Help me factorise Please help me factorize $9x^2-4a^2+4ay +y^2$. I took out "-" as a common factor from $(- 4a^2+4ay +y^2)$ but I cannot factorize it further using difference of 2 squares.
 A: Well, here goes nothing....
Let $(ux + vy + wa)(\alpha x+ \beta y + \gamma a) = 9x^2 -4a^2 +4ay + y^2$ so
So $u\alpha x^2 + (v\alpha + u\beta)xy + (u\gamma + w\alpha)xa +$
$v\beta y^2 + (w\beta + v\gamma)ay + w\gamma a^2=9x^2 -4a^2 +4ay + y^2$
So $u\alpha =9; v\alpha + u\beta =u\gamma + w\alpha = 0; v\beta =1; w\beta +v\gamma =4; w\gamma = -4$.
So $\beta = \frac 1v$ and
$u\alpha =9; v\alpha + \frac uv =u\gamma + w\alpha = 0;  \frac wv +v\gamma =4; w\gamma = -4$
Let $\gamma = -\frac 4w$ and
$u\alpha =9; v\alpha + \frac uv = w\alpha-\frac {4u}w = 0;  \frac wv -\frac {4v}w =4;$
Let $\alpha = \frac 9u$ and
$\color{red}{\frac {9v}u + \frac uv =0}; \frac {9w}u-\frac {4u}w = 0;  \frac wv -\frac {4v}w =4;$
And so $9w^2=4u^2; \color{red}{9v^2=-u^2}; w^2-4v^2-4wv=(w-2v)^2 =0$
But we have unresolvable error.  $\frac {9v}u$ and $\frac uv$ must be the same sign and can't add up to $0$ unless they are both $0$ which they can't be.
We could continue to solve this using complex numbers (That would require $3v = \pm ui$ and $w=2v$ etc....) but it can not be factored over the reals.
A: I believe the exact question is :
$$9x^2-4a^2+4ay-y^2$$ which is
$$=9x^2-(4a^2-4ay+y^2)=(3x)^2-(2a-y)^2=?$$
