Solve for v v^2+8v+16=0

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

Step 1

Factor using the perfect square rule.

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

$v^{2} + 8 v + 4^{2} = 0$

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

$8 v = 2 \cdot v \cdot 4$

Rewrite the polynomial.

$v^{2} + 2 \cdot v \cdot 4 + 4^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = v$ and $b = 4$ .

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

Step 2

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

$v + 4 = 0$

Step 3

Subtract $4$ from both sides of the equation.

$v = - 4$