According to the requirements of this problem, we can directly assume two linear factors:

(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. . . . .