Solve 20xy^2+y^2-x+20y^3+xy-y

$$20 x y ^ { 2 } + y ^ { 2 } - x + 20 y ^ { 3 } + x y - y$$

Factor

$$\left(5y-1\right)\left(4y+1\right)\left(x+y\right)$$

Solution Steps

Do the grouping 20xy^{2}+y^{2}-x+20y^{3}+xy-y=\left(20xy^{2}+xy-x\right)+\left(20y^{3}+y^{2}-y\right), and factor out x in the first and y in the second group.

$$x\left(20y^{2}+y-1\right)+y\left(20y^{2}+y-1\right)$$

Factor out common term 20y^{2}+y-1 by using distributive property.

$$\left(20y^{2}+y-1\right)\left(x+y\right)$$

Consider 20y^{2}+y-1. Factor the expression by grouping. First, the expression needs to be rewritten as 20y^{2}+ay+by-1. To find a and b, set up a system to be solved.

$$a+b=1 ab=20\left(-1\right)=-20$$

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 -20.

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

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite 20y^{2}+y-1 as \left(20y^{2}-4y\right)+\left(5y-1\right).

$$\left(20y^{2}-4y\right)+\left(5y-1\right)$$

Factor out 4y in 20y^{2}-4y.

$$4y\left(5y-1\right)+5y-1$$

Factor out common term 5y-1 by using distributive property.

$$\left(5y-1\right)\left(4y+1\right)$$

Rewrite the complete factored expression.

$$\left(5y-1\right)\left(4y+1\right)\left(x+y\right)$$

Evaluate

$$\left(5y-1\right)\left(4y+1\right)\left(x+y\right)$$