Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set: 13

Answer

$(x^2+9)(x+3)(x-3)$

Work Step by Step

The expressions $ x^4 $ and $ 81 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^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} (x^2)^2-(9)^2 \\\\= (x^2+9)(x^2-9) .\end{array} The expressions $ x^2 $ and $ 9 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^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} (x^2+9)(x^2-9) \\\\= (x^2+9)[(x)^2-(3)^2] \\\\= (x^2+9)[(x+3)(x-3)] \\\\= (x^2+9)(x+3)(x-3) .\end{array}
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