Solve by Factoring x^4+15x^2-16=0

$x^{4} + 15 x^{2} - 16 = 0$

Step 1

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

${(x^{2})}^{2} + 15 x^{2} - 16 = 0$

Step 2

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

$u^{2} + 15 u - 16 = 0$

Step 3

Factor $u^{2} + 15 u - 16$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 16$ and whose sum is $15$ .

$- 1, 16$

Write the factored form using these integers.

$(u - 1) (u + 16) = 0$

Step 4

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

$(x^{2} - 1) (x^{2} + 16) = 0$

Step 5

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

$(x^{2} - 1^{2}) (x^{2} + 16) = 0$

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 $b = 1$ .

$(x + 1) (x - 1) (x^{2} + 16) = 0$

Step 7

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

$x + 1 = 0$

$x - 1 = 0$

$x^{2} + 16 = 0$

Step 8

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 9

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 10

Set $x^{2} + 16$ equal to $0$ and solve for $x$ .

Set $x^{2} + 16$ equal to $0$ .

$x^{2} + 16 = 0$

Solve $x^{2} + 16 = 0$ for $x$ .

Subtract $16$ from both sides of the equation.

$x^{2} = - 16$

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

$x = \pm \sqrt{- 16}$

Simplify $\pm \sqrt{- 16}$ .

Rewrite $- 16$ as $- 1 (16)$ .

$x = \pm \sqrt{- 1 (16)}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{16}$

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

$x = \pm i \cdot \sqrt{16}$

Rewrite $16$ as $4^{2}$ .

$x = \pm i \cdot \sqrt{4^{2}}$

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

$x = \pm i \cdot 4$

Move $4$ to the left of $i$ .

$x = \pm 4 i$

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.

$x = 4 i$

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

$x = - 4 i$

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

$x = 4 i, - 4 i$

Step 11

The final solution is all the values that make $(x + 1) (x - 1) (x^{2} + 16) = 0$ true.

$x = - 1, 1, 4 i, - 4 i$

Solve by Factoring x^4+15x^2-16=0

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