## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises - Page 348: 65

#### Answer

$(3m^2-5)(7m^2+1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $21m^4-32m^2-5 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 21 ,b= -32 ,\text{ and } c= -5 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 21(-5)=-105$ and whose sum is $b$ are $\left\{ -35,3 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 21m^4-35m^2+3m^2-5 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (21m^4-35m^2)+(3m^2-5) \\\\= 7m^2(3m^2-5)+(3m^2-5) \\\\= (3m^2-5)(7m^2+1) .\end{array}

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