Solve for q 4q^2+8q+4=0

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

Factor the left side of the equation.

Factor $4$ out of $4 q^{2} + 8 q + 4$ .

Factor $4$ out of $4 q^{2}$ .

$4 (q^{2}) + 8 q + 4 = 0$

Factor $4$ out of $8 q$ .

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

Factor $4$ out of $4$ .

$4 (q^{2}) + 4 (2 q) + 4 (1) = 0$

Factor $4$ out of $4 (q^{2}) + 4 (2 q)$ .

$4 (q^{2} + 2 q) + 4 (1) = 0$

Factor $4$ out of $4 (q^{2} + 2 q) + 4 (1)$ .

$4 (q^{2} + 2 q + 1) = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$4 (q^{2} + 2 q + 1^{2}) = 0$

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

$2 q = 2 \cdot q \cdot 1$

Rewrite the polynomial.

$4 (q^{2} + 2 \cdot q \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = q$ and $b = 1$ .

$4 {(q + 1)}^{2} = 0$

Divide each term in $4 {(q + 1)}^{2} = 0$ by $4$ and simplify.

Divide each term in $4 {(q + 1)}^{2} = 0$ by $4$ .

$\frac{4 {(q + 1)}^{2}}{4} = \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 {(q + 1)}^{2}}{4} = \frac{0}{4}$

Divide ${(q + 1)}^{2}$ by $1$ .

${(q + 1)}^{2} = \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

${(q + 1)}^{2} = 0$

Set the $q + 1$ equal to $0$ .

$q + 1 = 0$

Subtract $1$ from both sides of the equation.

$q = - 1$