## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set: 81

#### Answer

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

#### Work Step by Step

The two numbers whose product is $ac= 2(-9)=-18$ and whose sum is $b= 3$ are $\{ 6,-3 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $2x^6+3x^3-9$, is \begin{array}{l}\require{cancel} 2x^6+6x^3-3x^3-9 \\\\= (2x^6+6x^3)-(3x^3+9) \\\\= 2x^3(x^3+3)-3(x^3+3) \\\\= (x^3+3)(2x^3-3) .\end{array}

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