Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 14

Answer

$3(9x^2+t^2)(3x+t)(3x-t)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 243x^4-3t^4 ,$ factor first the $GCF.$ Then, use the factoring of the difference of $2$ squares twice. $\bf{\text{Solution Details:}}$ Factoring the $GCF= 3 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3(81x^4-t^4) .\end{array} Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, the factored form of the expression above is \begin{array}{l}\require{cancel} 3(9x^2+t^2)(9x^2-t^2) .\end{array} The last factor is also a difference of $2$ squares. Using the factoring of the difference of $2$ squares again, the expression above is equivalent to \begin{array}{l}\require{cancel} 3(9x^2+t^2)(3x+t)(3x-t) .\end{array}
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