x + (a – 2)
x2 + 2x + 1  x3 + ax2 + bx + 3
                    x3 + 2x2 + x
                          (a – 2)x2 + (b – 1)x + 3
                          (a – 2)x2 + 2(a – 2)x + (a – 2)
                                            (b – 2a + 3)x + (5 – a)
Since (x+1)2 was a factor, then it implied that (b -2a+3) =0...........equation 1
and(5 – a) =0......equation (2). From equation (2), a=5. Substituting 5 for a in equation (1) gave b-2(5) +3=0. i.e. b=7. Hence the polynomial was
f(x) = x3+5x2+7x+3. Factors of f(x) were (x+1)2 and x+ (5-2) i.e. f(x)=(x+1)(x+1)(x+3). Therefore the zeroes of f(x) were -1,-1and -3.