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messersmith_power_intermediate_algebra_1e_ch4_7_10

Write out all of the steps as you are reading the problem. Solution a) Begin by grouping terms together so that each group has a common factor. T T Factor out a to get a(b 5). a(b 5) 3(b 5) Factor out 3 to get 3(b 5). (b 5)(a 3) Factor out (b 5). Check: (b 5)(a 3) ab 5a 3b 15 ✓ b) Group terms together so that each group has a common factor. T T Factor out r to get r(2p 5q). r(2p 5q) 3(2p 5q) Factor out 3 to get 3(2p 5q). (2p 5q)(r 3) Factor out (2p 5q). Check: (2p 5q)(r 3) 2pr 5qr 6p 15q ✓ c) Group terms together so that each group has a common factor. T T Factor out x2 to get x2(x 6). x2(x 6) 7(x 6) Factor out 7 to get 7(x 6). (x 6)(x2 7) Factor out (x 6). We must factor out 7 not 7 from the second group so that the binomial factors for both groups are the same! If we had factored out 7, then the factorization of the second group would have been 7(x 6). Check: (x 6)(x2 7) x3 6x2 7x 42 ✓ YOU TRY 6 Factor by grouping. ab 5a 3b 15 2pr 5qr 6p 15q x3 6x2 7x 42 a) 2cd 4d 5c 10 b) 4k2 36k km 9m c) h3 8h2 5h 40 Sometimes we have to rearrange the terms before we can factor. EXAMPLE 8 Factor 24y2 3z 4y 18yz completely. Solution Group terms together so that each group has a common factor. 24y2 3z 4y 18yz T T Factor out 3 to get 3(8y2 z). 3(8y2 z) 2y(2 9z) Factor out 2y to get 2y(2 9z). The groups do not have common factors! Let’s rearrange the terms in the original polynomial, and group the terms differently. 24y2 4y 18yz 3z T T Factor out 4y to get 4y(6y 1). 4y(6y 1) 3z(6y 1) Factor out 3z to get 3z(6y 1). (6y 1)(4y 3z) Factor out (6y 1). Check: (6y 1)(4y 3z) 24y2 3z 4y 18yz ✓ Often, there is more than one way that the terms can be rearranged so that the polynomial can be factored by grouping. 362 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith


messersmith_power_intermediate_algebra_1e_ch4_7_10
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