Question

find value of 'm' for which x²+3xy+x+my-m has two linear fraction in x an y, with integer coefficient with process

Answer 1

Let, x²+3xy+x+my-m=(ax+by+c)(ex+fy+g)
Since the given expression does not contain y² then one of the linear factors must not contain the term containing y. Therefore,
x²+3xy+x+my-m=(ax+by+c)(ex+g)
or, x²+3xy+x+my-m=aex²+bexy+cex+agx+bgy+cg
Equating the coefficients from both sides,
ae=1 ----------------------(1)
be=3 ----------------------(2)
ce+ag=1 -----------------(3)
bg=m ---------------------(4)
cg=-m --------------------(5)
Since all the coefficients are integers then from (1),
a=e=1
∴, from (2), b=3
Putting in(3),
c+g=1 ----------------(6)
Now dividing (4) by (5),
b/c=-1
or, c=-b
or, c=-3
∴, from (6),
-3+g=1
or, g=4
Putting in (5),
-m=-3×4
or, m=12 Ans.

Answer 2

Answer:

Step-by-step explanation:

Coefficient means

If 0, then we can multiple with any number then answer also 0

There are 6x and 3y and 3m s

Adding all these x+y+m: 12

So answer m: 0 or 12