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

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

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

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

Solve for $x$ .

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Factor the left side of the equation.

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

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

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

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

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

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

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

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

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

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

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

Factor.

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Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x = 0$

$x - 3 = 0$

$x + 4 = 0$

Set the first factor equal to $0$ .

$x = 0$

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

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Set the next factor equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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Set the next factor equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$x = 0, 3, - 4$

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

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