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 - Page 326: 78



Work Step by Step

Factoring the $GCF= a^7 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 40a^8+16a^7+25a^9 \\\\= 25a^9+40a^8+16a^7 \\\\= a^7(25a^2+40a+16) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} a^7(25a^2+40a+16) \end{array} has $ac= 25(16)=400 $ and $b= 40 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 20,20 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} a^7(25a^2+20a+20a+16) .\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^7[(25a^2+20a)+(20a+16)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a^7[5a(5a+4)+4(5a+4)] .\end{array} Factoring the $GCF= (5a+4) $ of the entire expression above results to \begin{array}{l}\require{cancel} a^7[(5a+4)(5a+4)] \\\\= a^7(5a+4)^2 .\end{array}
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