Solve x^3-x^2-12x

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

Factor

$$x\left(x-4\right)\left(x+3\right)$$

Solution Steps

Factor out x.

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

Consider x^{2}-x-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.

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

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 -1.

$$a=-4 b=3$$

Rewrite x^{2}-x-12 as \left(x^{2}-4x\right)+\left(3x-12\right).

$$\left(x^{2}-4x\right)+\left(3x-12\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-4\right)+3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+3\right)$$

Rewrite the complete factored expression.

$$x\left(x-4\right)\left(x+3\right)$$

Evaluate

$$x\left(x-4\right)\left(x+3\right)$$