Is there a way to factor $uv-u-v-1$? $uv-u-v-1$. I tried $(u+1)(v-1)$ and it's almost correct but I can't quite get it/ This is part of a limit problem I'm doing. Thanks!
 A: This polynomial in $u$ and $v$ is irreducible (not factorable). Observe that since it is degree 2, it can only factor nontrivially as the product of two linear polynomials in $u$ and $v$. So $$uv-u-v-1=(au+bv+c)(du+ev+f)$$ Without loss of generality we can rescale so that $a=1$. Then $d$ must be $0$, so that the product has no $u^2$ term. Since the product is nontrivial, $e$ must now be nonzero, and therefore $b=0$ so that there is no $v^2$ term. Now
$$uv-u-v-1=(u+c)(ev+f)$$ forcing $e$ to equal 1 to mach the coefficient of $uv$. Lastly, to match coefficients on $u$ and $v$, we have $c=f=-1$, but this is in contradiction with the constant term.
A: Condition for $ax^{2}+by^{2}+2hxy+2gx+2fy+c$ to have linear factors $\left|\array{a&h&g\\h&b&f\\g&f&c}\right|=0$.
In the given expression $a=0,b=0,h=\frac{1}{2},g=\frac{-1}{2},f=\frac{-1}{2},c=-1.$
The  expression does not satisfy the condition.
A: $$uv - u - v - 1 = (u-1)(v-1) - 2$$
or $$uv - u - v - 1 = u(v-1) - (v + 1)$$
...that's as good as it will get. It cannot be otherwise factored: that is, it cannot be written as a product of factors.
