Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises: 51

Answer

$2xy^3(x-12y)^2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, factor the GCF of the terms. Then, to factor the quadratic expression $x^2+bx+c,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$. Express the factored form as $GCF(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ Factoring the $GCF= 2xy^3 ,$ the given expression, $ 2x^3y^3-48x^2y^4+288xy^5 ,$ is equivalent to \begin{array}{l}\require{cancel} 2xy^3(x^2-24xy+144y^2) .\end{array} In the expression above, the value of $c$ is $ 144 $ and the value of $b$ is $ -24 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,144\}, \{2,72\}, \{3,48\}, \{4,38\}, \{6,24\}, \{8,18\}, \{9,16\}, \{12,12\}, \{-1,-144\}, \{-2,-72\}, \{-3,-48\}, \{-4,-38\}, \{-6,-24\}, \{-8,-18\}, \{-9,-16\}, \{-12,-12\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -12,-12 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} 2xy^3(x-12y)(x-12y) \\\\= 2xy^3(x-12y)^2 .\end{array}
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