Try 4 and 5. Do not put 4 in the same binomial as 2c since then it would be possible to factor out 2. 2c2 3c 20 (2c 5)(c 4) 5c 8c 3c This must equal 3c. Incorrect Only the sign of the sum is incorrect. Change the signs in the binomials to get the correct sum. 2c2 3c 20 (2c 5)(c 4) 5c (8c) 3c Correct Remember that we factored out 1 to begin the problem. 2c2 3c 20 1(2c2 3c 20) (2c 5)(c 4) Check by multiplying. YOU TRY 4 Factor completely. a) 10y2 58y 40 b) 4n2 5n 6 We have seen two methods for factoring ax2 bx c (a 1): factoring by grouping and factoring by trial and error. In either case, remember to begin by taking out a common factor from all terms whenever possible. 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; 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 whether 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 11x 4 2) 2x2 x 15 3) 5x2 6x 8 4) 2x2 5x 3 5) 4x2 3x 10 6) 14x2 x 4 410 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_introductory_algebra_1e_ch4_7_10
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