Solve x^3-x^2-20x

$$x ^ { 3 } - x ^ { 2 } - 20 x$$

Factor

$$x\left(x-5\right)\left(x+4\right)$$

Solution Steps

Factor out x.

$$x\left(x^{2}-x-20\right)$$

Consider x^{2}-x-20. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-20\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

Rewrite x^{2}-x-20 as \left(x^{2}-5x\right)+\left(4x-20\right).

$$\left(x^{2}-5x\right)+\left(4x-20\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x-5\right)+4\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$x\left(x-5\right)\left(x+4\right)$$

Evaluate

$$x\left(x-5\right)\left(x+4\right)$$