Precalculus Examples

$f (x) = x^{5} - x^{4} + x^{3} - 3 x^{2} - 6 x$

Step 1

Set $x^{5} - x^{4} + x^{3} - 3 x^{2} - 6 x$ equal to $0$ .

$x^{5} - x^{4} + x^{3} - 3 x^{2} - 6 x = 0$

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 $x^{5} - x^{4} + x^{3} - 3 x^{2} - 6 x$ .

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Step 2.1.1.1

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

$x \cdot x^{4} - x^{4} + x^{3} - 3 x^{2} - 6 x = 0$

Step 2.1.1.2

Factor $x$ out of $- x^{4}$ .

$x \cdot x^{4} + x (- x^{3}) + x^{3} - 3 x^{2} - 6 x = 0$

Step 2.1.1.3

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

$x \cdot x^{4} + x (- x^{3}) + x \cdot x^{2} - 3 x^{2} - 6 x = 0$

Step 2.1.1.4

Factor $x$ out of $- 3 x^{2}$ .

$x \cdot x^{4} + x (- x^{3}) + x \cdot x^{2} + x (- 3 x) - 6 x = 0$

Step 2.1.1.5

Factor $x$ out of $- 6 x$ .

$x \cdot x^{4} + x (- x^{3}) + x \cdot x^{2} + x (- 3 x) + x \cdot - 6 = 0$

Step 2.1.1.6

Factor $x$ out of $x \cdot x^{4} + x (- x^{3})$ .

$x (x^{4} - x^{3}) + x \cdot x^{2} + x (- 3 x) + x \cdot - 6 = 0$

Step 2.1.1.7

Factor $x$ out of $x (x^{4} - x^{3}) + x \cdot x^{2}$ .

$x (x^{4} - x^{3} + x^{2}) + x (- 3 x) + x \cdot - 6 = 0$

Step 2.1.1.8

Factor $x$ out of $x (x^{4} - x^{3} + x^{2}) + x (- 3 x)$ .

$x (x^{4} - x^{3} + x^{2} - 3 x) + x \cdot - 6 = 0$

Step 2.1.1.9

Factor $x$ out of $x (x^{4} - x^{3} + x^{2} - 3 x) + x \cdot - 6$ .

$x (x^{4} - x^{3} + x^{2} - 3 x - 6) = 0$

Step 2.1.2

Regroup terms.

$x (- x^{3} - 3 x + x^{4} + x^{2} - 6) = 0$

Step 2.1.3

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

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Step 2.1.3.1

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

$x (- x \cdot x^{2} - 3 x + x^{4} + x^{2} - 6) = 0$

Step 2.1.3.2

Factor $- x$ out of $- 3 x$ .

$x (- x \cdot x^{2} - x \cdot 3 + x^{4} + x^{2} - 6) = 0$

Step 2.1.3.3

Factor $- x$ out of $- x (x^{2}) - x (3)$ .

$x (- x (x^{2} + 3) + x^{4} + x^{2} - 6) = 0$

Step 2.1.4

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

$x (- x (x^{2} + 3) + {(x^{2})}^{2} + x^{2} - 6) = 0$

Step 2.1.5

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

$x (- x (x^{2} + 3) + u^{2} + u - 6) = 0$

Step 2.1.6

Factor $u^{2} + u - 6$ using the AC method.

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Step 2.1.6.1

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 2.1.6.2

Write the factored form using these integers.

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

Step 2.1.7

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

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

Step 2.1.8

Factor $x^{2} + 3$ out of $- x (x^{2} + 3) + (x^{2} - 2) (x^{2} + 3)$ .

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Step 2.1.8.1

Factor $x^{2} + 3$ out of $- x (x^{2} + 3)$ .

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

Step 2.1.8.2

Factor $x^{2} + 3$ out of $(x^{2} - 2) (x^{2} + 3)$ .

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

Step 2.1.8.3

Factor $x^{2} + 3$ out of $(x^{2} + 3) (- x) + (x^{2} + 3) (x^{2} - 2)$ .

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

Step 2.1.9

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

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

Step 2.1.10

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

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Step 2.1.10.1

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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 2.1.10.2

Write the factored form using these integers.

$x ((x^{2} + 3) ((u - 2) (u + 1))) = 0$

Step 2.1.11

Factor.

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Step 2.1.11.1

Factor.

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Step 2.1.11.1.1

Replace all occurrences of $u$ with $x$ .

$x ((x^{2} + 3) ((x - 2) (x + 1))) = 0$

Step 2.1.11.1.2

Remove unnecessary parentheses.

$x ((x^{2} + 3) (x - 2) (x + 1)) = 0$

Step 2.1.11.2

Remove unnecessary parentheses.

$x (x^{2} + 3) (x - 2) (x + 1) = 0$

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 $0$ .

$x = 0$

$x^{2} + 3 = 0$

$x - 2 = 0$

$x + 1 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

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

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Step 2.4.1

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

$x^{2} + 3 = 0$

Step 2.4.2

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

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Step 2.4.2.1

Subtract $3$ from both sides of the equation.

$x^{2} = - 3$

Step 2.4.2.2

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

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

Step 2.4.2.3

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

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Step 2.4.2.3.1

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

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

Step 2.4.2.3.2

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

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

Step 2.4.2.3.3

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

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

Step 2.4.2.4

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

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Step 2.4.2.4.1

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

$x = i \sqrt{3}$

Step 2.4.2.4.2

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

$x = - i \sqrt{3}$

Step 2.4.2.4.3

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

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

Step 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

The final solution is all the values that make $x (x^{2} + 3) (x - 2) (x + 1) = 0$ true.

$x = 0, i \sqrt{3}, - i \sqrt{3}, 2, - 1$

Step 3