Solve 2x^3+20x^2+48x

$$2 x ^ { 3 } + 20 x ^ { 2 } + 48 x$$

Factor

$$2x\left(x+4\right)\left(x+6\right)$$

Solution Steps

Factor out 2.

$$2\left(x^{3}+10x^{2}+24x\right)$$

Consider x^{3}+10x^{2}+24x. Factor out x.

$$x\left(x^{2}+10x+24\right)$$

Consider x^{2}+10x+24. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\times 24=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 10.

$$a=4 b=6$$

Rewrite x^{2}+10x+24 as \left(x^{2}+4x\right)+\left(6x+24\right).

$$\left(x^{2}+4x\right)+\left(6x+24\right)$$

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

$$x\left(x+4\right)+6\left(x+4\right)$$

Factor out common term x+4 by using distributive property.

$$\left(x+4\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$2x\left(x+4\right)\left(x+6\right)$$

Evaluate

$$2x\left(x+4\right)\left(x+6\right)$$