Solve by Factoring 4x^4-13x^2+9=0

$4 x^{4} - 13 x^{2} + 9 = 0$

Step 1

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

$4 {(x^{2})}^{2} - 13 x^{2} + 9 = 0$

Step 2

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

$4 u^{2} - 13 u + 9 = 0$

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 $a \cdot c = 4 \cdot 9 = 36$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 u$ .

$4 u^{2} - 13 u + 9 = 0$

Rewrite $- 13$ as $- 4$ plus $- 9$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$4 u (u - 1) - 9 (u - 1) = 0$

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

$(u - 1) (4 u - 9) = 0$

Step 4

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

$(x^{2} - 1) (4 x^{2} - 9) = 0$

Step 5

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

$(x^{2} - 1^{2}) (4 x^{2} - 9) = 0$

Step 6

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

$(x + 1) (x - 1) (4 x^{2} - 9) = 0$

Step 7

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

$(x + 1) (x - 1) ({(2 x)}^{2} - 9) = 0$

Step 8

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

$(x + 1) (x - 1) ({(2 x)}^{2} - 3^{2}) = 0$

Step 9

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 3$ .

$(x + 1) (x - 1) ((2 x + 3) (2 x - 3)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 1) (2 x + 3) (2 x - 3) = 0$

Step 10

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

$x + 1 = 0$

$x - 1 = 0$

$2 x + 3 = 0$

$2 x - 3 = 0$

Step 11

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 12

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 13

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

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

$2 x + 3 = 0$

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

Subtract $3$ from both sides of the equation.

$2 x = - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 14

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

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

$2 x - 3 = 0$

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

Add $3$ to both sides of the equation.

$2 x = 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 15

The final solution is all the values that make $(x + 1) (x - 1) (2 x + 3) (2 x - 3) = 0$ true.

$x = - 1, 1, - \frac{3}{2}, \frac{3}{2}$

Solve by Factoring 4x^4-13x^2+9=0

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