Factorization of $ x^2 +xy+5x+m+5 $ I want to factorize $ x^2 +xy+5x+m+5 $ . For what value of m ,  $ x^2 +xy+5x+m+5 $ can be resolved into linear factors ?
My try  : 
  $ x^2 +xy+5x+m+5 $ = $ x^2 +(5+y)x+(m+5) $ 
To get the linear factors , we must have the determinant of this eqn is >= 0 . 
D = $ (5+y)^2-4(m+5)$ . Then I can not proceed .  

Note that factoring we will expect the factors to be of the form:
$$(x+\alpha y + \beta)(x+\gamma y + \delta)$$
Comparing the coefficient of $y^2$ gives us $\alpha \gamma = 0$. Due to symmetry, let us set $\gamma =0$. Hence, we will expect the factors to be of the form:
$$(x+\alpha y + \beta)(x + \delta)$$
Now comparing the coefficient of $y$, we get that
$$\alpha \delta = 0$$
If $\alpha = 0$, there will be no $y$ term. Hence, $\delta = 0$. Hence, the factorization has to be of the form
$$(x+\alpha y + \beta)x$$
Hence, $m+5 = 0 \implies m=-5$.
