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: 35

Answer

$16(4b+5c)(4b-5c)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 256b^2-400c^2 ,$ 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 $ 16 ,$ 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} 16(16b^2-25c^2) .\end{array} The expressions $ 16b^2 $ and $ 25c^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 16b^2-25c^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} 16[(4b)^2-(5c)^2] \\\\= 16[(4b+5c)(4b-5c)] \\\\= 16(4b+5c)(4b-5c) .\end{array}
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