EXAMPLE 9 Factor 3(x 4)2 11(x 4) 10 completely. Solution Notice that the binomial x 4 appears as a squared quantity and a linear quantity. We will use another letter to represent this quantity. We can use any letter except x. Let’s use the letter u. Let u x 4. Then, u2 (x 4)2. Substitute the u and u2 into the original polynomial, and factor. 3(x 4)2 11(x 4) 10 3u2 11u 10 (3u 5)(u 2) 3(x 4) 5 (x 4) 2 Since the original polynomial was in terms of x, substitute x 4 for u. (3x 12 5)(x 4 2) Distribute. (3x 17)(x 6) Combine like terms. The final factorization is not the one containing the substitution variable, u. We must go back and replace u with the expression it represented so that the factorization is in terms of the original variable. YOU TRY 8 Factor 2(x 3)2 3(x 3) 35 completely. Using Technology We found some ways to narrow down the possibilities when factoring ax2 bx c (a 1) using the trial and error method. We can also use a graphing calculator to help with the process. Consider the trinomial 2x2 9x 35. Enter the trinomial into Y1 and press ZOOM and then enter 6 to display the graph in the standard viewing window. Look on the graph for the x-intercept (if any) that appears to be an integer. It appears that 7 is an x-intercept. To check if 7 is an x-intercept, press TRACE then 7 and press ENTER. As shown on the graph, when x 7, y 0, so 7 is an x-intercept. When an x-intercept is an integer, then x minus that x-intercept is a factor of the trinomial. In this case, x 7 is a factor of 2x2 9x 35. We can then complete the factoring as (2x 5)(x 7), since we must multiply 7 by 5 to obtain 35. Find an x-intercept using a graphing calculator, and factor the trinomial. 1) 3x2 13x 10 2) 2x2 x 36 3) 5x2 16x 3 4) 2x2 13x 21 5) 3x2 23x 8 6) 7x2 11x 6 www.mhhe.com/messersmith SECTION 7.2 Factoring Trinomials 375
messersmith_power_intermediate_algebra_1e_ch4_7_10
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