Solve 8x^4y+6x^3y-2x^2y

$$8 x ^ { 4 } y + 6 x ^ { 3 } y - 2 x ^ { 2 } y$$

Factor

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

Solution Steps

Factor out 2.

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

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

$$x^{2}y\left(4x^{2}+3x-1\right)$$

Consider 4x^{2}+3x-1. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=3 ab=4\left(-1\right)=-4$$

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 -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite 4x^{2}+3x-1 as \left(4x^{2}-x\right)+\left(4x-1\right).

$$\left(4x^{2}-x\right)+\left(4x-1\right)$$

Factor out x in 4x^{2}-x.

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

Factor out common term 4x-1 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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