5 3 Common Factors and Factoring by Grouping

Factoring is the reverse of multiplication. To factor an expression means to write an equivalent expression that is a product. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 2

Terms with Common Factors When factoring a polynomial, we look for factors common to every term and then use the distributive law. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 3

Example Factor out a common factor: 9 m 2 – 27. Solution 9 m 2 – 27 = 9(m 2) – 9(3) = 9(m 2 – 3) Noting that 9 is a common factor Using the distributive law Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 4

Example Write an equivalent expression by factoring: 25 x 2 y 5 + 35 x 6 y 3 – 15 x 3 y 4. Solution 25, 35, 15 x 2, x 6, x 3 Greatest common factor = 5. Greatest common factor = x 2. y 5, y 3, y 4 Greatest common factor = y 3. Thus 5 x 2 y 3 is the greatest common factor. 25 x 2 y 5 + 35 x 6 y 3 – 15 x 3 y 4 = 5 x 2 y 3(5 y 2 + 7 x 4 – 3 xy) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 5

Polynomials that cannot be factored further are said to be factored completely. The factors in the resulting factorization are said to be prime polynomials. When the leading coefficient is a negative number, we generally factor out a common factor with a negative coefficient, Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 6

Example Write an equivalent expression for – 4 x 2 – 16 x by factoring out a common factor with a negative coefficient. Solution Notice that – 4 x is the largest common factor. – 4 x 2 – 16 x = – 4 x(x + 4) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 7

Factoring by Grouping The largest common factor is sometimes a binomial. Often, in order to identify a common binomial factor, we must regroup into two groups of two terms each. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 8

Example Write an equivalent expression by factoring: (x + 4)m + (x + 4)(y – b). Solution Factor out x + 4 (x + 4)m + (x + 4)(y – b) = (x + 4)[m + y – b] Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 9

Example Write an equivalent expression by factoring: y 5 + 5 y 3 + 3 y 2 + 15. Solution y 5 + 5 y 3 + 3 y 2 + 15 = (y 5 + = y 3(y 2 5 y 3) + (3 y 2 + 5) + 3(y 2 = (y 2 + 5)(y 3 + 3) + 15) Each grouping has a common factor + 5) Factor a common term from each binomial Factor out y 2 + 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 10

Example Write an equivalent expression by factoring: 3 x – 3 y – ax + ay. Solution 3 x – 3 y – ax + ay = 3(x – y) – a(x – y) = (3 – a)(x – y) Notice that – a is factored out so that both terms have x – y Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 11

Some polynomials with four terms are prime. 3 + 2 x 2 + 2 x – 4 x Example No matter how we group terms, there is no common binomial factor. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 12