Vikas TU
Last Activity: 7 Years ago
Attempting to illuminate x2−3xy+ 2y2
=x2−2xy−xy+2y2
=x(x−2y)−y(x−2y)
=(x−y)(x−2y)
Presently other direct and consistent terms will be expected to an and b.
(x−y+a)(x−2y+b)=x2−3xy+2y2+x−2
or,bx+ax+ab−by−2ay=x−2
or,(a+b)x+ab−(2a+b)y=x−2
a+b=1(i)
2a+b=0 (ii)
Subtracting (i) from (ii)
a=−1
At that point, b=2
In this way,
(x−y−1)(x−2y+2)=0
Either,
x−y−1=0
or, on the other hand,
x−2y+2=0
On the off chance that both are zero all the while we can discover arrangement,
y−3=0
y=3 andx=4