Express $\frac{5x^3 - 4x^2 + 15x - 21}{(x-1)(x^2 + 4)}$ in partial fractions.

1. Recognise that the fraction is improper and set up the correct partial fraction form

The degree of the numerator (3) is equal to the degree of the denominator (3). The leading term of the numerator is $5x^3$ and the leading term of the denominator is $x^3$, so the quotient is 5.

Let $\frac{5x^3 - 4x^2 + 15x - 21}{(x-1)(x^2 + 4)} = 5 + \frac{A}{x-1} + \frac{Bx+C}{x^2+4}$

2. Form the basic identity by multiplying throughout by the denominator

$5x^3 - 4x^2 + 15x - 21 = 5(x-1)(x^2+4) + A(x^2+4) + (Bx+C)(x-1)$

3. Substitute a suitable value of $x$ to find $A$

Substitute $x = 1$:
$5(1)^3 - 4(1)^2 + 15(1) - 21 = 5(0) + A(1^2+4) + 0$
$5 - 4 + 15 - 21 = 5A$
$-5 = 5A \implies A = -1$

4. Substitute $x = 0$ to find $C$

Substitute $x = 0$ and $A = -1$:
$-21 = 5(-1)(4) + (-1)(4) + C(-1)$
$-21 = -20 - 4 - C$
$-21 = -24 - C \implies C = -3$

5. Compare coefficients to find $B$ and state the final answer

Compare coefficients of $x^2$:
$-4 = 5(-1) + A + B$
$-4 = -5 - 1 + B$
$-4 = -6 + B \implies B = 2$

Therefore, $\frac{5x^3 - 4x^2 + 15x - 21}{(x-1)(x^2 + 4)} = 5 - \frac{1}{x-1} + \frac{2x-3}{x^2+4}$