f(x) = ax³ - 11x² + bx + 4, where a and b are constants - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 5

To find the values of a and b, we will use the Remainder Theorem.

1. Calculate f(3): Since the remainder when f(x) is divided by (x - 3) is 55, we set: $f(3) = 6(3)^3 - 11(3)^2 + 3b + 4 = 55$
   Calculating gives: $6(27) - 11(9) + 3b + 4 = 55$
   Simplifying: $162 - 99 + 3b + 4 = 55$
   $3b + 67 = 55$
   $3b = 55 - 67$
   $3b = -12$
   $b = -4$.

2. Calculate f(-1): Since the remainder when f(x) is divided by (x + 1) is -9, we set: $f(-1) = 6(-1)^3 - 11(-1)^2 + b(-1) + 4 = -9$
   Calculating gives: $6(-1) - 11(1) - 4 + 4 = -9$
   Simplifying: $-6 - 11 - 4 + 4 = -9$,
   which is incorrect; simplifying correctly:
   $-6 - 11 - b + 4 = -9$
   thus, $-b - 13 = -9$:

   $$$$

ightarrow b = -4$$.

3. Finally, Solve for a using the values: We can substitute back: $6 + 4 = a$
   Finally, $a = 6$; hence, the values are: $a = 6, b = -4$.