Factoring Copyright Cengage Learning All rights reserved 6

Factoring Copyright © Cengage Learning. All rights reserved. 6

SECTION 6. 1 The Greatest Common Factor and Factoring by Grouping Copyright © Cengage Learning. All rights reserved.

Objectives A Factor the greatest common factor from a polynomial. B Factor by grouping. 3

A Factoring Out the Greatest Common Factor 4

Factoring Out the Greatest Common Factor In this section we will apply the distributive property to polynomials to factor from them what is called the greatest common factor. We use the term largest monomial to mean the monomial with the greatest coefficient and highest power of the variable. 5

Example 1 Find the greatest common factor for the following terms. 12 x 2, 18 x 3, 42 x Solution: The largest number that divides into 12, 18, and 42 is 6. The lowest power of x among 12 x 2, 18 x 3, 42 x is 1 (please remember that 42 x means the same as 42 x 1 ) The greatest common factor for the above three terms is 6 x 1 or simply 6 x 6

Example 2 Find the greatest common factor. Solution: The largest number that divides into 8, 16, and 20 is 4. The greatest common factor (GCF) for The greatest common factor for You can also think this way: 2 is the smallest power of a among the 3 terms, and 2 is also the smallest power of b in those 3 terms. is 4 a 2 b 2 b/c it is the largest monomial that divides of each term. 7

Example 4 Factoring is the reverse of distributing. When we expand 5(x + y) into 5 x + 5 y, we call it distributing. But when we go backwards and change 5 x + 5 y back to 5(x + y), we call it a factorization process. Factor out the greatest common factor. 15 x + 40 y Solution: The largest number that can divide into both 15 and 40 is 5, while x and y have nothing in common, so the GCF of 15 x and 40 y is just 5, . therefore 15 x + 40 y =5(3 x) + 5(8 y) = 5(3 x + 8 y) 8

Example 5 Factor out the greatest common factor from Solution: The greatest common factor is 4 x 3. 3 is the smallest power of x among the three terms. We rewrite the polynomial so we can see the greatest common factor 4 x 3 in each term. Then we apply the (reverse) distributive property to factor it out. 9

Example 6 Factor the greatest common factor from 3 a 2 b – 6 a 3 b 2 + 9 a 3 b 3 Solution: The greatest common factor is 3 a 2 b. We rewrite the polynomial so we can see the greatest common factor 3 a 2 b in each term. Then we apply the distributive property to factor it out. 3 a 2 b – 6 a 3 b 2 + 9 a 3 b 3 = 3 a 2 b(1 – 2 ab + 3 ab 2) 10

B Factoring by Grouping 11

Factoring by Grouping To develop our next method of factoring, called factoring by grouping, we start by examining the polynomial xc + yc. The greatest common factor for the two terms is c. Factoring c from each term we have But suppose that c itself was a more complicated expression, such as a + b, then the expression we were trying to factor was x(a + b) + y(a + b), instead of xc + yc. 12

Factoring by Grouping The greatest common factor for Factoring this common factor from each term looks like this: This last expression is in factored form. The next slide shows an example using the process called factoring by grouping. 13

Example 8 Factor 6 x 2 + 3 x + 4 x + 2 by grouping. Solution: The first two terms have 3 x in common, and the last two terms have a 2 in common. Suppose we factor 3 x from the first two terms and 2 from the last two terms. We get 6 x 2 + 3 x + 4 x + 2 = [6 x 2 + 3 x] + [4 x + 2] = 3 x(2 x + 1) + 2(2 x + 1) Now we see that there is a new GCF, the (2 x + 1), that we can factor out. 14

Example 8 – Solution cont’d Now we see that there is a new GCF, the (2 x + 1), that we can factor out. 6 x 2 + 3 x + 4 x + 2 = 3 x (2 x + 1) + 2 (2 x + 1) = ( ) This completes our factorization. 15

Example 9 Factor 2 x 2 + 5 ax – 2 xy – 5 ay by grouping. Solution: The first two terms have x in common, and the last two terms have -y in common. Suppose we factor x from the first two terms and -y from the last two terms. We get 2 x 2 + 5 ax – 2 xy – 5 ay = x(2 x + 5 a) – y(2 x + 5 a) = (2 x + 5 a)(x – y) 16

Example 10 Factor 3 xb – 4 b – 15 x + 20 by grouping. Solution: The first two terms have b in common, and the last two terms have 5 in common. Suppose we factor b from the first two terms and 5 from the last two terms. We get 3 xb – 4 b – 15 x + 20 = b(3 x – 4) + 5(-3 x + 4) but (3 x – 4) does not match (-3 x + 4), so we need to factor out -1 from (-3 x + 4) = b(3 x – 4) – 5(3 x – 4) = (b – 5)(3 x – 4) it doesn’t matter whether we put (3 x – 4) in the front or back. 17

Example 11 Factor 8 x 3 – 12 x 2 + 14 x – 21 by grouping. Solution: The first two terms have 4 x 2 in common, and the last two terms have 7 in common. Suppose we factor 4 x 2 from the first two terms and 7 from the last two terms. We get 8 x 3 – 12 x 2 + 14 x – 21 = [8 x 3 – 12 x 2] + [14 x – 21] = 4 x 2(2 x – 3) + 7(2 x – 3) = (4 x 2 + 7)(2 x – 3) 18