Solve by Factoring 3x^2+14x-40=0

$3 x^{2} + 14 x - 40 = 0$

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 $a \cdot c = 3 \cdot - 40 = - 120$ and whose sum is $b = 14$ .

Factor $14$ out of $14 x$ .

$3 x^{2} + 14 (x) - 40 = 0$

Rewrite $14$ as $- 6$ plus $20$

$3 x^{2} + (- 6 + 20) x - 40 = 0$

Apply the distributive property.

$3 x^{2} - 6 x + 20 x - 40 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 6 x) + 20 x - 40 = 0$

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

$3 x (x - 2) + 20 (x - 2) = 0$

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

$(x - 2) (3 x + 20) = 0$

Step 2

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

$x - 2 = 0$

$3 x + 20 = 0$

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 $3 x + 20$ equal to $0$ and solve for $x$ .

Set $3 x + 20$ equal to $0$ .

$3 x + 20 = 0$

Solve $3 x + 20 = 0$ for $x$ .

Subtract $20$ from both sides of the equation.

$3 x = - 20$

Divide each term in $3 x = - 20$ by $3$ and simplify.

Divide each term in $3 x = - 20$ by $3$ .

$\frac{3 x}{3} = \frac{- 20}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 20}{3}$

Divide $x$ by $1$ .

$x = \frac{- 20}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{20}{3}$

Step 5

The final solution is all the values that make $(x - 2) (3 x + 20) = 0$ true.

$x = 2, - \frac{20}{3}$

Solve by Factoring 3x^2+14x-40=0

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