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 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set: 102

Answer

$a(4a-3)(3a+1)$

Work Step by Step

Factoring the $GCF=a,$ then \begin{array}{l}\require{cancel} 12a^3-5a^2-3a \\\\= a(12a^2-5a-3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$, the value of $ac$ in the trinomial expression above is $ 12(-3)=-36 $ and the value of $b$ is $ -5 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -9,4 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} a(12a^2-9a+4a-3) .\end{array} Grouping the first and second terms and the third and fourth terms, the expression above is equivalent to \begin{array}{l}\require{cancel} a[(12a^2-9a)+(4a-3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a[3a(4a-3)+(4a-3)] .\end{array} Factoring the $GCF= (4a-3) $ of the entire expression above results to \begin{array}{l}\require{cancel} a[(4a-3)(3a+1)] \\\\= a(4a-3)(3a+1) .\end{array}
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