Solve by Factoring 4y^3-7y^2+28=16y

$4 y^{3} - 7 y^{2} + 28 = 16 y$

Step 1

Subtract $16 y$ from both sides of the equation.

$4 y^{3} - 7 y^{2} + 28 - 16 y = 0$

Step 2

Reorder terms.

$4 y^{3} - 7 y^{2} - 16 y + 28 = 0$

Step 3

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 y^{3} - 7 y^{2}) - 16 y + 28 = 0$

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

$y^{2} (4 y - 7) - 4 (4 y - 7) = 0$

Step 4

Factor the polynomial by factoring out the greatest common factor, $4 y - 7$ .

$(4 y - 7) (y^{2} - 4) = 0$

Step 5

Rewrite $4$ as $2^{2}$ .

$(4 y - 7) (y^{2} - 2^{2}) = 0$

Step 6

Factor.

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

$(4 y - 7) ((y + 2) (y - 2)) = 0$

Remove unnecessary parentheses.

$(4 y - 7) (y + 2) (y - 2) = 0$

Step 7

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

$4 y - 7 = 0$

$y + 2 = 0$

$y - 2 = 0$

Step 8

Set $4 y - 7$ equal to $0$ and solve for $y$ .

Set $4 y - 7$ equal to $0$ .

$4 y - 7 = 0$

Solve $4 y - 7 = 0$ for $y$ .

Add $7$ to both sides of the equation.

$4 y = 7$

Divide each term in $4 y = 7$ by $4$ and simplify.

Divide each term in $4 y = 7$ by $4$ .

$\frac{4 y}{4} = \frac{7}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{7}{4}$

Divide $y$ by $1$ .

$y = \frac{7}{4}$

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

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

$y - 2 = 0$

Add $2$ to both sides of the equation.

$y = 2$

Step 11

The final solution is all the values that make $(4 y - 7) (y + 2) (y - 2) = 0$ true.

$y = \frac{7}{4}, - 2, 2$

Solve by Factoring 4y^3-7y^2+28=16y

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