Solve 20xy^2-20xy-15x

$$20x { y }^{ 2 } -20xy-15x$$

Factor

$$5x\left(2y-3\right)\left(2y+1\right)$$

Solution Steps

Factor out 5.

$$5\left(4xy^{2}-4xy-3x\right)$$

Consider 4xy^{2}-4xy-3x. Factor out x.

$$x\left(4y^{2}-4y-3\right)$$

Consider 4y^{2}-4y-3. Factor the expression by grouping. First, the expression needs to be rewritten as 4y^{2}+ay+by-3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=4\left(-3\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite 4y^{2}-4y-3 as \left(4y^{2}-6y\right)+\left(2y-3\right).

$$\left(4y^{2}-6y\right)+\left(2y-3\right)$$

Factor out 2y in 4y^{2}-6y.

$$2y\left(2y-3\right)+2y-3$$

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(2y+1\right)$$

Rewrite the complete factored expression.

$$5x\left(2y-3\right)\left(2y+1\right)$$

Evaluate

$$5x\left(2y-3\right)\left(2y+1\right)$$