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: 27

Answer

$(25+x^2)(5+x)(5-x)$

Work Step by Step

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