Solve by Factoring 3y^3-14y^2-5y=0

$3 y^{3} - 14 y^{2} - 5 y = 0$

Step 1

Factor $y$ out of $3 y^{3} - 14 y^{2} - 5 y$ .

Factor $y$ out of $3 y^{3}$ .

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

Factor $y$ out of $- 14 y^{2}$ .

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

Factor $y$ out of $- 5 y$ .

$y (3 y^{2}) + y (- 14 y) + y \cdot - 5 = 0$

Factor $y$ out of $y (3 y^{2}) + y (- 14 y)$ .

$y (3 y^{2} - 14 y) + y \cdot - 5 = 0$

Factor $y$ out of $y (3 y^{2} - 14 y) + y \cdot - 5$ .

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

Step 2

Factor.

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 $- 14 y$ .

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

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

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

Apply the distributive property.

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

Multiply $y$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

Step 3

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

$y = 0$

$3 y + 1 = 0$

$y - 5 = 0$

Step 4

Set $y$ equal to $0$ .

$y = 0$

Step 5

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 $y$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 6

Set $y - 5$ equal to $0$ and solve for $y$ .

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

Add $5$ to both sides of the equation.

$y = 5$

Step 7

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

$y = 0, - \frac{1}{3}, 5$

Solve by Factoring 3y^3-14y^2-5y=0

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