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 - Section 5.7 - Polynomial Equations and Their Applications - Exercise Set - Page 391: 55



Work Step by Step

Given: $f(c)=c^2-4c-27$ Plug $f(c)=5$ then we have $c^2-4c-27=5$ This can be written as: $c^2-4c-32=0$ Need to use factorization to solve the quadratic equation . $c^2+4c-8c-32=5$ This implies $(c+4)(c-8)=0$ Now, $c+4=0 $ and $c-8=0$ Hence, $c=-4,8$
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