Solve by Factoring 9y^2+12y-5=0

$9 y^{2} + 12 y - 5 = 0$

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 $a \cdot c = 9 \cdot - 5 = - 45$ and whose sum is $b = 12$ .

Factor $12$ out of $12 y$ .

$9 y^{2} + 12 (y) - 5 = 0$

Rewrite $12$ as $- 3$ plus $15$

$9 y^{2} + (- 3 + 15) y - 5 = 0$

Apply the distributive property.

$9 y^{2} - 3 y + 15 y - 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(9 y^{2} - 3 y) + 15 y - 5 = 0$

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

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

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

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

Step 2

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

$3 y - 1 = 0$

$3 y + 5 = 0$

Step 3

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

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

$3 y - 1 = 0$

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

Add $1$ to both sides of the equation.

$3 y = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 4

Set $3 y + 5$ equal to $0$ and solve for $y$ .

Set $3 y + 5$ equal to $0$ .

$3 y + 5 = 0$

Solve $3 y + 5 = 0$ for $y$ .

Subtract $5$ from both sides of the equation.

$3 y = - 5$

Divide each term in $3 y = - 5$ by $3$ and simplify.

Divide each term in $3 y = - 5$ by $3$ .

$\frac{3 y}{3} = \frac{- 5}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{- 5}{3}$

Divide $y$ by $1$ .

$y = \frac{- 5}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{5}{3}$

Step 5

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

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

Solve by Factoring 9y^2+12y-5=0

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