$x^2 + xy + xz + yz$ can be factorised as $(x+y)(x+z)$. Is there a simple formula for factorising $ax^2 + bxy + cxz + dyz$ into $(a'x+b'y)(a'x+c'z)$, how can I get $a',b',c'$?
In general : Given a polynomial with all the monomials from $\prod_{i=1}^{m}(x_0+x_i)$ with real non-negative coefficients $\{k_l\}$, how can I get the factorisation $\prod_{i=1}^{m}(a_0x_0+a_ix_i)$ i.e. the values of $\{a_i\}$ in terms of $\{k_l\}$ ?
In general, no. It is easier to see this if we reduce the polynomial $ax^2+b x y + c x z + d y z$ into a monic one, $a(x^2+ p x y + q x z + r y z)$, where $p = b/a, q = c/a, r = d/a$ (assuming that $a\neq 0$). Then you can try to factor it into $(x + \alpha y) (x + \beta z)$. To express $\alpha$ and $\beta$ in terms of the original coefficients, we multiply the factorized form out and compare the coefficient of each term. In other words, in order that $$(x+\alpha y)(x+\beta z) = x^2 + \alpha xy + \beta xz + \alpha\beta yz = x^2+p xy + q xz + r yz\ ,$$ we would need $\alpha = p$, $\beta = q$, and $\alpha\beta = r$. However, since $r$ can in principle be unequal to $p q$, these relations cannot necessarily be satisfied. An example of such a polynomial that cannot be factorized into binomials is $x^2+xy + xz + 2 yz$.
In the general case where you have multiple monomial factors, the same problem persists (and gets even worse). Again, considering only the monic case (i.e. $a_0 = 1$), $a_i$ is determined by the coefficient of $x_0^{m-1} x_i$ in the polynomial. However, since the coefficients of other terms, say for example $x_0^{m-2} x_1 x_2$ now must be $a_1 a_2$, you see that the desired $a_i$'s cannot always be found.