Find the Roots (Zeros) f(x) equals three x to the power of three minus twelve x to the power of two minus fifteen x

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

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

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

Calculate $x$ .

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The left part of the equation should be factored.

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

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

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

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

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

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

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

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

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

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

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

Factor.

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

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Use the $x^{2} + b x + c$ form. Find a pair of integers whose product is $c$ and whose sum is $b$ . In such a case, whose product is $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Use the integers to write the factored form.

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

Delete unnecessary parentheses.

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

First, divide each term by $3$ and simplify.

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Divide each term in $3 x (x - 5) (x + 1) = 0$ by $3$ .

$\frac{3 x (x - 5) (x + 1)}{3} = \frac{0}{3}$

Clarify $\frac{3 x (x - 5) (x + 1)}{3}$ .

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Clarify terms.

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The common factor should be canceled of $3$ .

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Cancel the common factor.

$\frac{3 x (x - 5) (x + 1)}{3} = \frac{0}{3}$

Divide $(x (x - 5)) (x + 1)$ by $1$ .

$(x (x - 5)) (x + 1) = \frac{0}{3}$

Use the distributive law.

$(x \cdot x + x \cdot - 5) (x + 1) = \frac{0}{3}$

Clarify the expression.

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First, multiply $x$ by $x$ .

$(x^{2} + x \cdot - 5) (x + 1) = \frac{0}{3}$

Move $- 5$ to the left of $x$ .

$(x^{2} - 5 \cdot x) (x + 1) = \frac{0}{3}$

Expand $(x^{2} - 5 x) (x + 1)$ using the FOIL Method.

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Apply the distributive property.

$x^{2} (x + 1) - 5 x (x + 1) = \frac{0}{3}$

Apply the distributive property.

$x^{2} x + x^{2} \cdot 1 - 5 x (x + 1) = \frac{0}{3}$

Apply the distributive property.

$x^{2} x + x^{2} \cdot 1 - 5 x \cdot x - 5 x \cdot 1 = \frac{0}{3}$

Clarify and Add like terms.

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Clarify each term.

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Multiply $x^{2}$ by $x$ by adding the exponents.

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Multiply $x^{2}$ by $x$ .

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Raise $x$ to the power of $1$ .

$x^{2} x^{1} + x^{2} \cdot 1 - 5 x \cdot x - 5 x \cdot 1 = \frac{0}{3}$

Apply the power rule $a^{m} a^{n} = a^{m + n}$ to Add exponents.

$x^{2 + 1} + x^{2} \cdot 1 - 5 x \cdot x - 5 x \cdot 1 = \frac{0}{3}$

Add $2$ and $1$ .

$x^{3} + x^{2} \cdot 1 - 5 x \cdot x - 5 x \cdot 1 = \frac{0}{3}$

Multiply $x^{2}$ by $1$ .

$x^{3} + x^{2} - 5 x \cdot x - 5 x \cdot 1 = \frac{0}{3}$

Multiply $x$ by $x$ by adding the exponents.

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Put $x$ .

$x^{3} + x^{2} - 5 (x \cdot x) - 5 x \cdot 1 = \frac{0}{3}$

Multiply $x$ by $x$ .

$x^{3} + x^{2} - 5 x^{2} - 5 x \cdot 1 = \frac{0}{3}$

Multiply $- 5$ by $1$ .

$x^{3} + x^{2} - 5 x^{2} - 5 x = \frac{0}{3}$

Subtract $5 x^{2}$ from $x^{2}$ .

$x^{3} - 4 x^{2} - 5 x = \frac{0}{3}$

First, divide $0$ by $3$ .

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

Factor the left side of the equation.

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

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

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

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

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

Factor $x$ out of $- 5 x$ .

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

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

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

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

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

Factor.

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

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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 $- 4$ .

$- 5, 1$

Write the factored form using these integers.

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

Delete unnecessary parentheses.

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

In case any individual factor in the left part of the equation equals $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 5 = 0$

$x + 1 = 0$

Make the initial factor equal to $0$ .

$x = 0$

Make the following factor equal to $0$ and solve.

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Set the Then factor equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set the Then factor equal to $0$ and solve.

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Set the Then factor equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = 0, 5, - 1$

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