## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 3 - Exponents, Polynomials and Functions - 3.5 Special Factoring Techniques - 3.5 Exercises - Page 279: 32

#### Answer

$(9m^2+4p^2)(3m+2p)(3m-2p)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $81m^4-16p^4 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $81m^4$ and $16p^4$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $81m^4-16p^4 ,$ 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} (9m^2)^2-(4p^2)^2 \\\\= (9m^2+4p^2)(9m^2-4p^2) .\end{array} The expressions $9m^2$ and $4p^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $9m^2-4p^2 ,$ 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} (9m^2+4p^2)[(3m)^2-(2p)^2] \\\\= (9m^2+4p^2)(3m+2p)(3m-2p) .\end{array}

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