## College Algebra (6th Edition)

$x=-4$, $x=1$ and $x=-1$ satisfy the given conditions
$y=x^{3}+4x^{2}-x+6$ and $y=10$ Substitute $y$ by $10$: $x^{3}+4x^{2}-x+6=10$ Take all terms to the left side and simplify: $x^{3}+4x^{2}-x+6-10=0$ $x^{3}+4x^{2}-x-4=0$ Group the first two terms together and the last two terms together: $(x^{3}+4x^{2})-(x+4)=0$ Take out common factor $x^{2}$ from the first group: $x^{2}(x+4)-(x+4)=0$ Factor out $x+4$: $(x+4)(x^{2}-1)=0$ Factor $x^{2}-1$: $(x+4)(x-1)(x+1)=0$ Set all terms equal to $0$ and solve each individual equation for $x$: $x+4=0$ $x=-4$ $x-1=0$ $x=1$ $x+1=0$ $x=-1$ $x=-4$, $x=1$ and $x=-1$ satisfy the given conditions