TS EAMCET 5 May 2018 Shift 1 Solved Paper

The given equation can be expressed as,
x4+x2+1=x4+x2 +x2−x2+1
=(x2+x+1)(x2−x+1)
Again, the given equation can be expressed as,

x3−2x2+3x−4

x4+x2+1

=

Ax+B

x2+x+1

=

Cx+D

x2−x+1

x3−2x2+3x−4=[(A+C)x3 +x2(B−A+C+D) +x(A−B+C+D)+(B+D)]
Compare of like terms,
A+C=1 …… (1)
B−A+C+D=−2
A−B+C+D=3
B+D=−4 …… (2)
Add equation (1) and equation (2),
A+B+C+D=−3