Precalculus Examples

$f (x) = 3 x^{3} - 12 x^{2} - 15 x$

Step 1

Set $3 x^{3} - 12 x^{2} - 15 x$ equal to $0$ .

$3 x^{3} - 12 x^{2} - 15 x = 0$

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 $3 x$ out of $3 x^{3} - 12 x^{2} - 15 x$ .

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Step 2.1.1.1

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

$3 x (x^{2}) - 12 x^{2} - 15 x = 0$

Step 2.1.1.2

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

$3 x (x^{2}) + 3 x (- 4 x) - 15 x = 0$

Step 2.1.1.3

Factor $3 x$ out of $- 15 x$ .

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

Step 2.1.1.4

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

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

Step 2.1.1.5

Factor $3 x$ out of $3 x (x^{2} - 4 x) + 3 x (- 5)$ .

$3 x (x^{2} - 4 x - 5) = 0$

Step 2.1.2

Factor.

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Step 2.1.2.1

Factor $x^{2} - 4 x - 5$ 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 $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Step 2.1.2.1.2

Write the factored form using these integers.

$3 x ((x - 5) (x + 1)) = 0$

Step 2.1.2.2

Remove unnecessary parentheses.

$3 x (x - 5) (x + 1) = 0$

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 $0$ .

$x = 0$

$x - 5 = 0$

$x + 1 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.4.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.5

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

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Step 2.5.1

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

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

The final solution is all the values that make $3 x (x - 5) (x + 1) = 0$ true.

$x = 0, 5, - 1$

Step 3