Factoring, SolvingEquations, And Problem Solving 5

05-W4801-AM1.qxd8/19/088:45 PMPage 241REVISED PAGESFactoring, SolvingEquations, andProblem Solving55.1 Factoring by Using theDistributive Property5.2 Factoring the Differenceof Two Squares5.3 Factoring Trinomials ofthe Form x2 bx c5.4 Factoring Trinomials ofthe Form ax2 bx cPhotodisc/Getty Images5.5 Factoring, SolvingEquations, andProblem Solving Algebraic equations can be used to solve a large varietyof problems involving geometric relationships.Aflower garden is in the shape of a right triangle with one leg 7 meters longerthan the other leg and the hypotenuse 1 meter longer than the longer leg.Find the lengths of all three sides of the right triangle. A popular geometric formula,called the Pythagorean theorem, serves as a guideline for setting up an equation tosolve this problem. We can use the equation x2 1x 72 2 1x 82 2 to determinethat the sides of the right triangle are 5 meters, 12 meters, and 13 meters long.The distributive property has allowed us to combine similar terms and multiplypolynomials. In this chapter, we will see yet another use of the distributive propertyas we learn how to factor polynomials. Factoring polynomials will allow us to solveother kinds of equations, which will, in turn, help us to solve a greater variety of wordproblems.Exciting videos of all objective concepts are available in a variety of delivery models.241

05-W4801-AM1.qxd8/19/088:45 PMPage 242REVISED PAGESI N T E R N E TP R O J E C TPythagoras is widely known for the Pythagorean theorem pertaining to right triangles. Do anInternet search to determine at least two other fields where Pythagoras made significantcontributions. Pythagoras also founded a school. While conducting your search, find the namegiven to the students attending Pythagoras' school and some of the school rules for students. Canyou think of any modern-day schools that might have the same requirements?5.1Factoring by Using the Distributive PropertyOBJECTIVES1Find the Greatest Common Factor2Factor Out the Greatest Common Factor3Factor by Grouping4Solve Equations by Factoring5Solve Word Problems Using Factoring1 Find the Greatest Common FactorIn Chapter 1, we found the greatest common factor of two or more whole numbers byinspection or by using the prime factored form of the numbers. For example, by inspection we see that the greatest common factor of 8 and 12 is 4. This means that 4 isthe largest whole number that is a factor of both 8 and 12. If it is difficult to determinethe greatest common factor by inspection, then we can use the prime factorizationtechnique as follows:42 2 # 3 # 770 2 # 5 # 7We see that 2 # 7 14 is the greatest common factor of 42 and 70.It is meaningful to extend the concept of greatest common factor to monomials.Consider the next example.EXAMPLE 1Find the greatest common factor of 8x 2 and 12x 3.Solution8x2 2 # 2 # 2 # x # x12x3 2 # 2 # 3 # x # x # xTherefore, the greatest common factor is 2 # 2 # x # x 4x2. PRACTICE YOUR SKILLFind the greatest common factor of 14a2 and 7a5.7a2 By the greatest common factor of two or more monomials we mean the monomial with the largest numerical coefficient and highest power of the variables that is afactor of the given monomials.242

05-W4801-AM1.qxd8/19/088:45 PMPage 243REVISED PAGES5.1 Factoring by Using the Distributive PropertyEXAMPLE 2243Find the greatest common factor of 16x 2y, 24x 3y 2, and 32xy.Solution16x2y 2 # 2 # 2 # 2 # x # x # y24x3y2 2 # 2 # 2 # 3 # x # x # x # y # y32xy 2 # 2 # 2 # 2 # 2 # x # yTherefore, the greatest common factor is 2 # 2 # 2 # x # y 8xy. PRACTICE YOUR SKILLFind the greatest common factor of 18m2n4, 4m3n5, and 10m4n3.2m2n3 2 Factor Out the Greatest Common FactorWe have used the distributive property to multiply a polynomial by a monomial; forexample,3x1x 2 2 3x 2 6xSuppose we start with 3x 2 6x and want to express it in factored form. We use thedistributive property in the form ab ac a(b c).3x 2 6x 3x1x2 3x122 3x1x 223x is the greatest common factor of 3x2 and 6xUse the distributive propertyThe next four examples further illustrate this process of factoring out the greatest common monomial factor.EXAMPLE 3Factor 12x 3 8x 2.Solution12x3 8x2 4x2 13x2 4x2 122 4x2 13x 22ab ac a(b c) PRACTICE YOUR SKILLFactor 15a2 21a6.EXAMPLE 43a2 (5 7a4) Factor 12x 2y 18xy 2.Solution12x2y 18xy2 6xy12x2 6xy13y2 6xy12x 3y2 PRACTICE YOUR SKILLFactor 8m3n2 2m6n.2m3n(4n m3)

05-W4801-AM1.qxd2448/19/088:45 PMPage 244REVISED PAGESChapter 5 Factoring, Solving Equations, and Problem SolvingEXAMPLE 5Factor 24x 3 30x 4 42x 5.Solution24x 3 30x 4 42x 5 6x 3 142 6x 3 15x2 6x 3 17x 2 2 6x 3 14 5x 7x 2 2 PRACTICE YOUR SKILLFactor 48y8 16y6 24y4.EXAMPLE 68y 4(6y 4 2y 2 3) Factor 9x 2 9x.Solution9x 2 9x 9x1x2 9x112 9x1x 12 PRACTICE YOUR SKILLFactor 8b3 8b2.8b2(b 1) We want to emphasize the point made just before Example 3. It is important torealize that we are factoring out the greatest common monomial factor. We couldfactor an expression such as 9x 2 9x in Example 6 as 91x 2 x 2 , 313x 2 3x 2 ,13x13x 3 2 , or even 118x2 18x2 , but it is the form 9x1x 1 2 that we want. We2can accomplish this by factoring out the greatest common monomial factor; we sometimes refer to this process as factoring completely. A polynomial with integralcoefficients is in completely factored form if these conditions are met:1.It is expressed as a product of polynomials with integral coefficients.2.No polynomial, other than a monomial, within the factored form can befurther factored into polynomials with integral coefficients.Thus 91x 2 x 2 , 313x 2 3x 2 , and 3x13x 2 2 are not completely factored be1cause they violate condition 2. The form 118x2 18x2 violates both conditions21 and 2.3 Factor by GroupingSometimes there may be a common binomial factor rather than a common monomialfactor. For example, each of the two terms of x(y 2) z(y 2) has a commonbinomial factor of (y 2). Thus we can factor (y 2) from each term and getx(y 2) z(y 2) (y 2)(x z)Consider a few more examples involving a common binomial factor:a1b c2 d1b c2 1b c2 1a d2x1x 22 31x 22 1x 221x 32x1x 52 41x 52 1x 521x 42

05-W4801-AM1.qxd8/19/088:45 PMPage 245REVISED PAGES5.1 Factoring by Using the Distributive Property245It may be that the original polynomial exhibits no apparent common monomialor binomial factor, which is the case withab 3a bc 3cHowever, by factoring a from the first two terms and c from the last two terms, we seethatab 3a bc 3c a(b 3) c(b 3)Now a common binomial factor of (b 3) is obvious, and we can proceed as before:a(b 3) c(b 3) (b 3)(a c)This factoring process is called factoring by grouping. Let’s consider two more examples of factoring by grouping.EXAMPLE 7Factor each polynomial completely.(a) x2 x 5x 5(b) 6x2 4x 3x 2Solution(a) x2 x 5x 5 x1x 12 51x 12 1x 12 1x 52(b) 6x2 4x 3x 2 2x13x 22 113x 22 13x 22 12x 12Factor x from first twoterms and 5 from last twotermsFactor common binomialfactor of (x 1) from bothtermsFactor 2x from first twoterms and 1 from last twotermsFactor common binomialfactor of (3x 2) from bothterms PRACTICE YOUR SKILLFactor each polynomial completely.(a) ab 5a 3b 15(a) (a 3)(b 5)(b) xy 2x 4y 8 (b) (x 4)(y 2)4 Solve Equations by FactoringSuppose we are told that the product of two numbers is 0. What do we know about thenumbers? Do you agree we can conclude that at least one of the numbers must be 0?The next property formalizes this idea.Property 5.1For all real numbers a and b,ab 0 if and only if a 0 or b 0Property 5.1 provides us with another technique for solving equations.

05-W4801-AM1.qxd2468/19/088:45 PMPage 246REVISED PAGESChapter 5 Factoring, Solving Equations, and Problem SolvingEXAMPLE 8Solve x 2 6x 0.SolutionTo solve equations by applying Property 5.1, one side of the equation must be a product, and the other side of the equation must be zero. This equation already has zeroon the right-hand side of the equation, but the left-hand side of this equation is a sum.We will factor the left-hand side, x2 6x, to change the sum into a product.x 2 6x 0x1x 62 0x 0orx 0orx 6 0ab 0 if and only if a 0 or b 0x 6The solution set is { 6, 0}. (Be sure to check both values in the original equation.) PRACTICE YOUR SKILLSolve y2 7y 0.EXAMPLE 9 {0, 7}Solve x 2 12x.SolutionIn order to solve this equation by Property 5.1, we will first get zero on the right-handside of the equation by adding 12x to each side. Then we factor the expression onthe left-hand side of the equation.x 2 12xx 2 12x 0Added 12x to both sidesx1x 122 0x 0orx 0orx 12 0ab 0 if and only if a 0 or b 0x 12The solution set is {0, 12}. PRACTICE YOUR SKILLSolve a2 15a. {0,15}Remark: Note in Example 9 that we did not divide both sides of the original equationby x. Doing so would cause us to lose the solution of 0.EXAMPLE 10Solve 4x 2 3x 0.Solution4x2 3x 0x14x 32 0x 0or4x 3 0x 0or4x 3x 0orx The solution set is e 0,3f.434ab 0 if and only if a 0 or b 0

05-W4801-AM1.qxd8/19/088:45 PMPage 247REVISED PAGES5.1 Factoring by Using the Distributive Property247 PRACTICE YOUR SKILLSolve 5y2 2y 0.EXAMPLE 112e , 0 f5 Solve x1x 2 2 31x 2 2 0.SolutionIn order to solve this equation by Property 5.1, we will factor the left-hand side of theequation. The greatest common factor of the terms is (x 2).x1x 22 31x 22 01x 221x 32 0x 2 0x 2x 3 0ororThe solution set is 5 3, 26.ab 0 if and only if a 0 or b 0x 3 PRACTICE YOUR SKILLSolve m(m 2) 5(m 2) 0.{ 5, 2} 5 Solve Word Problems Using FactoringEach time we expand our equation-solving capabilities, we gain more techniquesfor solving word problems. Let’s solve a geometric problem with the ideas we learnedin this section.EXAMPLE 12Apply Your SkillThe area of a square is numerically equal to twice its perimeter. Find the length of aside of the square.SolutionssSketch a square and let s represent the length of each side (see Figure 5.1). Then thearea is represented by s 2 and the perimeter by 4s. Thuss 2 214s2ss 2 8ssFigure 5.1s 2 8s 0s1s 82 0s 0ors 8 0s 0ors 8Because 0 is not a reasonable answer to the problem, the solution is 8. (Be sure tocheck this solution in the original statement of the example!) PRACTICE YOUR SKILLThe area of a square is numerically equal to three times its perimeter. Find the lengthof a side of the square.The length is 12