Find the Roots (Zeros) f(x)=x^3-x^2-12x

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

Step 1

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

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

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

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

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

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

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

Factor $x$ out of $- 12 x$ .

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

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

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

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

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

Factor.

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

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 $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$x ((x - 4) (x + 3)) = 0$

Remove unnecessary parentheses.

$x (x - 4) (x + 3) = 0$

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

$x = 0$

$x - 4 = 0$

$x + 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = 0, 4, - 3$

Step 3

Find the Roots (Zeros) f(x)=x^3-x^2-12x

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