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: 77

Answer

$a^6(3a+4)(3a+2)$

Work Step by Step

Factoring the $GCF= a^6 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 18a^7+8a^6+9a^8 \\\\= 9a^8+18a^7+8a^6 \\\\= a^6(9a^2+18a+8) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} a^6(9a^2+18a+8) \end{array} has $ac= 9(8)=72 $ and $b= 18 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 12,6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} a^6(9a^2+12a+6a+8) .\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^6[(9a^2+12a)+(6a+8)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a^6[3a(3a+4)+2(3a+4)] .\end{array} Factoring the $GCF= (3a+4) $ of the entire expression above results to \begin{array}{l}\require{cancel} a^6[(3a+4)(3a+2)] \\\\= a^6(3a+4)(3a+2) .\end{array}
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