SECTION P.5 Factoring Polynomials

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BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 4848 Chapter P Prerequisites: Fundamental Concepts of AlgebraTechnology Exercises98. The common cold is caused by a rhinovirus. Thepolynomial-0.75x4 3x3 5describes the billions of viral particles in our bodies after xdays of invasion. Use a calculator to find the number of viralparticles after 0 days (the time of the cold’s onset), 1 day, 2days, 3 days, and 4 days.After how many days is the numberof viral particles at a maximum and consequently the day wefeel the sickest? By when should we feel completely better?99. Using data from the National Institute on DrugAbuse, the polynomialCritical Thinking ExercisesIn Exercises 100–103, perform the indicatedoperations.100. (x - y)2 - (x y)2101. [(7x 5) 4yD C(7x 5) - 4yD102. C(3x y) 1D 2103. (x y)(x - y)Ax2 y2 B104. Express the area of the plane figure shown as apolynomial in standard form.0.0032x3 0.0235x2 - 2.2477x 61.1998approximately describes the percentage of U.S. highschool seniors in the class of x who had ever usedmarijuana, where x is the number of years after 1980. Usea calculator to find the percentage of high school seniorsfrom the class of 1980 through the class of 2000 who hadused marijuana. Round to the nearest tenth of a percent.Describe the trend in the data.xxx 1x 3SECTION P.5 Factoring PolynomialsObjectives1. Factor out the greatestcommon factor of apolynomial.2. Factor by grouping.3. Factor trinomials.4. Factor the difference ofsquares.5. Factor perfect squaretrinomials.6. Factor the sum anddifference of cubes.7. Use a general strategy forfactoring polynomials.8. Factor algebraicexpressions containingfractional and negativeexponents.A two-year-old boy is asked, “Do you have a brother?” He answers, “Yes.” “Whatis your brother’s name?” “Tom.” Asked if Tom has a brother, the two-year-oldreplies, “No.” The child can go in the direction from self to brother, but he cannotreverse this direction and move from brother back to self.As our intellects develop, we learn to reverse the direction of our thinking.Reversibility of thought is found throughout algebra. For example, we canmultiply polynomials and show that(2x 1)(3x - 2) 6x 2 - x - 2.We can also reverse this process and express the resulting polynomial as6x2 - x - 2 (2x 1)(3x - 2).

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 49Section P.5 Factoring Polynomials 49Factoring is the process of writing a polynomial as the product of two ormore polynomials. The factors of 6x2 - x - 2 are 2x 1 and 3x - 2.In this section, we will be factoring over the set of integers, meaning thatthe coefficients in the factors are integers. Polynomials that cannot be factoredusing integer coefficients are called irreducible over the integers, or prime.The goal in factoring a polynomial is to use one or more factoringtechniques until each of the polynomial’s factors is prime or irreducible. In thissituation, the polynomial is said to be factored completely.We will now discuss basic techniques for factoring polynomials.1Factor out the greatestcommon factor of apolynomial.Common FactorsIn any factoring problem, the first step is to look for the greatest commonfactor. The greatest common factor, abbreviated GCF, is an expression of thehighest degree that divides each term of the polynomial. The distributiveproperty in the reverse directionab ac a(b c)can be used to factor out the greatest common factor.EXAMPLE 1Factoring out the Greatest Common FactorFactor: a. 18x3 27x2b. x2(x 3) 5(x 3).SolutionStudy TipThe variable part of the greatestcommon factor always containsthe smallest power of a variableor algebraic expression thatappears in all terms of thepolynomial.a. We begin by determining the greatest common factor. 9 is the greatestinteger that divides 18 and 27. Furthermore, x2 is the greatest expression thatdivides x3 and x2. Thus, the greatest common factor of the two terms in thepolynomial is 9x2.18x3 27x2 9x2(2x) 9x2(3)Express each term as the product of the greatestcommon factor and its other factor. 9x2(2x 3)Factor out the greatest common factor.b. In this situation, the greatest common factor is the common binomial factor(x 3). We factor out this common factor as follows:x2(x 3) 5(x 3) (x 3)Ax2 5B. Factor out the commonbinomial factor.CheckPoint12Factor by grouping.Factor:a. 10x3 - 4x2b. 2x(x - 7) 3(x - 7).Factoring by GroupingSome polynomials have only a greatest common factor of 1. However, by asuitable rearrangement of the terms, it still may be possible to factor. Thisprocess, called factoring by grouping, is illustrated in Example 2.

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 5050 Chapter P Prerequisites: Fundamental Concepts of AlgebraEXAMPLE 2Factoring by GroupingFactor: x3 4x2 3x 12.Solution Group terms that have a common factor:x3 4x2 Commonfactor is x2.Ax3 3xB A4x2 12B.Factor out the greatestcommon factor from eachgroup and complete thefactoring process. Describewhat happens. What can youconclude?We now factor the given polynomial as follows.x3 4x2 3x 12 (x3 4x2) (3x 12) x2(x 4) 3(x 4)Group terms with common factors. (x 4)(x2 3)Factor (x 4) out of both terms.Thus, x3 4x2 3x 12 (x 4)(x2 3). Check the factorization bymultiplying the right side of the equation using the FOIL method. If thefactorization is correct, you will obtain the original polynomial.CheckPoint2x3 5x2 - 2x - 10.Factoring TrinomialsTo factor a trinomial of the form ax2 bx c, a little trial and error may benecessary.A Strategy for Factoring ax 2 ⴙ bx ⴙ c(Assume, for the moment, that there is no greatest common factor.)1. Find two First terms whose product is ax2:(n x )(n x ) ax2 bx c.Î2. Find two Last terms whose product is c:(x n)(x n) ax 2 bx c.ÎÎFactor trinomials.Factor:Î3Factor out the greatest common factor from thegrouped terms. The remaining two terms have x 4as a common binomial factor.ÎIn Example 2, group theterms as follows:Commonfactor is 3.ÎDiscovery3x 12 .3. By trial and error, perform steps 1 and 2 until the sum of the Outsideproduct and Inside product is bx:Î(nx n)(nx n) ax2 bx c.IO(sum of O I)If no such combinations exist, the polynomial is prime.

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 51Section P.5 Factoring Polynomials 51EXAMPLE 3 Factoring Trinomials Whose LeadingCoefficients Are 1Factor: a. x2 6x 8b. x2 3x - 18.Solutiona. The factors of the first term are x and x :)(xx2 6x 8 (xFactorsof 88, 14, 2 -8, -1 -4, -2Sumof Factors96-9).To find the second term of each factor, we must find two numbers whoseproduct is 8 and whose sum is 6. From the table in the margin, we see that4 and 2 are the required integers. Thus,-6x2 6x 8 (x 4) (x 2) or (x 2) (x 4).This is thedesired sum.b. We begin withx2 3x - 18 (xFactorsof -1818, -1-18, 19, -2-9, 26, -3-6, 3Sumof factors17-177-73-3This is thedesired sum.CheckPoint3)(x).To find the second term of each factor, we must findtwo numbers whose product is -18 and whose sum is 3.From the table in the margin, we see that 6 and -3 arethe required integers. Thus,x2 3x - 18 (x 6)(x - 3)or (x - 3)(x 6).Factor:a. x2 13x 40 b. x2 - 5x - 14.EXAMPLE 4 Factoring a Trinomial Whose LeadingCoefficient Is Not 1Factor: 8x2 - 10x - 3.SolutionStep 1 Find two First terms whose product is 8x2.8x2 - 10x - 3 ⱨ (8x)(x)2ⱨ8x - 10x - 3 (4x)(2x)Step 2 Find two Last terms whose product is -3. The possible factorizationsare 1(-3) and -1(3).Step 3 Try various combinations of these factors. The correct factorization of8x2 - 10x - 3 is the one in which the sum of the Outside and Inside products isequal to -10x. Here is a list of the possible factorizations:Possible Factorizationsof 8x 2 - 10x - 3Sum of Outside and InsideProducts (Should Equal -10x)(8x 1)(x - 3)-24x x -23x(8x - 3)(x 1)8x - 3x 5x(8x - 1)(x 3)24x - x 23x(8x 3)(x - 1)-8x 3x -5x(4x 1)(2x - 3)-12x 2x -10x(4x - 3)(2x 1)4x - 6x -2x(4x - 1)(2x 3)12x - 2x 10x(4x 3)(2x - 1)-4x 6x 2xThis is the requiredmiddle term.

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 5252 Chapter P Prerequisites: Fundamental Concepts of AlgebraThus,8x2 - 10x - 3 (4x 1)(2x - 3) or (2x - 3)(4x 1).Show that this factorization is correct by multiplying the factors using the FOILmethod. You should obtain the original trinomial.CheckPoint44Factor the difference ofsquares.Factor: 6x2 19x - 7.Factoring the Difference of Two SquaresA method for factoring the difference of two squares is obtained by reversingthe special product for the sum and difference of two terms.The Difference of Two SquaresIf A and B are real numbers, variables, or algebraic expressions, thenA2 - B2 (A B)(A - B).In words: The difference of the squares of two terms factors as theproduct of a sum and a difference of those terms.EXAMPLE 5Factoring the Difference of Two SquaresFactor: a. x - 4b. 81x2 - 49.2Solution We must express each term as the square of some monomial. Thenwe use the formula for factoring A2 - B2.x2 - 4 x2 - 22 (x 2) (x - 2)a.A2 B 2 (A B) (A B)b. 81x2 - 49 (9x)2 - 72 (9x 7)(9x - 7)CheckPoint5Factor:a. x2 - 81b. 36x2 - 25.We have seen that a polynomial is factored completely when it is written asthe product of prime polynomials. To be sure that you have factored completely,check to see whether the factors can be factored.Study TipFactoring x4 - 81 asAx2 9B Ax2 - 9Bis not a complete factorization.The second factor, x2 - 9, isitself a difference of twosquares and can be factored.EXAMPLE 6A Repeated FactorizationFactor completely: x4 - 81.Solutionx4 - 81 Ax2 B 2 - 92 Ax2 9B Ax2 - 9BExpress as the difference of two squares.The factors are the sum and difference ofthe squared terms.

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 53Section P.5 Factoring Polynomials 53CheckPoint65Factor perfect squaretrinomials. Ax2 9B Ax2 - 32 BThe factor x2 - 9 is the difference of twosquares and can be factored. Ax2 9B(x 3)(x - 3)The factors of x2 - 9 are the sum anddifference of the squared terms.Factor completely: 81x4 - 16.Factoring Perfect Square TrinomialsOur next factoring technique is obtained by reversing the special products forsquaring binomials. The trinomials that are factored using this technique arecalled perfect square trinomials.Factoring Perfect Square TrinomialsLet A and B be real numbers, variables, or algebraic expressions.1. A2 2AB B2 (A B)2Same sign2. A - 2AB B2 (A - B)22Same signThe two items in the box show that perfect square trinomials come in twoforms: one in which the middle term is positive and one in which the middle termis negative. Here’s how to recognize a perfect square trinomial:1. The first and last terms are squares of monomials or integers.2. The middle term is twice the product of the expressions being squared inthe first and last terms.EXAMPLE 7Factoring Perfect Square TrinomialsFactor: a. x2 6x 9b. 25x2 - 60x 36.Solutiona. x2 6x 9 x2 2 ⴢ x ⴢ 3 32 (x 3)2 The middle term has apositive sign.A2 2 A B B 2 (A B)2b. We suspect that 25x2 - 60x 36 is a perfect square trinomial because25x2 (5x)2 and 36 62. The middle term can be expressed as twice theproduct of 5x and 6.25x2 - 60x 36 (5x)2 - 2 ⴢ 5x ⴢ 6 62 (5x - 6)2A2 2 A B B 2 (A B)2

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 5454 Chapter P Prerequisites: Fundamental Concepts of AlgebraCheckPoint76Factor the sum anddifference of cubes.Factor:a. x2 14x 49 b. 16x2 - 56x 49.Factoring the Sum and Difference of Two CubesWe can use the following formulas to factor the sum or the difference of twocubes:Factoring the Sum and Difference of Two Cubes1. Factoring the Sum of Two CubesA3 B3 (A B) AA2 - AB B2 B2. Factoring the Difference of Two CubesA3 - B3 (A - B) AA2 AB B2 BEXAMPLE 8Factoring Sums and Differences of Two CubesFactor: a. x3 8b. 64x3 - 125.Solutiona. x3 8 x3 2 3 (x 2)Ax2 - x ⴢ 2 2 2 B (x 2) Ax2 - 2x 4BA3 B 3 (A B)(A2 AB B 2)b. 64x3 - 125 (4x)3 - 53 (4x - 5) [(4x)2 (4x)(5) 52]A3 B 3 (A B)(A2 A B B2) (4x - 5) (16x2 20x 25)CheckPoint87Use a general strategy forfactoring polynomials.Factor:a. x3 1b. 125x3 - 8.A Strategy for Factoring PolynomialsIt is important to practice factoring a wide variety of polynomials so that youcan quickly select the appropriate technique. The polynomial is factoredcompletely when all its polynomial factors, except possibly for monomialfactors, are prime. Because of the commutative property, the order of thefactors does not matter.

BLITMCPB.QXP.131013599 48-74 tg1/13/0312:18 PMPage 55Section P.5 Factoring Polynomials 55A Strategy for Factoring a Polynomial1. If there is a common factor, factor out the GCF.2. Determine the number of terms in the polynomial and try factoring asfollows:a. If there are two terms, can the binomial be factored by one of thefollowing special forms?Difference of two squares: A2 - B2 (A B)(A - B)Sum of two cubes: A3 B3 (A B)(A2 - AB B2)Difference of two cubes: A3 - B3 (A - B)(A2 AB B2)b. If there are three terms, is the trinomial a perfect square trinomial? Ifso, factor by one of the following special forms:A2 2AB B2 (A B)2A2 - 2AB B2 (A - B)2.If the trinomial is not a perfect square trinomial, try factoring bytrial and error.c. If there are four or more terms, try factoring by grouping.3. Check to see if any factors with more than one term in the factoredpolynomial can be factored further. If so, factor completely.EXAMPLE 9Factoring a PolynomialFactor: 2x3 8x2 8x.SolutionStep 1 If there is a common factor, factor out the GCF. Because 2x is common toall terms, we factor it out.2x3 8x2 8x 2x(x2 4x 4) Factor out the GCF.Step 2 Determine the number of terms and factor accordingly. The factorx2 4x 4 has three terms and is a perfect square trinomial. We factor usingA2 2AB B2 (A B)2.2x3 8x2 8x 2x(x2 4x 4) 2x(x2 2 ⴢ x ⴢ 2 22)A2 2 A B B2 2x(x 2)2A2 2AB B2 (A B)2Step 3 Check to see if factors can be factored further.cannot. Thus,2x3 8x2 8x 2x(x 2)2.CheckPoint9Factor: 3x3 - 30x2 75x.In this problem, they

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 5656 Chapter P Prerequisites: Fundamental Concepts of AlgebraEXAMPLE 10Factoring a PolynomialFactor: x2 - 25a2 8x 16.SolutionStep 1 If there is a common factor, factor out the GCF. Other than 1 or –1, thereis no common factor.Step 2 Determine the number of terms and factor accordingly. There arefour terms. We try factoring by grouping. Grouping into two groups of twoterms does not result in a common binomial factor. Let’s try grouping as adifference of squares.x2 - 25a2 8x 16 (x2 8x 16) - 25a2 (x 4)2 - (5a)2 (x 4 5a)(x 4 - 5a)Rearrange terms and group as a perfectsquare trinomial minus 25a2 to obtain adifference of squares.Factor the perfect square trinomial.Factor the difference of squares. The factorsare the sum and difference of the expressionsbeing squared.Step 3 Check to see if factors can be factored further. In this case, they cannot, sowe have factored completely.CheckPoint108Factor algebraic expressionscontaining fractional andnegative exponents.Factor: x2 - 36a2 20x 100.Factoring Algebraic Expressions Containing Fractionaland Negative ExponentsAlthough expressions containing fractional and negative exponents are notpolynomials, they can be simplified using factoring techniques.EXAMPLE 11Factoring Involving Fractional and NegativeExponentsFactor and simplify: x(x 1)-3兾4 (x 1)1兾4.Solution The greatest common factor is x 1 with the smallest exponent inthe two terms. Thus, the greatest common factor is (x 1)-3兾4.x(x 1)-3兾4 (x 1)1兾4 (x 1)-3兾4x (x 1)-3兾4(x 1) Express each term as the product of thegreatest common factor and its other factor.-3兾4 (x 1) 2x 1(x 1)3兾4[x (x 1)]Factor out the greatest common factor.b-n 1bn

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 57Exercise Set P.5 57CheckPoint11Factor and simplify: x(x - 1)-1兾2 (x - 1)1兾2.EXERCISE SET P.5Practice ExercisesIn Exercises 1–10, factor out the greatest commonfactor.1.3.5.7.9.18x 272.3x2 6x4.6.9x4 - 18x3 27x28.x(x 5) 3(x 5)x2(x - 3) 12(x - 3) 10.16x - 244x2 - 8x6x4 - 18x3 12x2x(2x 1) 4(2x 1)x2(2x 5) 17(2x 5)In Exercises 11–16, factor by grouping.11. x3 - 2x2 5x - 1013. x3 - x2 2x - 215. 3x3 - 2x2 - 6x 412. x3 - 3x2 4x - 1214. x3 6x2 - 2x - 1216. x3 - x2 - 5x 5In Exercises 57–84, factor completely, or state that thepolynomial is prime.57. 3x3 - 3x58. 5x3 - 45x259. 4x - 4x - 2460. 6x2 - 18x - 60461. 2x - 16262. 7x4 - 73263. x 2x - 9x - 1864. x3 3x2 - 25x - 75265. 2x - 2x - 11266. 6x2 - 6x - 1267. x3 - 4x68. 9x3 - 9x69. x2 6470. x2 3671. x3 2x2 - 4x - 872. x3 2x2 - x - 2573. y - 81y74. y5 - 16y4275. 20y - 45y76. 48y4 - 3y277. x2 - 12x 36 - 49y2In Exercises 17–30, factor each trinomial, or state that thetrinomial is prime.17.19.21.23.25.27.29.x2 5x x2 - 2x x2 - 8x 3x2 - x 3x2 - 25x6x2 - 11x4x2 16x615152- 28 4 1518.20.22.24.26.28.30.x2 8x 15x2 - 4x - 5x2 - 14x 452x2 5x - 33x2 - 2x - 56x2 - 17x 128x2 33x 4In Exercises 31–40, factor the difference of two squares.31.33.35.37.39.x2 - 10036x2 - 499x2 - 25y 2x4 - 1616x4 - 8132.34.36.38.40.x2 64x236x2x4 81x4144- 81- 49y21- 1In Exercises 41–48, factor any perfect square trinomials, orstate that the polynomial is prime.41.43.45.47.x2 2x 1x2 - 14x 494x2 4x 19x2 - 6x 142.44.46.48.x2 x2 25x264x24x 410x 25 10x 1- 16x 1In Exercises 49–56, factor using the formula for the sum ordifference of two cubes.49.51.53.55.x3 27x3 - 648x3 - 164x3 2750.52.54.56.x3 64x3 - 2727x3 - 18x3 12578. x2 - 10x 25 - 36y279. 9b x - 16y - 16x 9b y2280. 16a2x - 25y - 25x 16a2y81. x2y - 16y 32 - 2x282. 12x2y - 27y - 4x2 983. 2x - 8a x 24x 72x32284. 2x3 - 98a2x 28x2 98xIn Exercises 85–94, factor and simplify each algebraicexpression.85. x3兾2 - x1兾286. x3兾4 - x1兾487. 4x-2兾3 8x1兾388. 12x-3兾4 6x1兾41兾23兾289. (x 3) - (x 3)90. (x2 4)3兾2 (x2 4)7兾291. (x 5)-1兾2 - (x 5)-3兾292. (x2 3)-2兾3 (x2 3)-5兾393. (4x - 1)1兾2 - 13(4x - 1)3兾294. -8(4x 3)-2 10(5x 1)(4x 3)-1Application Exercises95. Your computer store is having an incredible sale. Theprice on one model is reduced by 40%.Then the sale priceis reduced by another 40%. If x is the computer’s originalprice, the sale price can be represented by(x - 0.4x) - 0.4(x - 0.4x).a. Factor out (x - 0.4x) from each term. Then simplifythe resulting expression.b. Use the simplified expression from part (a) to answerthese questions: With a 40% reduction followed by a40% reduction, is the computer selling at 20% of itsoriginal price? If not, at what percentage of theoriginal price is it selling?

BLITMCPB.QXP.131013599 48-7412/30/0210:43 AMPage 5858 Chapter P Prerequisites: Fundamental Concepts of Algebra96. The polynomial 8x2 20x 2488 describes the number,in thousands, of high school graduates in the United Statesx years after 1993.a. According to this polynomial, how many students willgraduate from U.S. high schools in 2003?b. Factor the polynomial.c. Use the factored form of the polynomial in part (b) tofind the number of high school graduates in 2003. Doyou get the same answer as you did in part (a)? If so,does this prove that your factorization is correct?Explain.97. A rock is dropped from the top of a 256-foot cliff. Theheight, in feet, of the rock above the water after tseconds is described by the polynomial 256 - 16t2.Factor this expression completely.98. The amount of sheet metal needed to manufacture acylindrical tin can, that is, its surface area, S, isS 2 r2 2 rh. Express the surface area, S, infactored form.102. Suppose that a polynomial contains four terms. Explainhow

Section P.5 Factoring Polynomials 49 1 Factor out the greatest common factor of a polynomial. 2 Factor by grouping. Factoring is the process of writing a polynomial as the product of two or more polynomials.The factors of are and In this section, we will be factoring over the set of integers,meaning that the coefficients in the factors are integers.

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