Answer

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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

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