Solve x^3-4x^2-12x=

$$x ^ { 3 } - 4 x ^ { 2 } - 12 x =$$

Factor

$$x\left(x-6\right)\left(x+2\right)$$

Solution Steps

Factor out x.

$$x\left(x^{2}-4x-12\right)$$

Consider x^{2}-4x-12. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite x^{2}-4x-12 as \left(x^{2}-6x\right)+\left(2x-12\right).

$$\left(x^{2}-6x\right)+\left(2x-12\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-6\right)+2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$x\left(x-6\right)\left(x+2\right)$$

Evaluate

$$x\left(x-6\right)\left(x+2\right)$$