Solve by Factoring 4y^3-2=y-8y^2

$4 y^{3} - 2 = y - 8 y^{2}$

Step 1

Move all the expressions to the left side of the equation.

Subtract $y$ from both sides of the equation.

$4 y^{3} - 2 - y = - 8 y^{2}$

Add $8 y^{2}$ to both sides of the equation.

$4 y^{3} - 2 - y + 8 y^{2} = 0$

Step 2

Reorder terms.

$4 y^{3} + 8 y^{2} - y - 2 = 0$

Step 3

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 y^{3} + 8 y^{2}) - y - 2 = 0$

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

$4 y^{2} (y + 2) - (y + 2) = 0$

Step 4

Factor the polynomial by factoring out the greatest common factor, $y + 2$ .

$(y + 2) (4 y^{2} - 1) = 0$

Step 5

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

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

Step 6

Rewrite $1$ as $1^{2}$ .

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

Step 7

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 y$ and $b = 1$ .

$(y + 2) ((2 y + 1) (2 y - 1)) = 0$

Remove unnecessary parentheses.

$(y + 2) (2 y + 1) (2 y - 1) = 0$

Step 8

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

$y + 2 = 0$

$2 y + 1 = 0$

$2 y - 1 = 0$

Step 9

Set $y + 2$ equal to $0$ and solve for $y$ .

Set $y + 2$ equal to $0$ .

$y + 2 = 0$

Subtract $2$ from both sides of the equation.

$y = - 2$

Step 10

Set $2 y + 1$ equal to $0$ and solve for $y$ .

Set $2 y + 1$ equal to $0$ .

$2 y + 1 = 0$

Solve $2 y + 1 = 0$ for $y$ .

Subtract $1$ from both sides of the equation.

$2 y = - 1$

Divide each term in $2 y = - 1$ by $2$ and simplify.

Divide each term in $2 y = - 1$ by $2$ .

$\frac{2 y}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{- 1}{2}$

Divide $y$ by $1$ .

$y = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{1}{2}$

Step 11

Set $2 y - 1$ equal to $0$ and solve for $y$ .

Set $2 y - 1$ equal to $0$ .

$2 y - 1 = 0$

Solve $2 y - 1 = 0$ for $y$ .

Add $1$ to both sides of the equation.

$2 y = 1$

Divide each term in $2 y = 1$ by $2$ and simplify.

Divide each term in $2 y = 1$ by $2$ .

$\frac{2 y}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{1}{2}$

Divide $y$ by $1$ .

$y = \frac{1}{2}$

Step 12

The final solution is all the values that make $(y + 2) (2 y + 1) (2 y - 1) = 0$ true.

$y = - 2, - \frac{1}{2}, \frac{1}{2}$

Solve by Factoring 4y^3-2=y-8y^2

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