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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set: 98

Answer

$a(a^{n}-1)^2$

Work Step by Step

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