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: 43

Answer

$(4t-3)(2t-7)$

Work Step by Step

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 8t^2-6t-28t+21 \\\\= (8t^2-6t)-(28t-21) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2t(4t-3)-7(4t-3) .\end{array} Factoring the $GCF= (4t-3) $ of the entire expression above results to \begin{array}{l}\require{cancel} (4t-3)(2t-7) .\end{array}
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