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

Published by Pearson

# Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 327: 97

#### Answer

$(a^{3n}-1)(3a^{3n}+1)$

#### Work Step by Step

Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3a^{6n}-2a^{3n}-1 \end{array} has $ac= 3(-1)=-3$ and $b= -2 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -3,1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3a^{6n}-3a^{3n}+1a^{3n}-1 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (3a^{6n}-3a^{3n})+(a^{3n}-1) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3a^{3n}(a^{3n}-1)+(a^{3n}-1) .\end{array} Factoring the $GCF= (a^{3n}-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (a^{3n}-1)(3a^{3n}+1) .\end{array}

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