Without multiplying we know that this choice is incorrect. How? In the factor (2r 20), a 2 can be factored out to get 2(r 10). But, we said that we could not factor out a GCF from the original polynomial, 2r2 13r 20. Therefore, it will not be possible to take out a common factor from one of the binomial factors. Note If you cannot factor out a GCF from the original polynomial, then you cannot take out a factor from one of the binomial factors either. For the same reason, the pair 2 and 10 will not give us the correct factorization: 2r2 13r 20 (2r 2)(r 10) 2 can be factored out of (2r 2). 2r2 13r 20 (2r 10)(r 2) 2 can be factored out of (2r 10). Try using 4 and 5: 2r2 13r 20 (2r 4)(r 5) 2 can be factored out of (2r 4). Switch the 4 and 5: 2r2 13r 20 (2r 5)(r 4) Correct! 5r These must (8r) both be 13r. 13r The factorization of 2r2 13r 20 is (2r 5)(r 4). Check by multiplying. YOU TRY 6 Factor completely. a) 2m2 11m 12 b) 3v2 28v 9 EXAMPLE 8 Factor completely. a) 12d2 46d 8 b) 2h2 9h 56 Solution a) Ask yourself, “Can I take out a common factor?” Yes! The GCF is 2. 12d 2 46d 8 2(6d 2 23d 4) Now, try to factor 6d 2 23d 4. To get a product of 6d 2, we can try either 6d and d or 3d and 2d. Let’s start by trying 6d and d. 6d 2 23d 4 (6d )(d ) List pairs of integers that multiply to 4: 4 and 1, 4 and 1, 2 and 2. Try 4 and 1. Do not put 4 in the same binomial as 6d since then it would be possible to factor out 2. But, a 2 does not factor out of 6d 2 23d 4. Put the 4 in the same binomial as d. 6d 2 23d 4 (6d 1)(d 4) d 24d 23d Correct! www.mhhe.com/messersmith SECTION 7.2 Factoring Trinomials 373
messersmith_power_intermediate_algebra_1e_ch4_7_10
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