Solve by Factoring 3x^2+13x-38=0

$3 x^{2} + 13 x - 38 = 0$

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 $a \cdot c = 3 \cdot - 38 = - 114$ and whose sum is $b = 13$ .

Factor $13$ out of $13 x$ .

$3 x^{2} + 13 (x) - 38 = 0$

Rewrite $13$ as $- 6$ plus $19$

$3 x^{2} + (- 6 + 19) x - 38 = 0$

Apply the distributive property.

$3 x^{2} - 6 x + 19 x - 38 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 6 x) + 19 x - 38 = 0$

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

$3 x (x - 2) + 19 (x - 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 2$ .

$(x - 2) (3 x + 19) = 0$

Step 2

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

$x - 2 = 0$

$3 x + 19 = 0$

Step 3

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 4

Set $3 x + 19$ equal to $0$ and solve for $x$ .

Set $3 x + 19$ equal to $0$ .

$3 x + 19 = 0$

Solve $3 x + 19 = 0$ for $x$ .

Subtract $19$ from both sides of the equation.

$3 x = - 19$

Divide each term in $3 x = - 19$ by $3$ and simplify.

Divide each term in $3 x = - 19$ by $3$ .

$\frac{3 x}{3} = \frac{- 19}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 19}{3}$

Divide $x$ by $1$ .

$x = \frac{- 19}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{19}{3}$

Step 5

The final solution is all the values that make $(x - 2) (3 x + 19) = 0$ true.

$x = 2, - \frac{19}{3}$

Solve by Factoring 3x^2+13x-38=0

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