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: 69

Answer

$2(s+t)(4s+7t)$

Work Step by Step

Factoring the $GCF= 2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 8s^2+22st+14t^2 \\\\= 2(4s^2+11st+7t^2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 2(4s^2+11st+7t^2) \end{array} has $ac= 4(7)=28 $ and $b= 11 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 4,7 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2(4s^2+4st+7st+7t^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} 2[(4s^2+4st)+(7st+7t^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2[4s(s+t)+7t(s+t)] .\end{array} Factoring the $GCF= (s+t) $ of the entire expression above results to \begin{array}{l}\require{cancel} 2[(s+t)(4s+7t)] \\\\= 2(s+t)(4s+7t) .\end{array}
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