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 - 3.5 Special Factoring Techniques - 3.5 Exercises: 48

Answer

$9abc(a^2b+3c)(a^2b-c)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 9a^5b^3c+18a^3b^2c^2-27abc^3 ,$ factor first the $GCF.$ 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. $\bf{\text{Solution Details:}}$ The $GCF$ of the terms is $GCF= 9abc $ since it is the highest expression that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 9abc(a^4b^2+2a^2bc-3c^2) .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 1(-3)=-3 $ and the value of $b$ is $ 2 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 3,-1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 9abc(a^4b^2+3a^2bc-1a^2bc-3c^2) .\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} 9abc[(a^4b^2+3a^2bc)-(1a^2bc+3c^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 9abc[a^2b(a^2b+3c)-c(a^2b+3c)] .\end{array} Factoring the $GCF= (a^2b+3c) $ of the entire expression above results to \begin{array}{l}\require{cancel} 9abc[(a^2b+3c)(a^2b-c)] \\\\= 9abc(a^2b+3c)(a^2b-c) .\end{array}
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