## College Algebra (6th Edition)

$x=-\dfrac{1}{2}$, $x=2$ and $x=-2$ satisfy the given conditions
$y=2x^{3}+x^{2}-8x+2$ 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