Solve 30x^3y+125x^2y+120xy

$$30 x ^ { 3 } y + 125 x ^ { 2 } y + 120 x y$$

Factor

$$5xy\left(2x+3\right)\left(3x+8\right)$$

Solution Steps

Factor out 5.

$$5\left(6x^{3}y+25x^{2}y+24xy\right)$$

Consider 6x^{3}y+25x^{2}y+24xy. Factor out xy.

$$xy\left(6x^{2}+25x+24\right)$$

Consider 6x^{2}+25x+24. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=25 ab=6\times 24=144$$

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 144.

$$1,144 2,72 3,48 4,36 6,24 8,18 9,16 12,12$$

Calculate the sum for each pair.

$$1+144=145 2+72=74 3+48=51 4+36=40 6+24=30 8+18=26 9+16=25 12+12=24$$

The solution is the pair that gives sum 25.

$$a=9 b=16$$

Rewrite 6x^{2}+25x+24 as \left(6x^{2}+9x\right)+\left(16x+24\right).

$$\left(6x^{2}+9x\right)+\left(16x+24\right)$$

Factor out 3x in the first and 8 in the second group.

$$3x\left(2x+3\right)+8\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(3x+8\right)$$

Rewrite the complete factored expression.

$$5xy\left(2x+3\right)\left(3x+8\right)$$

Evaluate

$$5xy\left(2x+3\right)\left(3x+8\right)$$