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: 46

Answer

$6(a^2+4b^2)(a+2b)(a-2b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 6a^4-96b^4 ,$ factor first the $GCF.$ Then use the factoring of the sum or difference of $2$ squares. $\bf{\text{Solution Details:}}$ The $GCF$ of the terms is $GCF= 6 $ since it is the highest number that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6(a^4-16b^4) .\end{array} The expressions $ a^4 $ and $ 16b^4 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ a^4-16b^4 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares, which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6[(a^2)^2-(4b^2)^2] \\\\= 6(a^2+4b^2)(a^2-4b^2) .\end{array} The expressions $ a^2 $ and $ 4b^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ a^2-4b^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares, which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6(a^2+4b^2)[(a)^2-(2b)^2] \\\\= 6(a^2+4b^2)(a+2b)(a-2b) .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.