Precalculus Examples

$f (x) = x^{4} - x^{3} - 12 x^{2}$

Step 1

Set $x^{4} - x^{3} - 12 x^{2}$ equal to $0$ .

$x^{4} - x^{3} - 12 x^{2} = 0$

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 $x^{2}$ out of $x^{4} - x^{3} - 12 x^{2}$ .

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Step 2.1.1.1

Factor $x^{2}$ out of $x^{4}$ .

$x^{2} x^{2} - x^{3} - 12 x^{2} = 0$

Step 2.1.1.2

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

$x^{2} x^{2} + x^{2} (- x) - 12 x^{2} = 0$

Step 2.1.1.3

Factor $x^{2}$ out of $- 12 x^{2}$ .

$x^{2} x^{2} + x^{2} (- x) + x^{2} \cdot - 12 = 0$

Step 2.1.1.4

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

$x^{2} (x^{2} - x) + x^{2} \cdot - 12 = 0$

Step 2.1.1.5

Factor $x^{2}$ out of $x^{2} (x^{2} - x) + x^{2} \cdot - 12$ .

$x^{2} (x^{2} - x - 12) = 0$

Step 2.1.2

Factor.

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Step 2.1.2.1

Factor $x^{2} - x - 12$ using the AC method.

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Step 2.1.2.1.1

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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Step 2.1.2.1.2

Write the factored form using these integers.

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

Step 2.1.2.2

Remove unnecessary parentheses.

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

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 $0$ .

$x^{2} = 0$

$x - 4 = 0$

$x + 3 = 0$

Step 2.3

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

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Step 2.3.1

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

$x^{2} = 0$

Step 2.3.2

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

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Step 2.3.2.1

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

$x = \pm \sqrt{0}$

Step 2.3.2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.3.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 2.3.2.2.2

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

$x = \pm 0$

Step 2.3.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.4

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 2.4.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.6

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

$x = 0, 4, - 3$

Step 3