Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Common Factors and Factoring by Grouping Terms with Common Factors Factoring by Grouping 5.3

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

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

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

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

Slide 5- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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,

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

Slide 5- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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.

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

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

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

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