Question 4397
 Let a,b,c be the three roots of a cubic polynomial (monic) equation f(x) = 0.
 Then f(x) = (x-a)(x-b)(x-c)
 By the relationship of roots and coefficients, we have the polyonmial is:
 x^3 - 9x^2 + 26x - 24 = 0,
 Or x^3 - 2x^2 - 7x^2 + 26x - 24 = 0,
 Or x^2(x-2) - (x - 2)(x - 12) = 0
 So, (x-2)(x^2 - x + 12) = 0 (Or divide by (x-2) directly)
 We have (x-2)(x-3)(x-4) = 0 
 
 Hence,2,3 & 4 are roots of f(x) = x^3 - 9x^2 + 26x - 24 = 0.
 This means [a,b,c] = [2,3,4] (as a set)

 Kenny