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: 57



Work Step by Step

Given: $f(c)=2c^3+c^2-8c+2$ Plug $f(c)=6$ then we have $2c^3+c^2-8c+2=6$ This can be written as: $2c^3+c^2-8c-4=0$ Need to use factorization to solve the quadratic equation . or, $(2c+1)(c^2-2^2)=0$ $(2c+1)(c+2)(c-2)=0$ This implies $(2c+1)=0$ Now, $c=\dfrac{-1}{2} $ and $c=-2,2$ Hence, $c=\dfrac{-1}{2},-2,2$
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