Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Review Exercises - Page 400: 28



Work Step by Step

Given:$f(x) \cdot g(x)=(x-3)(2x+5)$ or,$=x(2x+5)-3(2x+5)$ or, $=2x^2+5x-6x-15$ Thus, $f(x) \cdot g(x)=2x^2-x-15$ Plug $x=-4$ in above equation, we have $f(-4) \cdot g(-4)=2(-4)^2-(-4)-15=36-15=21$
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