Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapters R-5 - Cumulative Review Exercises: 29

Answer

$(10x^2+9 )(10x^2-9 )$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 100x^4-81 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $ 100x^4 $ and $ 81 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 100x^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} (10x^2)^2-(9 )^2 \\\\= (10x^2+9 )(10x^2-9 ) .\end{array}
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