Precalculus Examples

$x^{5} + x^{3} - 20 x = 0$

Step 1

Factor the left side of the equation.

Tap for more steps...

Step 1.1

Factor $x$ out of $x^{5} + x^{3} - 20 x$ .

Tap for more steps...

Step 1.1.1

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

$x \cdot x^{4} + x^{3} - 20 x = 0$

Step 1.1.2

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

$x \cdot x^{4} + x \cdot x^{2} - 20 x = 0$

Step 1.1.3

Factor $x$ out of $- 20 x$ .

$x \cdot x^{4} + x \cdot x^{2} + x \cdot - 20 = 0$

Step 1.1.4

Factor $x$ out of $x \cdot x^{4} + x \cdot x^{2}$ .

$x (x^{4} + x^{2}) + x \cdot - 20 = 0$

Step 1.1.5

Factor $x$ out of $x (x^{4} + x^{2}) + x \cdot - 20$ .

$x (x^{4} + x^{2} - 20) = 0$

Step 1.2

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

$x ({(x^{2})}^{2} + x^{2} - 20) = 0$

Step 1.3

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

$x (u^{2} + u - 20) = 0$

Step 1.4

Factor $u^{2} + u - 20$ using the AC method.

Tap for more steps...

Step 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 $- 20$ and whose sum is $1$ .

$- 4, 5$

Step 1.4.2

Write the factored form using these integers.

$x ((u - 4) (u + 5)) = 0$

Step 1.5

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

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

Step 1.6

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

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

Step 1.7

Factor.

Tap for more steps...

Step 1.7.1

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 $b = 2$ .

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

Step 1.7.2

Remove unnecessary parentheses.

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

Step 2

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

$x = 0$

$x + 2 = 0$

$x - 2 = 0$

$x^{2} + 5 = 0$

Step 3

Set $x$ equal to $0$ .

$x = 0$

Step 4

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

Tap for more steps...

Step 4.1

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

$x + 2 = 0$

Step 4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

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

Tap for more steps...

Step 5.1

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

$x - 2 = 0$

Step 5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 6

Set $x^{2} + 5$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 6.1

Set $x^{2} + 5$ equal to $0$ .

$x^{2} + 5 = 0$

Step 6.2

Solve $x^{2} + 5 = 0$ for $x$ .

Tap for more steps...

Step 6.2.1

Subtract $5$ from both sides of the equation.

$x^{2} = - 5$

Step 6.2.2

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

$x = \pm \sqrt{- 5}$

Step 6.2.3

Simplify $\pm \sqrt{- 5}$ .

Tap for more steps...

Step 6.2.3.1

Rewrite $- 5$ as $- 1 (5)$ .

$x = \pm \sqrt{- 1 (5)}$

Step 6.2.3.2

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{5}$

Step 6.2.3.3

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

$x = \pm i \sqrt{5}$

Step 6.2.4

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

Tap for more steps...

Step 6.2.4.1

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

$x = i \sqrt{5}$

Step 6.2.4.2

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

$x = - i \sqrt{5}$

Step 6.2.4.3

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

$x = i \sqrt{5}, - i \sqrt{5}$

Step 7

The final solution is all the values that make $x (x + 2) (x - 2) (x^{2} + 5) = 0$ true.

$x = 0, - 2, 2, i \sqrt{5}, - i \sqrt{5}$