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: 68

Answer

$(p-6q)(3p+2q)$

Work Step by Step

Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3p^2-16pq-12q^2 \end{array} has $ac= 3(-12)=-36 $ and $b= -16 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -18,2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3p^2-18pq+2pq-12q^2 .\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} (3p^2-18pq)+(2pq-12q^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3p(p-6q)+2q(p-6q) .\end{array} Factoring the $GCF= (p-6q) $ of the entire expression above results to \begin{array}{l}\require{cancel} (p-6q)(3p+2q) .\end{array}
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