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: 32

Answer

$(2z-k)(7z+2k)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 14z^2-3zk-2k^2 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 14 ,b= -3 ,\text{ and } c= -2 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 14(-2)=-28 $ and whose sum is $b$ are $\left\{ -7,4 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 14z^2-7zk+4zk-2k^2 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (14z^2-7zk)+(4zk-2k^2) \\\\= 7z(2z-k)+2k(2z-k) \\\\= (2z-k)(7z+2k) .\end{array}
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