Solve by Factoring 9x^2+24x+16=0

$9 x^{2} + 24 x + 16 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

${(3 x)}^{2} + 24 x + 16 = 0$

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

${(3 x)}^{2} + 24 x + 4^{2} = 0$

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

$24 x = 2 \cdot (3 x) \cdot 4$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 3 x$ and $b = 4$ .

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

Step 2

Set the $3 x + 4$ equal to $0$ .

$3 x + 4 = 0$

Step 3

Solve for $x$ .

Subtract $4$ from both sides of the equation.

$3 x = - 4$

Divide each term in $3 x = - 4$ by $3$ and simplify.

Divide each term in $3 x = - 4$ by $3$ .

$\frac{3 x}{3} = \frac{- 4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 4}{3}$

Divide $x$ by $1$ .

$x = \frac{- 4}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{4}{3}$