Factorising a quadratic with two variables with separate $x$ and $y$ terms

Question

I know that the following can be expressed as a product of two factors: $x^2 + x - y^2 + y$ but I don't know where to start with factorising it.

Usually I am able to match up the terms to the formula for quadratics, like $ax^2 + bx + c$, for example with $4a^2 - 8ab + 3b^2$ it's clear that it will factor to $(2a - b)(2a - 3b)$.

How does one go about solving these kind of problems? Before I could do the "matching up" method to see what would occupy the second term in each set of parentheses, but I've been hinted that $x^2 + x - y^2 + y$ will have three terms in one parenthesis.

I don't understand how the $+ x+ y$ in the middle can arise. Starting off I can guess it'll factorise to something like $(x ... y)(x ... y)$ because that provides both the $x^2$ term and the $y^2$ term.

Please help me understand how to solve this kind of problem.

Answer 1

$$ x^2 + x - y^2 + y = (x^2 - y^2) + (x + y) = (x-y)(x+y) + (x+y) = (x-y+1) (x+y) $$

Answer 2

we have $$x^2-y^2+x+y=(x-y)(x+y)+x+y=(x+y)(x-y+1)$$

Answer 3

Completing the square usually helps a lot:

$$x^2+x-y^2+y=\left(x+\frac12\right)^2-\frac14-\left(y-\frac12\right)^2+\frac14=$$

$$\left(x+\frac12\right)^2-\left(y-\frac12\right)^2\stackrel{\text{Diff. of squares}}=\left(x+\frac12-y+\frac12\right)\left(x+\frac12+y-\frac12\right)=$$

$$=(x-y+1)(x+y)$$