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 - Page 279: 47



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 5w^5x^3z+25w^3x^2z^2-120wxz^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= 5wxz $ 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} 5wxz(w^4x^2+5w^2xz-24z^2) .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 1(-24)=-24 $ and the value of $b$ is $ 5 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 8,-3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5wxz(w^4x^2+8w^2xz-3w^2xz-24z^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} 5wxz[(w^4x^2+8w^2xz)-(3w^2xz+24z^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5wxz[w^2x(w^2x+8z)-3z(w^2x+8z)] .\end{array} Factoring the $GCF= (w^2x+8z) $ of the entire expression above results to \begin{array}{l}\require{cancel} 5wxz[(w^2x+8z)(w^2x-3z)] \\\\= 5wxz(w^2x+8z)(w^2x-3z) .\end{array}
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