Complete each You Try by referring to the example before it. We must factor out 5, not 5, from the second group so that the binomial factors for both groups are the same! (If we had factored out 5, then the factorization of the second group would have been 5(n 8).) Check: (n 8)(n2 5) n3 8n2 5n 40 ✓ YOU TRY 6 Factor by grouping. a) xy 4x 10y 40 b) 5pr 8qr 10p 16q c) w3 9w2 6w 54 Sometimes we have to rearrange the terms before we can factor. EXAMPLE 8 In-Class Example 8 Factor 12a2 15b 18a 10ab completely. Answer: (2a 3)(6a 5b) Factor 12c2 2d 3c 8cd completely. Solution Group terms together so that each group has a common factor. 12c2 2d 3c 8cd d d T T Factor out 2 to get 2(6c2 d). 2(6c2 d ) c(3 8d ) Factor out c to get c(3 8d). These groups do not have common factors! Let’s rearrange the terms in the original polynomial and group the terms differently. 12c2 3c 8cd 2d T T d Factor out 3c to get 3c(4c 1). 3c(4c 1) 2d(4c 1) Factor out 2d to get 2d(4c 1). (4c 1)(3c 2d ) Factor out (4c 1). 12c2 2d 3c 8cd factors to (4c 1)(3c 2d). The check is left to the student. Note Often, there is more than one way to rearrange the terms so that the polynomial can be factored by grouping. YOU TRY 7 Factor 3k2 48m 8k 18km completely. Often, we have to combine the two factoring techniques we have learned here. That is, we begin by factoring out the GCF, and then we factor by grouping. Let’s summarize how to factor a polynomial by grouping and then look at another example. 394 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_introductory_algebra_1e_ch4_7_10
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