Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 6 - Algebra: Equations and Inequalities - 6.5 Quadratic Equations - Exercise Set 6.5 - Page 399: 23



Work Step by Step

$2x^{2}$ - 17x + 30 step 1. Find two first terms whose product is $2x^{2}$ $2x^{2}$ -17x +30 =(2x__)(x__) Step 2. To find the second term of each factor, we must find two integers whose product is 30 and whose sum is -17 List pairs of factors of the constant, 30 (1,30)(-1,-30)(2,15)(-2,-15)(5,6)(-5,-6) step 3. The correct factorization of $2x^{2}$ -17x +30 is the one in which the sum of the Outside and Inside products is equal to -17x list of the possible factorization : (x-1)(2x-30)= $2x^{2}$ -32x +30 (x-2)(2x-15) = $2x^{2}$ -19x +30 (x-6)(2x-5) = $2x^{2}$ -17x +30 So, (x-6)(2x-5) is the solution Verification using FOIL Two binomials can be quickly multiplied by using the FOIL method, in which F represents the product of the first terms in each binomial, O represents the product of the outside terms, I represents the product of the two inside terms, and L represents the product of the last, (x-6)(2x-5) F = x.2x = $2x^{2}$ O = x.-5 = -5x I = -6.2x = -12x L = -6.-5= 30 (x-6)(2x-5) = $2x^{2}$ -5x -12x +30 = $2x^{2}$ -17x +30
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