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: 16

Answer

$t(t-3)(5t-6)$

Work Step by Step

Factoring the $GCF= t ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 5t^3-21t^2+18t \\\\= t(5t^2-21t+18) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} t(5t^2-21t+18) \end{array} has $ac= 5(18)=90 $ and $b= -21 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -15,-6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} t(5t^2-15t-6t+18) .\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[(5t^2-15t)-(6t-18)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} t[5t(t-3)-6(t-3)] .\end{array} Factoring the $GCF= (t-3) $ of the entire expression above results to \begin{array}{l}\require{cancel} t[(t-3)(5t-6)] \\\\= t(t-3)(5t-6) .\end{array}
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