EXAMPLE 6 Notice that the trial and error method skips both the process of rewriting the middle term and factoring by grouping. EXAMPLE 7 Factor 3x2 10x 8 completely. Solution Can we factor out a GCF? No. So try to factor 3x2 10x 8 as the product of two binomials. Notice that all terms are positive, so all factors will be positive. Begin with the squared term, 3x2. Which two expressions with integer coeffi cients can we multiply to get 3x2? 3x and x. Put these in the binomials. 3x2 10x 8 (3x )(x ) 3x x 3x2 Next, look at the last term, 8. What are the pairs of positive integers that multiply to 8? They are 8 and 1 as well as 4 and 2. Try these numbers as the last terms of the binomials. The middle term, 10x, comes from fi nding the sum of the products of the outer terms and inner terms. First Try 3x2 10x 8 (3x 8)(x 1) Incorrect 8x These must 3x both be 10x. 11x Switch the 8 and 1. 3x2 10x 8 (3x 1)(x 8) Incorrect 1x These must 24x both be 10x. 25x Try using 4 and 2. 3x2 10x 8 (3x 4)(x 2) Correct! 4x These must 6x both be 10x. 10x The factorization of 3x2 10x 8 is (3x 4)(x 2). Check by multiplying. Factor 2r2 13r 20 completely. Solution Can we factor out a GCF? No. To get a product of 2r2, we will use 2r and r. 2r2 13r 20 (2r )(r ) 2r r 2r2 Since the last term is positive and the middle term is negative, we want pairs of negative integers that multiply to 20. The pairs are 1 and 20, 2 and 10, and 4 and 5. Try these numbers as the last terms of the binomials. The middle term, 13r, comes from fi nding the sum of the products of the outer terms and inner terms. 2r2 13r 20 (2r 1)(r 20) Incorrect r These must (40r) both be 13r. 41r Switch the 1 and 20: 2r2 13r 20 (2r 20)(r 1) 372 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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