- Find the values of 'm' for which x2 + 3xy + x + my - m has two linear factors in x and y, with integer coefficients.

$$\begin{gathered} Given expression is x^2+3xy+x+my-m \\ Let the two linear factors in xand y are (ax+by+c) and (dx+ey+f) where a,b,c,d,e,f are integers \\ Now \sin ce the given expression has no y^2 term hence one of the factors must not contain y \\ Hence factors are (ax+by+c) and (dx+f) \\ So (ax+by+c)(dx+f) = x^2+3xy+x+my-m \\ \Rightarrow ad{x}^2+bdxy+cdx+afx+bfy+cf = x^2+3xy+x+my-m \\ \Rightarrow ad{x}^2+bdxy+(cd+af)x+bfy+cf = x^2+3xy+x+my-m \\ Now on comparing the coefficients we get \\ ad=1.......\left(1\right) \\ bd =3.......\left(2\right) \\ cd+af =1........\left(3\right) \\ bf=m.........\left(4\right) \\ cf=-m........\left(5\right) \\ From \left(1\right) we can say that as a and d are integers so possible values of a and d are a=1 and d=1 \\ Hence a=1,d=1 \\ Hence from \left(2\right) ,b=3 \\ Putting the values of a and d in \left(3\right) we get c+f =1........\left(6\right) \\ On dividing \left(5\right) by \left(4\right) ,we get \frac{c}{b}=-1\Rightarrow c=-b \Rightarrow c=-3 \\ Putting the values of c in \left(6\right) we get f =1-(-3) =4 \\ Hence from \left(4\right) , m = 3\times 4=12 \end{gathered}$$

Hence the value of m=12 and the two linear factors are $(x+3y-3) and (x+4)$

Note :- One more value of m is possible i.e m =0 .In this case two linear factors are $\left(x\right) and (x+3y+1)$