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# Find the values of ‘m’ for which ${{x}^{2}}+3xy+x+my-m$ has two linear factors in x and y, with integer coefficients. (a) m = 1 or m = 11(b) m = 0 or m = 12(c) m = 10 or m = 0(d) None  Verified
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Hint: First of all, the given equation with a general 2-degree equation in x and y that is $a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0$. Get the values of a, b, c, f, g and h and substitute these in condition to resolve the equation into linear factors that is in $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$ to get the value of m.

We are given an expression ${{x}^{2}}+3xy+x+my-m$ that has 2 linear factors in x and y with the integer coefficient. Here, we have to find the value of ‘m’.
We know that general two-degree equation in x and y is:
$f\left( x,y \right)=a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0....\left( i \right)$
We know that to resolve the above equation into linear factors of x and y, it must satisfy the following condition:
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0....\left( ii \right)$
Now, let us consider the expression given in the question, that is,
$g\left( x,y \right)={{x}^{2}}+3xy+x+my-m$
Let us equate it to 0 to find its linear factors. So, we get,
${{x}^{2}}+3xy+x+my-m=0$
As we know that the above equation is also a 2-degree equation in x and y. So now, we compare the coefficients of the above equation with that of equation (i), we get
Value of a = 1
Value of b = 0
Value of 2h = 3 or $h=\dfrac{3}{2}$
Value of 2g = 1 or $g=\dfrac{1}{2}$
Value of 2f = m or $f=\dfrac{m}{2}$
Value of c = –m
By substituting these values of a, b, c, f, g and h in equation (ii), we get,
$\left( 1 \right)\left( 0 \right)\left( -m \right)+2.\left( \dfrac{m}{2} \right)\left( \dfrac{1}{2} \right)\left( \dfrac{3}{2} \right)-1{{\left( \dfrac{m}{2} \right)}^{2}}-0{{\left( \dfrac{1}{2} \right)}^{2}}-\left( -m \right){{\left( \dfrac{3}{2} \right)}^{2}}=0$
By simplifying the above equation, we get,
$0+\dfrac{3m}{4}-\dfrac{{{m}^{2}}}{4}-0+\dfrac{9}{4}m=0$
Or, $\dfrac{3m}{4}-\dfrac{{{m}^{2}}}{4}+\dfrac{9}{4}m=0$
By multiplying (– 4) on both sides of the above equation, we get,
${{m}^{2}}-9m-3m=0$
Or, ${{m}^{2}}-12m=0$
By taking m common in the above equation, we get,
$m\left( m-12 \right)=0$
So, we get, m = 0 and m = 12.
Hence we get 2 values of m that are 0 and 12 such that the given expression would be resolved into linear factors of x and y.
Therefore, option (b) is correct.

Note: Here, we can cross-check if these values of m are resolving the given expression into the linear factors or not as follows:
Let us take our given expression, that is,
$g\left( x,y \right)={{x}^{2}}+3xy+x+my-m=0$
Let us substitute m = 0 in the above equation, we get,
${{x}^{2}}+3xy+x=0$
By taking x common, we get,
$\left( x \right)\left( x+3y+1 \right)=0$
Hence, we get 2 linear factors as x and (x + 3y + 1). Similarly, we can check for m = 12.