Solve x^{2}y^2-5x^2y+4x^2-y^2+5y-4

$$x ^ { 2 } y ^ { 2 } - 5 x ^ { 2 } y + 4 x ^ { 2 } - y ^ { 2 } + 5 y - 4$$

Factor

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

Solution Steps

Do the grouping x^{2}y^{2}-5x^{2}y+4x^{2}-y^{2}+5y-4=\left(x^{2}y^{2}-5x^{2}y+4x^{2}\right)+\left(-y^{2}+5y-4\right), and factor out x^{2} in the first and -1 in the second group.

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

Factor out common term y^{2}-5y+4 by using distributive property.

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

Consider y^{2}-5y+4. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 4=4$$

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 4.

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

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

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

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

Factor out y in the first and -1 in the second group.

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

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

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

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Rewrite the complete factored expression.

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

Evaluate

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