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

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

Step 1

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

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

Step 2

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 $0$ .

$x + 3 = 0$

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

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.

$x = - 3$

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

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

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

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

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.

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

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

$x - 4 = 0$

$x + 3 = 0$

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

$x = 4, - 3$

The final solution is all the values that make $(x + 3) (x^{2} - x - 12) = 0$ true.

$x = - 3, 4$

Step 3

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

Related Questions

Support LCMGCF.com by disabling your adblocker.