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 - Mid-Chapter Review - Mixed Review - Page 335: 12



Work Step by Step

Factoring the $GCF= a ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} a-10a^2+25a^3 \\\\= a(1-10a+25a^2) \\\\= a(25a^2-10a+1) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} a(25a^2-10a+1) \end{array} has $ac= 25(1)=25 $ and $b= -10 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -5,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} a(25a^2-5a-5a+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[(25a^2-5a)-(5a-1)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a[5a(5a-1)-(5a-1)] .\end{array} Factoring the $GCF= (a^{n}-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} a[(5a-1)(5a-1)] \\\\= a(5a-1)^2 .\end{array}
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