Given that $$\frac{3x^4 - 2x^3 - 5x^2 - 4}{x^2 - 4} \equiv ax^2 + bx + c + \frac{dx + e}{x^2 - 4}, \quad x \neq \pm 2$$ find the values of the constants a, b, c, d and e. - Edexcel - A-Level Maths Pure - Question 3 - 2013 - Paper 7

To find the constants a, b, c, d, and e, we will perform polynomial long division of the numerator by the denominator:

1. Divide the leading term of the numerator, $3x^4$, by the leading term of the denominator, $x^2$, which gives us $3x^2$.

2. Multiply $3x^2$ by the entire denominator $x^2 - 4$, resulting in $3x^4 - 12x^2$.

3. Subtract this from the original numerator:

   $\left(3x^4 - 2x^3 - 5x^2 - 4\right) - \left(3x^4 - 12x^2\right) = -2x^3 + 7x^2 - 4$

4. Next, repeat the process: Divide $-2x^3$ by $x^2$, yielding $-2x$.

5. Multiply $-2x$ by $x^2 - 4$ resulting in $-2x^3 + 8x$.

6. Subtract this:

   $\left(-2x^3 + 7x^2 - 4\right) - \left(-2x^3 + 8x\right) = 7x^2 - 8x - 4$

7. Finally, divide $7x^2$ by $x^2$ giving $7$.

8. Multiply $7$ by $x^2 - 4$, resulting in $7x^2 - 28$.

9. Subtract:

   $\left(7x^2 - 8x - 4\right) - \left(7x^2 - 28\right) = -8x + 24$

10. Now we have:

$\frac{3x^4 - 2x^3 - 5x^2 - 4}{x^2 - 4} = 3x^2 - 2x + 7 + \frac{-8x + 24}{x^2 - 4}$