Solve by Factoring 4y^4+7y^2-36=0

$4 y^{4} + 7 y^{2} - 36 = 0$

Step 1

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

$4 {(y^{2})}^{2} + 7 y^{2} - 36 = 0$

Step 2

Let $u = y^{2}$ . Substitute $u$ for all occurrences of $y^{2}$ .

$4 u^{2} + 7 u - 36 = 0$

Step 3

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 $a \cdot c = 4 \cdot - 36 = - 144$ and whose sum is $b = 7$ .

Factor $7$ out of $7 u$ .

$4 u^{2} + 7 (u) - 36 = 0$

Rewrite $7$ as $- 9$ plus $16$

$4 u^{2} + (- 9 + 16) u - 36 = 0$

Apply the distributive property.

$4 u^{2} - 9 u + 16 u - 36 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 u^{2} - 9 u) + 16 u - 36 = 0$

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

$u (4 u - 9) + 4 (4 u - 9) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 u - 9$ .

$(4 u - 9) (u + 4) = 0$

Step 4

Replace all occurrences of $u$ with $y^{2}$ .

$(4 y^{2} - 9) (y^{2} + 4) = 0$

Step 5

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

$({(2 y)}^{2} - 9) (y^{2} + 4) = 0$

Step 6

Rewrite $9$ as $3^{2}$ .

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

Step 7

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 $b = 3$ .

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

Step 8

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

$2 y + 3 = 0$

$2 y - 3 = 0$

$y^{2} + 4 = 0$

Step 9

Set $2 y + 3$ equal to $0$ and solve for $y$ .

Set $2 y + 3$ equal to $0$ .

$2 y + 3 = 0$

Solve $2 y + 3 = 0$ for $y$ .

Subtract $3$ from both sides of the equation.

$2 y = - 3$

Divide each term in $2 y = - 3$ by $2$ and simplify.

Divide each term in $2 y = - 3$ by $2$ .

$\frac{2 y}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{- 3}{2}$

Divide $y$ by $1$ .

$y = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{3}{2}$

Step 10

Set $2 y - 3$ equal to $0$ and solve for $y$ .

Set $2 y - 3$ equal to $0$ .

$2 y - 3 = 0$

Solve $2 y - 3 = 0$ for $y$ .

Add $3$ to both sides of the equation.

$2 y = 3$

Divide each term in $2 y = 3$ by $2$ and simplify.

Divide each term in $2 y = 3$ by $2$ .

$\frac{2 y}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{3}{2}$

Divide $y$ by $1$ .

$y = \frac{3}{2}$

Step 11

Set $y^{2} + 4$ equal to $0$ and solve for $y$ .

Set $y^{2} + 4$ equal to $0$ .

$y^{2} + 4 = 0$

Solve $y^{2} + 4 = 0$ for $y$ .

Subtract $4$ from both sides of the equation.

$y^{2} = - 4$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$y = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$y = \pm \sqrt{- 1} \cdot \sqrt{4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \pm i \cdot \sqrt{4}$

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

$y = \pm i \cdot \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$y = \pm 2 i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = 2 i$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 2 i$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 2 i, - 2 i$

Step 12

The final solution is all the values that make $(2 y + 3) (2 y - 3) (y^{2} + 4) = 0$ true.

$y = - \frac{3}{2}, \frac{3}{2}, 2 i, - 2 i$

Solve by Factoring 4y^4+7y^2-36=0

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