Solve x^3-3x^2-40x

$$x ^ { 3 } - 3 x ^ { 2 } - 40 x$$

Factor

$$x\left(x-8\right)\left(x+5\right)$$

Solution Steps

Factor out x.

$$x\left(x^{2}-3x-40\right)$$

Consider x^{2}-3x-40. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-40\right)=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=-8 b=5$$

Rewrite x^{2}-3x-40 as \left(x^{2}-8x\right)+\left(5x-40\right).

$$\left(x^{2}-8x\right)+\left(5x-40\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-8\right)+5\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+5\right)$$

Rewrite the complete factored expression.

$$x\left(x-8\right)\left(x+5\right)$$

Evaluate

$$x\left(x-8\right)\left(x+5\right)$$