Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 9

Answer

$(b-3)(6b+1)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 6b^2-17b-3 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 6 ,b= -17 ,\text{ and } c= -3 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 6(-3)=-18 $ and whose sum is $b$ are $\left\{ -18,1 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 6b^2-18b+b-3 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (6b^2-18b)+(b-3) \\\\= 6b(b-3)+(b-3) \\\\= (b-3)(6b+1) .\end{array}
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