f(x) = 2x^3 + ax^2 + bx - 6 where a and b are constants - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 4

To find the values of a and b, we will use the given conditions of the polynomial.

1. Using the first condition: When dividing by (2x - 1), the remainder is -5. Let's substitute x = \frac{1}{2} into f(x):

   $$f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) - 6 = -5$$

   Simplifying:

   $$f\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{8} + a \cdot \frac{1}{4} + b \cdot \frac{1}{2} - 6 = -5$$ $$\frac{1}{4} + \frac{a}{4} + \frac{b}{2} - 6 = -5$$

   Rearranging gives:

   $$\frac{a + 2b - 24}{4} = -5 \Rightarrow a + 2b - 24 = -20 \Rightarrow a + 2b = 4 \quad \text{(1)}$$

2. Using the second condition: When dividing by (x + 2), there is no remainder. This means f(-2) = 0:

   $$f(-2) = 2(-2)^3 + a(-2)^2 + b(-2) - 6 = 0$$

   Simplifying:

   $$f(-2) = 2(-8) + 4a - 2b - 6 = 0$$ $$-16 + 4a - 2b - 6 = 0 \Rightarrow 4a - 2b - 22 = 0 \Rightarrow 4a - 2b = 22 \quad \text{(2)}$$

3. Solving the system of equations: We now have the two equations:

   (1) a + 2b = 4
   (2) 4a - 2b = 22

   We can solve these equations simultaneously.

   From equation (1): a = 4 - 2b

   Substituting into (2):

   $$4(4 - 2b) - 2b = 22 \Rightarrow 16 - 8b - 2b = 22 \Rightarrow 16 - 10b = 22 \Rightarrow -10b = 6 \Rightarrow b = -\frac{3}{5}$$

   Now substitute b back into (1):

   $$a + 2\left(-\frac{3}{5}\right) = 4 \Rightarrow a - \frac{6}{5} = 4 \Rightarrow a = 4 + \frac{6}{5} = \frac{26}{5}.$$

   Thus, the values are:

   $$a = \frac{26}{5}, \quad b = -\frac{3}{5}.$$