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