Solve f(x)=x^3+2x^2-8x

$$f ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 8 x$$

Factor

$$x\left(x-2\right)\left(x+4\right)$$

Solution Steps

Factor out x.

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

Consider x^{2}+2x-8. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-8\right)=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite x^{2}+2x-8 as \left(x^{2}-2x\right)+\left(4x-8\right).

$$\left(x^{2}-2x\right)+\left(4x-8\right)$$

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

$$x\left(x-2\right)+4\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$x\left(x-2\right)\left(x+4\right)$$

Evaluate

$$x\left(x-2\right)\left(x+4\right)$$