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: 24

Answer

$(8t-9)(t-1)$

Work Step by Step

Rearranging the terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 9+8t^2-18t \\\\= 8t^2-18t+9 .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 8t^2-18t+9 \end{array} has $ac= 8(9)=72 $ and $b= -18 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -9,-8 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 8t^2-9t-8t+9 .\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} (8t^2-9t)-(8t-9) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} t(8t-9)-(8t-9) .\end{array} Factoring the $GCF= (8t-9) $ of the entire expression above results to \begin{array}{l}\require{cancel} (8t-9)(t-1) .\end{array}
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