Solve 120x^3+244x^2+120x=

$$120 x ^ { 3 } + 244 x ^ { 2 } + 120 x =$$

Factor

$$4x\left(5x+6\right)\left(6x+5\right)$$

Solution Steps

Factor out 4.

$$4\left(30x^{3}+61x^{2}+30x\right)$$

Consider 30x^{3}+61x^{2}+30x. Factor out x.

$$x\left(30x^{2}+61x+30\right)$$

Consider 30x^{2}+61x+30. Factor the expression by grouping. First, the expression needs to be rewritten as 30x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=61 ab=30\times 30=900$$

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 900.

$$1,900 2,450 3,300 4,225 5,180 6,150 9,100 10,90 12,75 15,60 18,50 20,45 25,36 30,30$$

Calculate the sum for each pair.

$$1+900=901 2+450=452 3+300=303 4+225=229 5+180=185 6+150=156 9+100=109 10+90=100 12+75=87 15+60=75 18+50=68 20+45=65 25+36=61 30+30=60$$

The solution is the pair that gives sum 61.

$$a=25 b=36$$

Rewrite 30x^{2}+61x+30 as \left(30x^{2}+25x\right)+\left(36x+30\right).

$$\left(30x^{2}+25x\right)+\left(36x+30\right)$$

Factor out 5x in the first and 6 in the second group.

$$5x\left(6x+5\right)+6\left(6x+5\right)$$

Factor out common term 6x+5 by using distributive property.

$$\left(6x+5\right)\left(5x+6\right)$$

Rewrite the complete factored expression.

$$4x\left(6x+5\right)\left(5x+6\right)$$

Evaluate

$$4x\left(5x+6\right)\left(6x+5\right)$$