Solve 13x^{3}y^3+39x^3y^2-520x^3y

$$13 x ^ { 3 } y ^ { 3 } + 39 x ^ { 3 } y ^ { 2 } - 520 x ^ { 3 } y$$

Factor

$$13y\left(y-5\right)\left(y+8\right)x^{3}$$

Solution Steps

Factor out 13.

$$13\left(x^{3}y^{3}+3x^{3}y^{2}-40x^{3}y\right)$$

Consider x^{3}y^{3}+3x^{3}y^{2}-40x^{3}y. Factor out x^{3}y.

$$x^{3}y\left(y^{2}+3y-40\right)$$

Consider y^{2}+3y-40. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-40. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-40\right)=-40$$

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 -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=-5 b=8$$

Rewrite y^{2}+3y-40 as \left(y^{2}-5y\right)+\left(8y-40\right).

$$\left(y^{2}-5y\right)+\left(8y-40\right)$$

Factor out y in the first and 8 in the second group.

$$y\left(y-5\right)+8\left(y-5\right)$$

Factor out common term y-5 by using distributive property.

$$\left(y-5\right)\left(y+8\right)$$

Rewrite the complete factored expression.

$$13x^{3}y\left(y-5\right)\left(y+8\right)$$

Evaluate

$$13y\left(y-5\right)\left(y+8\right)x^{3}$$