Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 43

Answer

$3(4k^2+9)(2k+3)(2k-3)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 48k^4-243 ,$ factor first the $GCF.$ Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $ 3 ,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 3(16k^4-81) .\end{array} The expressions $ 16k^4 $ and $ 81 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 16k^4-81 ,$ 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} 3[(4k^2)^2-(9)^2] \\\\= 3[(4k^2+9)(4k^2-9)] \\\\= 3(4k^2+9)(4k^2-9) .\end{array} The expressions $ 4k^2 $ and $ 9 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 4k^2-9 ,$ 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} 3(4k^2+9)[(2k)^2-(3)^2] \\\\= 3(4k^2+9)[(2k+3)(2k-3)] \\\\= 3(4k^2+9)(2k+3)(2k-3) .\end{array}
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