Solve by Factoring 3x^2+14x-5=0

$3 x^{2} + 14 x - 5 = 0$

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 $a \cdot c = 3 \cdot - 5 = - 15$ and whose sum is $b = 14$ .

Factor $14$ out of $14 x$ .

$3 x^{2} + 14 (x) - 5 = 0$

Rewrite $14$ as $- 1$ plus $15$

$3 x^{2} + (- 1 + 15) x - 5 = 0$

Apply the distributive property.

$3 x^{2} - 1 x + 15 x - 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) + 15 x - 5 = 0$

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

$x (3 x - 1) + 5 (3 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(3 x - 1) (x + 5) = 0$

Step 2

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

$3 x - 1 = 0$

$x + 5 = 0$

Step 3

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 4

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 5

The final solution is all the values that make $(3 x - 1) (x + 5) = 0$ true.

$x = \frac{1}{3}, - 5$

Solve by Factoring 3x^2+14x-5=0

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