How can $f(x,y)= x^4+x^3y+x^2y^2+xy^3+y^4$ be factorized into a product of two polynomials? Let $x,y$ be 2 coprime integers. I assume the following polynomial:$$f(x,y)= x^4+x^3y+x^2y^2+xy^3+y^4$$ is not irreducible. So there must be at least 2 other polynomials of degree $\leq 4$ such that: $$f(x,y)=g(x,y)h(x,y)$$
How can one find $g(x,y)$ and $h(x,y)$?

Here's an efficient route, in the spirit of the remark in André Nicolas' answer:
On one hand, the product of $f(x, y)$ with $(x - y)$ telescopes:
$$(x - y) (x^4 + x^3 y + x^2 y^2 + x y^3 + y^4) = x^5 - y^5.$$
On the other,
$$x^5 - y^5 = \prod_{k = 0}^4 (x - \zeta^k y),$$
where $\zeta := e^{2 \pi i / 5},$ and so over $\Bbb C$ (in fact, over $\Bbb Q[\zeta]$), $f$ factors as
$$f(x, y) = \prod_{k = 1}^4 (x - \zeta^k y).$$
(Note that the lower limit in this product is not the same as in the earlier one.)
The irreducible factors of $f$ over $\Bbb R$ are thus the (quadratic) products of the conjugate factors, namely,
\begin{align}(x - \zeta^2 y) (x - \bar{\zeta}^2 y) = (x - \zeta^2 y) (x - \zeta^3 y) = x^2 - 2 (\Re (\zeta^2)) xy + y^2 &= x^2 + \phi xy + y^2 \\ (x - \zeta y) (x - \bar{\zeta} y) = (x - \zeta y) (x - \zeta^4 y) = x^2 - 2 (\Re \zeta) xy + y^2 &= x^2 + \tilde{\phi} xy + y^2 . \end{align}
Here, $\phi := \frac{1}{2}(1 + \sqrt{5})$ is the Golden Ratio and $\tilde{\phi} := \frac{1}{2}(1 - \sqrt{5})$ is its conjugate (in $\Bbb Q[\sqrt{5}]$). In particular, since some of the irreducible coefficients of the factors of $f$ over $\Bbb R$ are irrational, $f$ is irreducible over $\Bbb Q$.
