Precalculus Examples

$x^{8} - 32 x^{4} + 256 = 0$

Step 1

Rewrite $x^{8}$ as ${(x^{4})}^{2}$ .

${(x^{4})}^{2} - 32 x^{4} + 256 = 0$

Step 2

Let $u = x^{4}$ . Substitute $u$ for all occurrences of $x^{4}$ .

$u^{2} - 32 u + 256 = 0$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $256$ as $16^{2}$ .

$u^{2} - 32 u + 16^{2} = 0$

Step 3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$32 u = 2 \cdot u \cdot 16$

Step 3.3

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 16 + 16^{2} = 0$

Step 3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 16$ .

${(u - 16)}^{2} = 0$

Step 4

Replace all occurrences of $u$ with $x^{4}$ .

${(x^{4} - 16)}^{2} = 0$

Step 5

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

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

Step 6

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

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

Step 7

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 4$ .

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

Step 8

Simplify.

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Step 8.1

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

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

Step 8.2

Factor.

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Step 8.2.1

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 $b = 2$ .

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

Step 8.2.2

Remove unnecessary parentheses.

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

Step 9

Apply the product rule to $(x^{2} + 4) (x + 2) (x - 2)$ .

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

Step 10

Apply the product rule to $(x^{2} + 4) (x + 2)$ .

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

Step 11

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

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

${(x + 2)}^{2} = 0$

${(x - 2)}^{2} = 0$

Step 12

Set ${(x^{2} + 4)}^{2}$ equal to $0$ and solve for $x$ .

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Step 12.1

Set ${(x^{2} + 4)}^{2}$ equal to $0$ .

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

Step 12.2

Solve ${(x^{2} + 4)}^{2} = 0$ for $x$ .

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Step 12.2.1

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

$x^{2} + 4 = 0$

Step 12.2.2

Solve for $x$ .

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Step 12.2.2.1

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Step 12.2.2.2

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

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

Step 12.2.2.3

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

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Step 12.2.2.3.1

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

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

Step 12.2.2.3.2

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

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

Step 12.2.2.3.3

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

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

Step 12.2.2.3.4

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

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

Step 12.2.2.3.5

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

$x = \pm i \cdot 2$

Step 12.2.2.3.6

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

$x = \pm 2 i$

Step 12.2.2.4

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

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Step 12.2.2.4.1

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

$x = 2 i$

Step 12.2.2.4.2

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

$x = - 2 i$

Step 12.2.2.4.3

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

$x = 2 i, - 2 i$

Step 13

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

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Step 13.1

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

${(x + 2)}^{2} = 0$

Step 13.2

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

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Step 13.2.1

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

$x + 2 = 0$

Step 13.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 14

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

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Step 14.1

Set ${(x - 2)}^{2}$ equal to $0$ .

${(x - 2)}^{2} = 0$

Step 14.2

Solve ${(x - 2)}^{2} = 0$ for $x$ .

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Step 14.2.1

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

$x - 2 = 0$

Step 14.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 15

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

$x = 2 i, - 2 i, - 2, 2$