Find the Roots (Zeros) x^3+4x^2-9x-36

$x^{3} + 4 x^{2} - 9 x - 36$

Step 1

Set $x^{3} + 4 x^{2} - 9 x - 36$ equal to $0$ .

$x^{3} + 4 x^{2} - 9 x - 36 = 0$

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.

$(x^{3} + 4 x^{2}) - 9 x - 36 = 0$

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

$x^{2} (x + 4) - 9 (x + 4) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 4$ .

$(x + 4) (x^{2} - 9) = 0$

Rewrite $9$ as $3^{2}$ .

$(x + 4) (x^{2} - 3^{2}) = 0$

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 $b = 3$ .

$(x + 4) ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x + 4) (x + 3) (x - 3) = 0$

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

$x + 4 = 0$

$x + 3 = 0$

$x - 3 = 0$

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 $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x + 4) (x + 3) (x - 3) = 0$ true.

$x = - 4, - 3, 3$

Step 3