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

Published by Pearson

# Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 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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