Solve x^3+3x^2-18x

$$x ^ { 3 } + 3 x ^ { 2 } - 18 x$$

Factor

$$x\left(x-3\right)\left(x+6\right)$$

Solution Steps

Factor out x.

$$x\left(x^{2}+3x-18\right)$$

Consider x^{2}+3x-18. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-18\right)=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite x^{2}+3x-18 as \left(x^{2}-3x\right)+\left(6x-18\right).

$$\left(x^{2}-3x\right)+\left(6x-18\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-3\right)+6\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$x\left(x-3\right)\left(x+6\right)$$

Evaluate

$$x\left(x-3\right)\left(x+6\right)$$