## Intermediate Algebra for College Students (7th Edition)

$c=\dfrac{-1}{2},-2,2$
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, $c=\dfrac{-1}{2},-2,2$