Solve by Factoring 3x^3+4x^2=12x+16

$3 x^{3} + 4 x^{2} = 12 x + 16$

Step 1

Move all the expressions to the left side of the equation.

Subtract $12 x$ from both sides of the equation.

$3 x^{3} + 4 x^{2} - 12 x = 16$

Subtract $16$ from both sides of the equation.

$3 x^{3} + 4 x^{2} - 12 x - 16 = 0$

Step 2

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{3} + 4 x^{2}) - 12 x - 16 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Step 3

Factor the polynomial by factoring out the greatest common factor, $3 x + 4$ .

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

Step 4

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

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

Step 5

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$(3 x + 4) ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$(3 x + 4) (x + 2) (x - 2) = 0$

Step 6

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

$3 x + 4 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 7

Set $3 x + 4$ equal to $0$ and solve for $x$ .

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

$3 x + 4 = 0$

Solve $3 x + 4 = 0$ 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.

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

Step 8

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 9

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 10

The final solution is all the values that make $(3 x + 4) (x + 2) (x - 2) = 0$ true.

$x = - \frac{4}{3}, - 2, 2$

Solve by Factoring 3x^3+4x^2=12x+16

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