\(\frac{x^3-6x^2+2x-4}{1-3x}\) = \(-\frac{1}{3}x^2\) + \(\frac{17}{9}x - \frac{1}{27} + \frac{(\frac{-107}{27})}{1-3x}\)
Hence, the remainder when p(x) = x3 - 6x2 + 2x - 4 is divided by 1- 3x is \(\frac{-107}{27}\).
Varification :- \(\big(-\frac{1}{3}x^2 + \frac{17}{9}x- \frac{1}{27}\big)\)\(\big(1-3x\big)\) - \(\frac{107}{27}\)
= x3 - \(\frac{17}{3}x^2 - \frac{1}{3}x^2 + \frac{1}{9}x + \frac{17}{9}x - \frac{1}{27} - \frac{107}{27}\)
= x3 - 6x2 + 2x - 4
