Find the Roots (Zeros) f(x)=x^3+2x^2-16x-32

$f (x) = x^{3} + 2 x^{2} - 16 x - 32$

Step 1

Set $x^{3} + 2 x^{2} - 16 x - 32$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$x^{2} (x + 2) - 16 (x + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 2$ .

$(x + 2) (x^{2} - 16) = 0$

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

$(x + 2) (x^{2} - 4^{2}) = 0$

Factor.

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

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

Remove unnecessary parentheses.

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

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

$x + 2 = 0$

$x + 4 = 0$

$x - 4 = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

$x = - 2, - 4, 4$

Step 3