Factoring the polynomial $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$ 
can you help me factoring the following polynomial?
$6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$
I don't know where to start from.
 A: Regroup first,
\begin{align}
  (6a^4+5a^2b^2-b^4)+(a^3b+ab^3) &= (6a^2-b^2)(a^2+b^2)+ab(a^2+b^2) \\
  &= (6a^2+ab-b^2)(a^2+b^2) \\
  &= (3a-b)(2a+b)(a^2+b^2)
\end{align}
A: May be easier to follow if you define $t= a/b$ and, after dividing by $b^4$, write it as: $$6t^4 + t^3 +5t^2 + t - 1$$
At this point you can use the rational root theorem to find two of the roots, and in the end factor it as $(3 t - 1) (2 t + 1) (t^2 + 1)\,$. After that, you can multiply back by $b^4$ 
to get the form in $a,b$.
A: This factorises nicely as
$$
 (a^2 + b^2)(3a - b)(2a + b).
$$
To find it, one could see that setting $b=-2a$ gives $0$, as well as $b=3a$.
A: you can rewrite this as $6a^4+6a^2b^2+a^3b+ab^3-b^4-a^2b^2$. Can you finish?
A: Note that $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$ is homogenous. So set $u=b/a$. Then $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4 = a^4(6 + u +5u^2 + u^3 - u^4)$.
We try integer roots of $6 + u +5u^2 + u^3 - u^4$ by considering the divisors of $6$. We find $6 + u +5u^2 + u^3 - u^4=-(u + 2) (u - 3) (u^2 + 1)$.
This gives
$
6a^4 + a^3b +5a^2b^2 + ab^3 - b^4 = -(2 a + b) (b-3 a) (a^2 + b^2)
$.
