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: 17

Answer

$t(t+2)(7t+1)$

Work Step by Step

Factoring the $GCF= t ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 7t^3+15t^2+2t \\\\= t(7t^2+15t+2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} t(7t^2+15t+2) \end{array} has $ac= 7(2)=14 $ and $b= 15 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 14,1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} t(7t^2+14t+1t+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} t[(7t^2+14t)+(1t+2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} t[7t(t+2)+(t+2)] .\end{array} Factoring the $GCF= (t+2) $ of the entire expression above results to \begin{array}{l}\require{cancel} t[(t+2)(7t+1)] \\\\= t(t+2)(7t+1) .\end{array}
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