What is the remainder of (x^100 - 4x^98 + 5x + 6) divided by (x³ - 2x² - x + 2)?

The remainder shall be quadratic say f(x) = ax^2 + bx + c

Now (x³ - 2x² - x + 2) = (x-2)(x-1)(x+1)
Let g(x) = (x^100 - 4x^98 + 5x + 6)- (ax^2 + bx + c)
g(1) , g(2), g(-1) all are 0
g(1) = 1- 4 + 5 + 6 – (a + b + c) = 0 or a + b+ c = 8 ..1
g(-1) = 1 – 4 – 5 + 6 – (a-b+ c) = 0 or a-b+c = -2 ..2
g(2) = 2^100- 2^100 + 10 + 6 – (4a + 2b+c) = 0 or 4a + 2b + c = 16 ..3

subtract (2) from (1) 2b = 10 or b = 5
so a + c = 3 and 4a + c = 6=> a = 1 and c = 2

remainder = x^2 + 5x + 2