Because both terms in 6d 2 42d are divisible by 6, we could have started part b) by dividing by 6: 2 6 6d 42d 6 Divide by 6. d2 7d d2 7d 0 Write in standard form. d (d 7) 0 Factor. b R d 0 or d 7 0 Set each factor equal to zero. d 7 Solve. The solution set is {7, 0}. We get the same result. We cannot divide by d even though each term contains a factor of d. Doing so would eliminate the solution of zero. In general, we can divide an equation by a nonzero real number but we cannot divide an equation by a variable because we may eliminate a solution, and we may be dividing by zero. c) To solve k2 12(k 3), begin by writing the equation in standard form. k2 12k 36 Distribute. k2 12k 36 0 Write in standard form. (k 6)2 0 Factor. Because (k 6)2 0 means (k 6)(k 6) 0, setting each factor equal to zero will result in the same value for k. k 6 0 Set k 6 0. k 6 Solve. Check. The solution set is 566. d) We will have to perform several steps to write the equation in standard form. 2(x2 5) 5x 6x(x 1) 16 2x2 10 5x 6x2 6x 16 Distribute. Move the terms on the left side of the equation to the right side so that the coeffi cient of x2 is positive. 0 4x2 11x 6 Write in standard form. 0 (4x 3)(x 2) Factor. b R 4x 3 0 or x 2 0 Set each factor equal to zero. 4x 3 x 3 4 or x 2 Solve. The check is left to the student. The solution set is e 3 4 , 2 f . 396 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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