Solve 2x^2y^2-7xy^2+3y^2

$$2 x ^ { 2 } y ^ { 2 } - 7 x y ^ { 2 } + 3 y ^ { 2 }$$

Factor

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

Solution Steps

Factor out y^{2}.

$$y^{2}\left(2x^{2}-7x+3\right)$$

Consider 2x^{2}-7x+3. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\times 3=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite 2x^{2}-7x+3 as \left(2x^{2}-6x\right)+\left(-x+3\right).

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

Factor out 2x in the first and -1 in the second group.

$$2x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(2x-1\right)$$

Rewrite the complete factored expression.

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

Evaluate

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