Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapter 5 Test: 4

Answer

$-(x-4)(2x+9)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ -2x^2-x+36 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ To factor the trinomial expression above, note that the value of $ac$ is $ -2(36)=-72 $ and the value of $b$ is $ -1 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-72 \}, \{ 2,-36 \}, \{ 3,-24 \}, \{ 4,-18 \}, \{ 6,-12 \}, \{ 8,-9 \}, \\ \{ -1,72 \}, \{ -2,36 \}, \{ -3,24 \}, \{ -4,18 \}, \{ -6,12 \}, \{ -8,9 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 8,-9 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -2x^2+8x-9x+36 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (-2x^2+8x)-(9x-36) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -2x(x-4)-9(x-4) .\end{array} Factoring the $GCF= (x-4) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x-4)(-2x-9) \\\\= (x-4)(-1)(2x+9) \\\\= -(x-4)(2x+9) .\end{array}
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