Solve by Factoring 3x^2-5x-12=0

$3 x^{2} - 5 x - 12 = 0$

Step 1

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 12 = - 36$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$3 x^{2} - 5 x - 12 = 0$

Rewrite $- 5$ as $4$ plus $- 9$

$3 x^{2} + (4 - 9) x - 12 = 0$

Apply the distributive property.

$3 x^{2} + 4 x - 9 x - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} + 4 x) - 9 x - 12 = 0$

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

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

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

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

Step 2

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

$3 x + 4 = 0$

$x - 3 = 0$

Step 3

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 4

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

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

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

Solve by Factoring 3x^2-5x-12=0

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