Solve for x x^2+16x+64=0

$x^{2} + 16 x + 64 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$x^{2} + 16 x + 8^{2} = 0$

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

$16 x = 2 \cdot x \cdot 8$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 8 + 8^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 8$ .

${(x + 8)}^{2} = 0$

Step 2

Set the $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 3

Subtract $8$ from both sides of the equation.

$x = - 8$