Solve 4xy^2+4xy-80x

$$4 x y ^ { 2 } + 4 x y - 80 x$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(xy^{2}+xy-20x\right)$$

Consider xy^{2}+xy-20x. Factor out x.

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

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

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

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

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

Factor out y in the first and 5 in the second group.

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

Factor out common term y-4 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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