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: 58



Work Step by Step

Given: $f(c)=c^3+4c^2-c+6$ Plug $f(c)=10$ then we have $c^3+4c^2-c+6=10$ This can be written as: $c^3+4c^2-c-4=0$ Need to use factorization to solve the quadratic equation . or, $(c+4)(c^2-1)=0$ $(c+4)(c+1)(c-1)=0$ This implies $(c+4)=0$ Now, $c=-4 $ and $c=-1,1$ Hence, $c=-4,-1,1$
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