f(x) = 2x³ + ax² + bx - 6 where a and b are constants - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 4

To find the values of a and b, we will use the information about the remainders when f(x) is divided by (2x - 1) and (x + 2).

1. Remainder when divided by (2x - 1): We substitute the value determined by setting the divisor to zero:

   Set 2x - 1 = 0, hence x = \frac{1}{2}.

   Evaluate f(\frac{1}{2}):

   $f(\frac{1}{2}) = 2(\frac{1}{2})^3 + a(\frac{1}{2})^2 + b(\frac{1}{2}) - 6$

   Simplifying:

   $= 2(\frac{1}{8}) + a(\frac{1}{4}) + b(\frac{1}{2}) - 6$ $= \frac{1}{4} + \frac{a}{4} + \frac{b}{2} - 6$

   This must equal -5: $\frac{1 + a + 2b - 24}{4} = -5$

   Therefore:

   $1 + a + 2b = -20\quad(1)$

2. Remainder when divided by (x + 2):

   Set x + 2 = 0, hence x = -2.

   Evaluate f(-2):

   $f(-2) = 2(-2)^3 + a(-2)^2 + b(-2) - 6$ $= -16 + 4a - 2b - 6$ $= 4a - 2b - 22$

   Since there is no remainder: $4a - 2b - 22 = 0\quad(2)$

3. Solving the system of equations:

   From equations (1) and (2):

   1. $1 + a + 2b = -20$

   2. $4a - 2b = 22$

   We can rearrange the first equation: $a + 2b = -21\quad(3)$

   Now we can express a in terms of b: $a = -21 - 2b\quad(4)$

   Substituting (4) into (2):

   $4(-21 - 2b) - 2b - 22 = 0$ $-84 - 8b - 2b - 22 = 0$ $-10b = 106\quad(5)$ $b = -10.6$

   Substituting value of b back into (4): $a = -21 - 2(-10.6) = -21 + 21.2 = 0.2$

   Thus, the values are:

   • $a = 0.2$
   • $b = -10.6$.