(x+ay+b)(x+cy+d)=x^2+acy^2+(a+c)xy+(b+d)x+(ad+bc)y+bd
According to the original formula x 2-2xy-ky 2+3x-5y+2, we can get:
ac=-k
a+c=2
b+d=3
AD+BC =-5 years
bd=2
Push
b=2/d
2/d+d=3
From this, the values of b and d are obtained, and then the values of a and c can be deduced, and the value of k comes out.
This is the general idea. . . . .