How to factor the quadratic polynomial $2x^2-5xy-y^2$?

How do I factor this polynomial: $2x^2-5xy-y^2$ ?

EDIT: Your example cannot be factored over the integers, because $25 + 8 = 33$ is not a perfect square. That is, taking $$ a x^2 + b x y + c y^2 = 2 x^2 - 5 x y - y^2, $$ we get $$ a = 2, \; b = -5, \; c = -1, \; \Delta = b^2 - 4 a c = (-5)^2 - 4 \cdot 2 \cdot (-1)= 25 + 8 = 33, $$ and $33$ is nonnegative but is not a square.

ORIGINAL: Either the one-variable function $$ a x^2 + b x + c $$ or the quadratic form $$ a x^2 + b x y + c y^2, $$ with integers $a,b,c,$ factor over the rational numbers if and only if the discriminant $$ \Delta = b^2 - 4 a c $$ is a square.

You are familiar with this because of the quadratic formula $$ \frac{-b \pm \sqrt \Delta}{2a} $$ which gives the roots $x$ for $ a x^2 + b x + c =0. $

Only a little changes when inserting the letter $y,$ giving $ a x^2 + b x y + c y^2. $ First, if both $a,c$ are $0,$ then we have $bxy$ which is already factored. So, let me show the traditional case, when $a \neq 0.$ Also let $$ \Delta = \delta^2, $$ say with integer $\delta \geq 0.$

$$ a x^2 + b x y + c y^2 = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + 4 a c y^2 \right) $$ $$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - b^2 y^2 + 4 a c y^2 \right) $$ $$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - (b^2 y^2 - 4 a c y^2) \right) $$ $$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - (b^2 - 4 a c ) y^2 \right) $$ $$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - \Delta y^2 \right) $$ $$ = \frac{1}{4a} \left( (2ax+by)^2 - \Delta y^2 \right) $$ $$ = \frac{1}{4a} \left( (2ax+by)^2 - \delta^2 y^2 \right) $$ $$ = \frac{1}{4a} \left( \; (2ax+by + \delta y) \; (2ax+by - \delta y) \; \right) $$ $$ = \frac{1}{4a} \left( \; (2ax+ (b + \delta) y) \; (2ax+ (b - \delta) y) \; \right) $$ Now, either $b,\delta$ are both even or both odd. Either way, we may absorb a factor of $4$ into $$ = \frac{1}{a} \left( \; (ax+ \frac{(b + \delta)}{2} y) \; (ax+ \frac{(b - \delta)}{2} y) \; \right) $$ Finally, since $$ \frac{(b + \delta)}{2} \frac{(b - \delta)}{2} = \frac{b^2 - \Delta}{4} = ac $$ is divisible by $a,$ by unique factorization we may write $$a = a_1 a_2$$ with $a_1$ dividing the first fraction and $a_2$ the second, thus finally getting $$ a x^2 + b x y + c y^2 = \; \left(a_2x+ \left( \frac{b + \delta}{2a_1} \right) y \right) \; \; \left(a_1x+ \left( \frac{b - \delta}{2a_2} \right) y \right) \; $$ in integers.

NOTE, October 3, 2013: we may simply take $$ a_1 = \gcd \left( a, \frac{(b + \delta)}{2} \right) ; \; \; \; \; \; a_2 = \frac{a}{a_1} $$ without paying attention to any prime factorizations.

How would you factor it if the $y$ weren't there? Just drop the $y$s from the relevant terms to get $2x^2-5x-1$, factor that into the form $2(x-a)(x+b)$, then you'll readjoin the $y$ to get $2(x-ay)(x+by)$. You'll find it gives just what you wanted.

Edit: As Michael points out, $a,b$ will not be integers. The quadratic formula will let you find the zeroes of $2x^2-5x-1$, and you can use that to factor.