Find the Roots (Zeros) x(x^2+4x-12)

$x (x^{2} + 4 x - 12)$

Set $x (x^{2} + 4 x - 12)$ equal to $0$ .

$x (x^{2} + 4 x - 12) = 0$

Solve for $x$ .

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

$x = 0$

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

Set $x$ equal to $0$ .

$x = 0$

Set $x^{2} + 4 x - 12$ equal to $0$ and solve for $x$ .

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

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

Solve $x^{2} + 4 x - 12 = 0$ for $x$ .

Factor $x^{2} + 4 x - 12$ 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 $4$ .

$- 2, 6$

Write the factored form using these integers.

$(x - 2) (x + 6) = 0$

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

$x - 2 = 0$

$x + 6 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x - 2) (x + 6) = 0$ true.

$x = 2, - 6$

The final solution is all the values that make $x (x^{2} + 4 x - 12) = 0$ true.

$x = 0, 2, - 6$

Find the Roots (Zeros) x(x^2+4x-12)

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