Doubt

Here we are given (x-2) as one of the factor and sum of the roots of quadratic factor additionaly we know that in cubic expression constant term can only come from product of two constants whereas the x^3 term can only come from product of x^2 and x . By considering the logic we can form quadratic equation (x^2 - x(sum of roots) + product of root) = (x^2 - 3x + 3) , we get +3 here bcz we have -6 in orgininal eqation and x^2 bcz we have x^3 in original cubic equation. now we just multiply both linear and qadratic equation and get x^3 - 5x^2 + 9x -6 . so a=5 , b=9 , a-b = -4.

put x=2 in the given equation: u'll get a relation (4-2a+b=3 OR 2a-b=1). then using the long division method u'll get the quadratic factor (x^2+(2-a)x+3), sum of roots (-b/a), in this case is a-2 =3 , which gives a=5. putting a=5 in the relation, we get b=9. ans: a-b= -4.