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: 31

Answer

$(4b^2+25c^2)(2b+5c)(2b-5c)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 16b^4-625c^4 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ 16b^4 $ and $ 625c^4 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 16b^4-625c^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} (4b^2)^2-(25c^2)^2 \\\\= (4b^2+25c^2)(4b^2-25c^2) .\end{array} The expressions $ 4b^2 $ and $ 25c^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 4b^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} (4b^2+25c^2)[(2b)^2-(5c)^2] \\\\= (4b^2+25c^2)(2b+5c)(2b-5c) .\end{array}
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