Solve for x (x^2-9)(x^2+12x+36)=0

$(x^{2} - 9) (x^{2} + 12 x + 36) = 0$

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

$x^{2} - 9 = 0$

$x^{2} + 12 x + 36 = 0$

Set $x^{2} - 9$ equal to $0$ and solve for $x$ .

Set $x^{2} - 9$ equal to $0$ .

$x^{2} - 9 = 0$

Solve $x^{2} - 9 = 0$ for $x$ .

Add $9$ to both sides of the equation.

$x^{2} = 9$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

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

$x = \pm \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

Set $x^{2} + 12 x + 36$ equal to $0$ and solve for $x$ .

Set $x^{2} + 12 x + 36$ equal to $0$ .

$x^{2} + 12 x + 36 = 0$

Solve $x^{2} + 12 x + 36 = 0$ for $x$ .

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$x^{2} + 12 x + 6^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 6 + 6^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 6$ .

${(x + 6)}^{2} = 0$

Set the $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x^{2} - 9) (x^{2} + 12 x + 36) = 0$ true.

$x = 3, - 3, - 6$