Answer
$x=-\dfrac{1}{2}$, $x=2$ and $x=-2$ satisfy the given conditions
Work Step by Step
$y=2x^{3}+x^{2}-8x+2$ and $y=6$
Substitute $y$ by $6$:
$2x^{3}+x^{2}-8x+2=6$
Take the $6$ to the left side and simplify:
$2x^{3}+x^{2}-8x+2-6=0$
$2x^{3}+x^{2}-8x-4=0$
Group the first two terms together and the last two terms together:
$(2x^{3}+x^{2})-(8x+4)=0$
Take out common factor $x^{2}$ from the first group and common factor $4$ from the second group:
$x^{2}(2x+1)-4(2x+1)=0$
Factor out $2x+1$:
$(2x+1)(x^{2}-4)=0$
Factor $x^{2}-4$, which is a difference of squares:
$(2x+1)(x-2)(x+2)=0$
Set all factors equal to $0$ and solve each individual equation for $x$:
$2x+1=0$
$2x=-1$
$x=-\dfrac{1}{2}$
$x-2=0$
$x=2$
$x+2=0$
$x=-2$
$x=-\dfrac{1}{2}$, $x=2$ and $x=-2$ satisfy the given conditions
