divisability problem

a=44
b=-11

Its by simple long division method.

Let f(x) = x^2+2x+b

x^2 leaves a remainder of -(2x+b) on divn by f(x).

Since x(x^2+2x+b)=x^3+2x^2+bx is divisble by f(x), x^3 leaves a residue of -(2x^2+bx) and so a remainder of 2(2x+b)-bx = (4-b)x+2b

Hence the remainder when the given polynomial is divided by f(x) is

2(2x+b)-bx = (4-b)x+2b-6(2x+b)-3x-a = -(11+b)x-(a+4b)

Since the remainder is zero, we must have b=-11, a=44