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 .  

Since now you want to have the discriminant greater than $0$ which in itself is a quadratic polynomial ie $y^2+10y+(5-4m)$ . We see that this will be always greater than $0$ . If its discriminant is always less than $0$ . Because if a quadratic has real roots than it will be positive and negative . 
The determinant of $y^2+10y+(5-4m)$ is $ D = 100 -4(5-4m) $ = $80+16m$ 
Now $D=0$ => $ 80+16m = 0 $ =>$ m = -5 $
