The correct option is C (−2,1)
LHS 2x+1(x+a)(bx−1) ;RHS =5(bx−1)−3(x+a)(x+a)(bx−1)
LHS = RHS 2x+1=5(bx−1)−3(x+a) ------------(1)
Since this is the equation it must be true for any value of x
⇒ let x=0 we get from (1)
we get 1=5(−1)−3(a)
6=−3(a) ⇒ a=−2 -----------(1)
5=5(2b−1)
⇒2b=2
b=1
So, (a,b)=(−2,1)