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 - 5.6 Factoring: A General Strategy - 5.6 Exercise Set: 67

Answer

$(2t+5)(4t-3)$

Work Step by Step

The given expression, $ 14t+8t^2-15 ,$ can be re-written as \begin{array}{l} 8t^2+14t-15 .\end{array} The two numbers whose product is $ac= 8(-15)=-120 $ and whose sum is $b= 14 $ are $\{ 20,-6 \}$. Using these two numbers to decompose the middle term of the expression, $ 8t^2+14t-15 ,$ then the factored form is \begin{array}{l} 8t^2+20t-6t-15 \\\\= (8t^2+20t)-(6t+15) \\\\= 4t(2t+5)-3(2t+5) \\\\= (2t+5)(4t-3) .\end{array}
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