Solve by Factoring v^3-2v^2-16v=-32

$v^{3} - 2 v^{2} - 16 v = - 32$

Add $32$ to both sides of the equation.

$v^{3} - 2 v^{2} - 16 v + 32 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(v^{3} - 2 v^{2}) - 16 v + 32 = 0$

Factor out the greatest common factor (GCF) from each group.

$v^{2} (v - 2) - 16 (v - 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $v - 2$ .

$(v - 2) (v^{2} - 16) = 0$

Rewrite $16$ as $4^{2}$ .

$(v - 2) (v^{2} - 4^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = v$ and $b = 4$ .

$(v - 2) ((v + 4) (v - 4)) = 0$

Remove unnecessary parentheses.

$(v - 2) (v + 4) (v - 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$v - 2 = 0$

$v + 4 = 0$

$v - 4 = 0$

Set $v - 2$ equal to $0$ and solve for $v$ .

Set $v - 2$ equal to $0$ .

$v - 2 = 0$

Add $2$ to both sides of the equation.

$v = 2$

Set $v + 4$ equal to $0$ and solve for $v$ .

Set $v + 4$ equal to $0$ .

$v + 4 = 0$

Subtract $4$ from both sides of the equation.

$v = - 4$

Set $v - 4$ equal to $0$ and solve for $v$ .

Set $v - 4$ equal to $0$ .

$v - 4 = 0$

Add $4$ to both sides of the equation.

$v = 4$

The final solution is all the values that make $(v - 2) (v + 4) (v - 4) = 0$ true.

$v = 2, - 4, 4$