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: 52

Answer

$6mn^2(m-5)^2$

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 $GCF= 6mn^2 ,$ the given expression, $ 6m^3n^2-60m^2n^3+150mn^4 ,$ is equivalent to \begin{array}{l}\require{cancel} 6mn^2(m^2-10mn+25n^2) .\end{array} In the expression above, the value of $c$ is $ 25 $ and the value of $b$ is $ -10 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,25\}, \{5,5\}, \{-1,-25\}, \{-5,-5\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -5,-5 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} 6mn^2(m-5)(m-5) \\\\= 6mn^2(m-5)^2 .\end{array}
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