Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.4 Factoring Polynomials - 3.4 Exercises: 72

Answer

$-3a(2b+5)(b-15)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ -6ab^2+75ab-225a ,$ factor first the negative $GCF.$ Then 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. $\bf{\text{Solution Details:}}$ Factoring the negative $GCF$ equal $ -3a ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -3a(2b^2-25b-75) .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 2(-75)=-150 $ and the value of $b$ is $ -25 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 5,-30 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -3a(2b^2+5b-30b-75) .\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} -3a[(2b^2+5b)-(30b+75)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -3a[b(2b+5)-15(2b+5)] .\end{array} Factoring the $GCF= (2b+5) $ of the entire expression above results to \begin{array}{l}\require{cancel} -3a[(2b+5)(b-15)] \\\\= -3a(2b+5)(b-15) .\end{array}
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