Question

find the values of m for which X2+3xy+x+my-m has two linear factors in x and y, with integer coefficient

Answer 1

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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.

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Answer 2

$helleo \: mate$


$here \: is \: ur \: answer$

Given⤵
${x}^{2} + 3xy + x + my - m$

$now$

let x²+3xy+x+my-m=(ax+by+c)(ex+fy+g)

→since,the given expression does not contain y²,

thus,one of the linear factor(let it be "fy")must not contain the term containing y.

$hence$

→x²+3xy+x+my-m=(ax+by+c)(ex+g)

→x²+3xy+x+my-m=aex²+bexy+cex+agx+bgy+cg

↪equating coefficients from both sides↩

we get,

→
$ae = 1...(1)$
→
$be = 3...(2)$
→
$ce + ag = 1...(3)$
→
$bg = m...(4)$
→
$cg = - m...(5)$

since,all the coefficients are integers

from equation ( 1 )

→
$a = e = 1$

also,from (2)

→
$b = 3$

putting in ( 3 )

→
$c + g = 1...(6)$

dividing ( 4 ) by ( 5 )

→
$\frac{b}{c} = - 1$
→
$c = - b$

→
$c = - 3$

from ( 6 )

→
$- 3 + g = 1$
→
$g = 4$

putting in ( 5 )

→
$- m = - 3 \times 4$
→
$m = 12$

↪

$answer \: is \: 12 \: .$
↩

$thanks$

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