Precalculus Examples

$x^{5} - 5 x^{3} + 4 x$

Step 1

Set $x^{5} - 5 x^{3} + 4 x$ equal to $0$ .

$x^{5} - 5 x^{3} + 4 x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor $x$ out of $x^{5} - 5 x^{3} + 4 x$ .

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Step 2.1.1.1

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

$x \cdot x^{4} - 5 x^{3} + 4 x = 0$

Step 2.1.1.2

Factor $x$ out of $- 5 x^{3}$ .

$x \cdot x^{4} + x (- 5 x^{2}) + 4 x = 0$

Step 2.1.1.3

Factor $x$ out of $4 x$ .

$x \cdot x^{4} + x (- 5 x^{2}) + x \cdot 4 = 0$

Step 2.1.1.4

Factor $x$ out of $x \cdot x^{4} + x (- 5 x^{2})$ .

$x (x^{4} - 5 x^{2}) + x \cdot 4 = 0$

Step 2.1.1.5

Factor $x$ out of $x (x^{4} - 5 x^{2}) + x \cdot 4$ .

$x (x^{4} - 5 x^{2} + 4) = 0$

Step 2.1.2

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

$x ({(x^{2})}^{2} - 5 x^{2} + 4) = 0$

Step 2.1.3

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

$x (u^{2} - 5 u + 4) = 0$

Step 2.1.4

Factor $u^{2} - 5 u + 4$ using the AC method.

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Step 2.1.4.1

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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Step 2.1.4.2

Write the factored form using these integers.

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

Step 2.1.5

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

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

Step 2.1.6

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

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

Step 2.1.7

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

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

Step 2.1.8

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

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

Step 2.1.9

Factor.

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Step 2.1.9.1

Factor.

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Step 2.1.9.1.1

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

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

Step 2.1.9.1.2

Remove unnecessary parentheses.

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

Step 2.1.9.2

Remove unnecessary parentheses.

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

Step 2.2

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

$x = 0$

$x + 2 = 0$

$x - 2 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

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

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Step 2.4.1

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

$x + 2 = 0$

Step 2.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.5

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

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Step 2.5.1

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

$x - 2 = 0$

Step 2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.6

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

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Step 2.6.1

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

$x + 1 = 0$

Step 2.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.7

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

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Step 2.7.1

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

$x - 1 = 0$

Step 2.7.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.8

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

$x = 0, - 2, 2, - 1, 1$

Step 3