Precalculus Examples

$x^{9} - 25 x^{5} + 144 x = 0$

Step 1

Factor $x$ out of $x^{9} - 25 x^{5} + 144 x$ .

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Step 1.1

Factor $x$ out of $x^{9}$ .

$x \cdot x^{8} - 25 x^{5} + 144 x = 0$

Step 1.2

Factor $x$ out of $- 25 x^{5}$ .

$x \cdot x^{8} + x (- 25 x^{4}) + 144 x = 0$

Step 1.3

Factor $x$ out of $144 x$ .

$x \cdot x^{8} + x (- 25 x^{4}) + x \cdot 144 = 0$

Step 1.4

Factor $x$ out of $x \cdot x^{8} + x (- 25 x^{4})$ .

$x (x^{8} - 25 x^{4}) + x \cdot 144 = 0$

Step 1.5

Factor $x$ out of $x (x^{8} - 25 x^{4}) + x \cdot 144$ .

$x (x^{8} - 25 x^{4} + 144) = 0$

Step 2

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

$x ({(x^{4})}^{2} - 25 x^{4} + 144) = 0$

Step 3

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

$x (u^{2} - 25 u + 144) = 0$

Step 4

Factor $u^{2} - 25 u + 144$ using the AC method.

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Step 4.1

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 $144$ and whose sum is $- 25$ .

$- 16, - 9$

Step 4.2

Write the factored form using these integers.

$x ((u - 16) (u - 9)) = 0$

Step 5

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

$x ((x^{4} - 16) (x^{4} - 9)) = 0$

Step 6

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

$x (({(x^{2})}^{2} - 16) (x^{4} - 9)) = 0$

Step 7

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

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

Step 8

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 $b = 4$ .

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

Step 9

Simplify.

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Step 9.1

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

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

Step 9.2

Factor.

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Step 9.2.1

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 $b = 2$ .

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

Step 9.2.2

Remove unnecessary parentheses.

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

Step 10

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

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

Step 11

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

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

Step 12

Factor.

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Step 12.1

Factor.

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Step 12.1.1

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

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

Step 12.1.2

Remove unnecessary parentheses.

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

Step 12.2

Remove unnecessary parentheses.

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

Step 13

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

$x = 0$

$x^{2} + 4 = 0$

$x + 2 = 0$

$x - 2 = 0$

$x^{2} + 3 = 0$

$x^{2} - 3 = 0$

Step 14

Set $x$ equal to $0$ .

$x = 0$

Step 15

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

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Step 15.1

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

$x^{2} + 4 = 0$

Step 15.2

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

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Step 15.2.1

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Step 15.2.2

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

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

Step 15.2.3

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

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Step 15.2.3.1

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

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

Step 15.2.3.2

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

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

Step 15.2.3.3

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

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

Step 15.2.3.4

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

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

Step 15.2.3.5

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

$x = \pm i \cdot 2$

Step 15.2.3.6

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

$x = \pm 2 i$

Step 15.2.4

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

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Step 15.2.4.1

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

$x = 2 i$

Step 15.2.4.2

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

$x = - 2 i$

Step 15.2.4.3

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

$x = 2 i, - 2 i$

Step 16

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

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Step 16.1

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

$x + 2 = 0$

Step 16.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 17

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

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Step 17.1

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

$x - 2 = 0$

Step 17.2

Add $2$ to both sides of the equation.

$x = 2$

Step 18

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

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Step 18.1

Set $x^{2} + 3$ equal to $0$ .

$x^{2} + 3 = 0$

Step 18.2

Solve $x^{2} + 3 = 0$ for $x$ .

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Step 18.2.1

Subtract $3$ from both sides of the equation.

$x^{2} = - 3$

Step 18.2.2

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

$x = \pm \sqrt{- 3}$

Step 18.2.3

Simplify $\pm \sqrt{- 3}$ .

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Step 18.2.3.1

Rewrite $- 3$ as $- 1 (3)$ .

$x = \pm \sqrt{- 1 (3)}$

Step 18.2.3.2

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{3}$

Step 18.2.3.3

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

$x = \pm i \sqrt{3}$

Step 18.2.4

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

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Step 18.2.4.1

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

$x = i \sqrt{3}$

Step 18.2.4.2

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

$x = - i \sqrt{3}$

Step 18.2.4.3

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

$x = i \sqrt{3}, - i \sqrt{3}$

Step 19

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

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Step 19.1

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Step 19.2

Solve $x^{2} - 3 = 0$ for $x$ .

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Step 19.2.1

Add $3$ to both sides of the equation.

$x^{2} = 3$

Step 19.2.2

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

$x = \pm \sqrt{3}$

Step 19.2.3

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

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Step 19.2.3.1

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

$x = \sqrt{3}$

Step 19.2.3.2

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

$x = - \sqrt{3}$

Step 19.2.3.3

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

$x = \sqrt{3}, - \sqrt{3}$

Step 20

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

$x = 0, 2 i, - 2 i, - 2, 2, i \sqrt{3}, - i \sqrt{3}, \sqrt{3}, - \sqrt{3}$