factor a polynomial by You can also factor a polynomial by factoring out the Greatest Common factoring out the Greatest Common Monomial Factor. Monomial Factor.

Ex. 2xEx. 2x44-4x-4x33+8x+8x22

Step 1 – Find the GCF, which here it is Step 1 – Find the GCF, which here it is 2x2x22

Step 2 – Divide out 2xStep 2 – Divide out 2x22 from each term. from each term.

Step 3 – Write out your answer. Step 3 – Write out your answer.

2x2x22(x(x22-2x+4)-2x+4)

Common Numerical Powers Activity


• One last way to factor a polynomial is to One last way to factor a polynomial is to rearrange and group terms that you might rearrange and group terms that you might be able to find a GCF from.be able to find a GCF from.Ex. 3xy-4-6x+2yEx. 3xy-4-6x+2y

If we group the 1If we group the 1stst and 3 and 3rdrd terms terms together they have a common factor of 3x together they have a common factor of 3x and then grouping the 2and then grouping the 2ndnd and 4 and 4thth terms terms they have a common factor of 2.they have a common factor of 2.(3xy-6x)+(2y-4) -> 3x(y-2)+2(y-2)(3xy-6x)+(2y-4) -> 3x(y-2)+2(y-2)Now we have a common factor of (y-2) Now we have a common factor of (y-2) that we can pull out and we are left with that we can pull out and we are left with (y-2)(3x+2).(y-2)(3x+2).

Section 6 – Factoring Quadratic Section 6 – Factoring Quadratic PolynomialsPolynomials

• Factoring a Quadratic Polynomial – Factoring a Quadratic Polynomial –

axax22+bx+c can be factored into +bx+c can be factored into the product (px+q)(rx+s) where p, the product (px+q)(rx+s) where p, q, r, and s are integers.q, r, and s are integers.

Then axThen ax22+bx+c = prx+bx+c = prx22++(ps+qr)x+qs.(ps+qr)x+qs.

So a = pr, b=(ps+qr), and c = qsSo a = pr, b=(ps+qr), and c = qs


• Lets look at an example of how to factor Lets look at an example of how to factor a Quadratic Polynomial.a Quadratic Polynomial.

15t15t22-16t+4 -> 1. Lets begin by looking at -16t+4 -> 1. Lets begin by looking at the pr term which is 15 here. There are the pr term which is 15 here. There are two ways in which to factor 15ttwo ways in which to factor 15t22. They . They are (t___)(15t____) or (3t____)(5t____).are (t___)(15t____) or (3t____)(5t____).


• 2. We need to look at the qs term of 4. 2. We need to look at the qs term of 4. Since 4 is positive, both factors are Since 4 is positive, both factors are either positive or negative. To either positive or negative. To determine this we look at the middle determine this we look at the middle term or -16t. Since this is negative both term or -16t. Since this is negative both of our factors of 4 will be negative. of our factors of 4 will be negative. Thus there are 3 negative combinations Thus there are 3 negative combinations that factor 4. They are (___-1)(_-4), (_-that factor 4. They are (___-1)(_-4), (_-2)(_-2), or (_-4)(__-1).2)(_-2), or (_-4)(__-1).


• If we take our two If we take our two possible answers possible answers for the first terms for the first terms and our three for and our three for our second term, our second term, we find we have 6 we find we have 6 possible factoring possible factoring possibilities. possibilities. These are shown These are shown in the table.in the table.

(t-1)(15t-(t-1)(15t-4)4)

(3t-1)(5t-(3t-1)(5t-4)4)

(t-2)(15t-(t-2)(15t-2)2)

(3t-2)(5t-(3t-2)(5t-2)2)

(t-4)(15t-(t-4)(15t-1)1)

(3t-4)(5t-(3t-4)(5t-1)1)


• To find out which To find out which of these of these possibilities possibilities work, we check work, we check each of them by each of them by doing the O and I doing the O and I part of FOIL on part of FOIL on each pair to find each pair to find the grouping that the grouping that gives us -16t.gives us -16t.

(t-1)(15t-(t-1)(15t-4)4)

-15t-4t=--15t-4t=-19t19t

(t-2)(15t-(t-2)(15t-2)2)

-30t-2t=--30t-2t=-32t32t

(t-4)(15t-(t-4)(15t-1)1)

-60t-t=-61t-60t-t=-61t

(3t-1)(5t-(3t-1)(5t-4)4)

-5t-12t=--5t-12t=-17t17t

(3t-2)(5t-(3t-2)(5t-2)2)

-10t-6t=--10t-6t=-16t16t

(3t-4)(5t-(3t-4)(5t-1)1)

-20t-3t=--20t-3t=-23t23t


• Therefore our quadratic polynomial Therefore our quadratic polynomial 15t15t22-16t+4 factors down into (3t-2)(5t-2).-16t+4 factors down into (3t-2)(5t-2).

• This polynomial is now This polynomial is now factored factored completelycompletely as it is written as the product as it is written as the product of factors that are prime polynomials. It of factors that are prime polynomials. It could also be factored completely if we could also be factored completely if we had monomials 2x(3y), or a power of a had monomials 2x(3y), or a power of a prime polynomial like (x-2)prime polynomial like (x-2)22 . .


• Prime PolynomialPrime Polynomial – a polynomial that is not – a polynomial that is not reducible and the greatest common factor of reducible and the greatest common factor of its coefficients is 1.its coefficients is 1.

Ex. XEx. X22+4x-3 is a Prime Polynomial, since it can +4x-3 is a Prime Polynomial, since it can not be factored into two lower degree not be factored into two lower degree polynomials with integers and the GCF of its polynomials with integers and the GCF of its coefficients is 1.coefficients is 1.

2x2x22+8x-6 is not Prime because the GCF of its +8x-6 is not Prime because the GCF of its coefficients is 2. Thus a 2 can be factored coefficients is 2. Thus a 2 can be factored out.out.


• Irreducible polynomialIrreducible polynomial – a polynomial that – a polynomial that contains more than one term and cannot contains more than one term and cannot be expressed as a product of lower degree be expressed as a product of lower degree polynomials taken from its factor set.polynomials taken from its factor set.

Ex. XEx. X22+4x-3+4x-3Why? Factor sets are (x-1)(x+3) and (x+1)Why? Factor sets are (x-1)(x+3) and (x+1)(x-3) and neither of these sets gives a (x-3) and neither of these sets gives a middle term of +4. (They give +2x and -middle term of +4. (They give +2x and -2x.)2x.)

Section 7 – Solving Polynomial Section 7 – Solving Polynomial EquationsEquations

• Polynomial Equation –Polynomial Equation – an equation that an equation that is equivalent to an equation with a is equivalent to an equation with a polynomial as one side and 0 as the other.polynomial as one side and 0 as the other.

Ex. xEx. x22-5x-24=0 and x-5x-24=0 and x22=5x+24=5x+24• Root –Root – is a solution to the polynomial is a solution to the polynomial

equation.equation.• Solution –Solution – a value of the variable that a value of the variable that

satisfies the equation.satisfies the equation.Ex. Roots and solutions for the above Ex. Roots and solutions for the above

problem are -3 and 8.problem are -3 and 8.


• To solve a polynomial equation we use To solve a polynomial equation we use the the Zero-Product PropertyZero-Product Property..

• Step 1 –Step 1 – Write the equation in Zero Write the equation in Zero Equals Form. Equals Form.

Ex. If we take the problem Ex. If we take the problem xx22=5x+24, we need to rewrite it as x=5x+24, we need to rewrite it as x22--5x-24=05x-24=0


• Step 2 –Step 2 – Factor the polynomial side of the Factor the polynomial side of the equation.equation.

Ex. From our equation xEx. From our equation x22-5x-24=0 we -5x-24=0 we factor the left side into (x-8)(x+3)=0factor the left side into (x-8)(x+3)=0

• Step 3 –Step 3 – Solve the equation by setting Solve the equation by setting each factor equal to zero.each factor equal to zero.

Ex. Taking our factor of (x-8)(x+3) we Ex. Taking our factor of (x-8)(x+3) we set each equal to zero and solve for x.set each equal to zero and solve for x.

x-8=0 and x+3=0 -> x=8 and x=-3x-8=0 and x+3=0 -> x=8 and x=-3


• Double Root or Double Zero –Double Root or Double Zero – of an of an equation is a solution of the equation equation is a solution of the equation that appears twice.that appears twice.

Ex. If we have (x-2)(x-2) as factors Ex. If we have (x-2)(x-2) as factors we have the solution of 2 twice. This we have the solution of 2 twice. This means that 2 is a double root of that means that 2 is a double root of that equation and can be written x=2d.r.equation and can be written x=2d.r.


• Multiple Root or Multiple Zero -Multiple Root or Multiple Zero - of of an equation is a solution of the an equation is a solution of the equation that appears more than equation that appears more than twice.twice.

Ex. If we have (t-1)(t-1)(t-1)(t-1) Ex. If we have (t-1)(t-1)(t-1)(t-1) as factors we have the solution of 1 as factors we have the solution of 1 four times. This means that 1 is a four times. This means that 1 is a multiple root of that equation and can multiple root of that equation and can be written t=1m.r.4.be written t=1m.r.4.


• Using the Zero-Product Property we Using the Zero-Product Property we can also find the zeros of a can also find the zeros of a polynomial function. That means polynomial function. That means find the values for when f(x)=0find the values for when f(x)=0

Ex. f(x)=xEx. f(x)=x22-9 following the steps -9 following the steps we factor the polynomial side to (x-3)we factor the polynomial side to (x-3)(x+3) (special products) and then (x+3) (special products) and then solve to x=3 and x=-3.solve to x=3 and x=-3.

Section 8 – Problem Solving Section 8 – Problem Solving Using Polynomial EquationsUsing Polynomial Equations

• The seven steps of word problems are:The seven steps of word problems are:1. Read and Write the problem1. Read and Write the problem2. Draw a picture2. Draw a picture3. Define the variables3. Define the variables4. Re-Read the problem and set up the 4. Re-Read the problem and set up the

basic problembasic problem5. Fill-in values from the problem5. Fill-in values from the problem6. Solve the equation6. Solve the equation7. Check your solutions (this is critical 7. Check your solutions (this is critical

now because not all values will be accurate now because not all values will be accurate solutions)solutions)


• Example – Step 1 – Read and Write the Example – Step 1 – Read and Write the problem. - A graphic artist is designing problem. - A graphic artist is designing a poster that consists of a rectangular a poster that consists of a rectangular prism with a uniform border. The print prism with a uniform border. The print is to be twice as tall as it is wide, and is to be twice as tall as it is wide, and the border is to be 3 inches wide. If the the border is to be 3 inches wide. If the area of the poster is to be 680 inches area of the poster is to be 680 inches squared, find the dimensions of the squared, find the dimensions of the print. print.


• Step 2 – Draw a picture – See overhead.Step 2 – Draw a picture – See overhead.

• Step 3 – Define the variable – w = width Step 3 – Define the variable – w = width of printof print

• Step 4 – Label drawing – See OverheadStep 4 – Label drawing – See Overhead

• Step 5 – Create equation – using area Step 5 – Create equation – using area formula – A = L*W -> 680 = (w+6)formula – A = L*W -> 680 = (w+6)(2w+6)(2w+6)


• Step 6 – Solve for variable.Step 6 – Solve for variable.680 = (w+6)(2w+6) -> 680 = 2w680 = (w+6)(2w+6) -> 680 = 2w22+18w +36+18w +36

-> 0 = 2w-> 0 = 2w22+18w-644 -> Divide by 2+18w-644 -> Divide by 2-> 0 = w-> 0 = w22 +9w -322 -> (w-14)(w+23) = +9w -322 -> (w-14)(w+23) =

00-> w-14 = 0 and w+23 = 0-> w-14 = 0 and w+23 = 0 -> w = 14 or w = -23-> w = 14 or w = -23

• Step 7 – Check solutions – We cannot have Step 7 – Check solutions – We cannot have a negative length thus x = -23 does not work a negative length thus x = -23 does not work and our solution is x = 14.and our solution is x = 14.

Section 9 – Solving Polynomial Section 9 – Solving Polynomial InequalitiesInequalities

• Polynomial Inequality –Polynomial Inequality – an an inequality that is equivalent to an inequality that is equivalent to an inequality with a polynomial as one inequality with a polynomial as one side and zero as the other side. side and zero as the other side.

Ex. xEx. x22-x-6>0 and x-x-6>0 and x22>x+6>x+6


• To solve a polynomial inequality To solve a polynomial inequality we start by following the first two we start by following the first two steps of that of an equation. steps of that of an equation. Step 1Step 1 - We make one side zero - We make one side zero and the other side the polynomial.and the other side the polynomial.

Ex. xEx. x22>x+6 becomes x>x+6 becomes x22-x--x-6>06>0


Step 2Step 2 - We factor the polynomial side of - We factor the polynomial side of the inequality.the inequality.

Ex. xEx. x22-x-6>0 becomes (x-3)-x-6>0 becomes (x-3)(x+2)>0.(x+2)>0.

Step 3Step 3 is actually a two part step. is actually a two part step.Part APart A – if the factors of the – if the factors of the

polynomial are >0 we will use the polynomial are >0 we will use the same inequality sign in Step 4.same inequality sign in Step 4.

Part BPart B – if the factors of the – if the factors of the polynomial are <0 we will use opposite polynomial are <0 we will use opposite inequality signs in Step 4.inequality signs in Step 4.


Step 3 cont.Step 3 cont. – Ex. (x-3)(x+2)>0 our factors are – Ex. (x-3)(x+2)>0 our factors are >0 and thus for Step 4 we will use the same >0 and thus for Step 4 we will use the same inequality sign.inequality sign.

Step 4Step 4 – is another two part process. If you – is another two part process. If you remember when we first talked about remember when we first talked about inequalities we had the terms inequalities we had the terms ConjunctionConjunction and and DisjunctionDisjunction, well they return here in this , well they return here in this step when we set the inequality factors to step when we set the inequality factors to zero.zero.


Step 4 cont.Step 4 cont. – In this step we have – In this step we have two conjunctions joined together by two conjunctions joined together by a disjunction. So in our example, a disjunction. So in our example,

(x-3)(x+2)>0 is possible only if both (x-3)(x+2)>0 is possible only if both factors are positive (conjunction) or factors are positive (conjunction) or (disjunction) both factors are (disjunction) both factors are negative (conjunction). negative (conjunction).


Step 4 cont.Step 4 cont. – Ex. we have the following – Ex. we have the following possibilities…possibilities…

x-3>0 and x+2>0 OR x-3<0 and x-3>0 and x+2>0 OR x-3<0 and x+2<0x+2<0

Step 5 Step 5 – solve for x…– solve for x…

x>3 and x>-2 OR x<3 and x<-2 so…x>3 and x>-2 OR x<3 and x<-2 so…

x>3 OR x<-2.x>3 OR x<-2.

Step 6Step 6 – Graph the solution. – Graph the solution.


Example 2 – 3t<4 - tExample 2 – 3t<4 - t22

Step 1 – tStep 1 – t22 + 3t – 4 <0 + 3t – 4 <0Step 2 – (t+4)(t-1)<0Step 2 – (t+4)(t-1)<0Step 3 – Use Part B as factors are <0.Step 3 – Use Part B as factors are <0.Step 4 – Step 4 – (t+4)>0 and (t-1)<0 OR (t+4)<0 and (t-(t+4)>0 and (t-1)<0 OR (t+4)<0 and (t-1)>01)>0Step 5 – t>-4 and t <1 OR t<-4 and t >1 Step 5 – t>-4 and t <1 OR t<-4 and t >1

-4<t<1 OR no solution-4<t<1 OR no solutionStep 6 graph the one remaining solution.Step 6 graph the one remaining solution.

Chapter 4 Polynomials TSWBAT determine if an expression is a monomial, binomial, or polynomial;...

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Transcript of Chapter 4 Polynomials TSWBAT determine if an expression is a monomial, binomial, or polynomial;...