# Find the values of `lambda` for which the expression f(x,y)=2x^2+`lambda` ` ` xy+3y^2-5y-2 can be written as the product of two linear factors

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`f(x,y)=2x^2+lambda xy+3y^2-5y-2`

If you can write the above as a product of two linear factors then;

`2x^2+lambda xy+3y^2-5y-2 = (ax+by+c)(px+qy+r)` where a,b,c,p,q and r are whole numbers.

When x = 0

`2x^2+lambda xy+3y^2-5y-2 = (ax+by+c)(px+qy+r) `

`3y^2-5y-2 = (by+c)(qy+r)`

But we can factorize `3y^2-5y-2` .

`3y^2-5y-2`

`= 3y^2-6y+y-2`

`= 3y(y-2)+(y-2)`

`= (y-2)(3y+1)`

`(y-2)(3y+1) = (by+c)(qy+r)`

So we can say;

b = 1

c = -2

q = 3

r = 1

`2x^2+lambda xy+3y^2-5y-2 = (ax+y-2)(px+3y+1)` for all x and y.

`2x^2+lambda xy+3y^2-5y-2 = (ax+y-2)(px+3y+1)`

When y = 0

`2x^2-2 = (ax-2)(px+1)`

`2x^2-2 = apx^2+(a-2p)x-2`

Comparing components will give us;

`x^2 rarr 2 = ap ---(1)`

`x rarr 0 = a-2p---(2)`

Solving (1) and (2) will give you;

`p = +-1`

`a = +-2`

Now we can say;

`2x^2+lambda xy+3y^2-5y-2 = (2x+y-2)(x+3y+1)`

OR

`2x^2+lambda xy+3y^2-5y-2 = (-2x+y-2)(-x+3y+1)`

Now we can compare the corresponding component of xy to find lambda;

`2x^2+lambda xy+3y^2-5y-2 = (2x+y-2)(x+3y+1)`

`xy rarr lambda = (6+1) = 7`

`2x^2+lambda xy+3y^2-5y-2 = (-2x+y-2)(-x+3y+1)`

`xy rarr lambda = (-6-1) = -7`

** So the answer is** `lambda = +-7`