How to factorise quadratic equation in the form of $x^2+6xy+y^2$? My Text book has some questions of quadratic equations in the form of 
$x^2+xy+y^2$
1)Is there any strategy or proper way to factorise quadratic equations in such way?
2)Is that a type of quadratic trinomials?if yes why?
 A: Hint: let $z = \frac{x}{y}$ then $a x^2+b xy+c y^2=y^2(a z^2+b z+c)$. The latter is a quadratic in $z$. Factoring it, then substituting $\frac{x}{y}$ back gives the factorization in terms of $x,y$.

[ EDIT ] Here is the fully worked out example for factoring $8x^2+6xy+y^2$.


*

*Let $z=\frac{x}{y}$ then write $8x^2+6xy+y^2 = y^2(8 z^2 + 6 z + 1)$.

*Factor $8 z^2 + 6 z + 1$ (either by inspection, or) by actually calculating its roots using the quadratic formula:
$$z_{1,2} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 8}}{16}= \frac{-6 \pm 2}{16} = \frac{-3 \pm 1}{8}$$
so $z_1 = -\frac{1}{2}$, $z_2=-\frac{1}{4}$

*It follows that $8 z^2 + 6 z + 1 = 8(z-z_1)(z-z_2) = 8(z+\frac{1}{2})(z+\frac{1}{4})=(2 z+1)(4 z+1)$

*Substituting back $z=\frac{x}{y}$:
$$y^2(8 z^2 + 6 z + 1)=y^2(2 \frac{x}{y}+1)(4 \frac{x}{y}+1)=(2x+y)(4x+y)$$
A: Yes, there is a strategy. We can use Eisenstein's criterion to show that $x^2+xy+y^2\in K[x,y]=K[x][y]$ is irreducible, hence does not factorize. 
We see, that also
$$
x^2+6xy+y^2
$$
does not factorize. 
If it is reducible, we may use factoring algorithms for polynomials in $R[y]$ with a principal ideal ring $R=K[x]$, if $K$ is a field. This works perfectly for
$$
8x^2+6xy+y^2=(2x+y)(4x+y).
$$
