Solve x^3+7x^2+12x

$$x ^ { 3 } + 7 x ^ { 2 } + 12 x$$

Factor

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

Solution Steps

Factor out x.

$$x\left(x^{2}+7x+12\right)$$

Consider x^{2}+7x+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.

$$a+b=7 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite x^{2}+7x+12 as \left(x^{2}+3x\right)+\left(4x+12\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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