Show that $x - 2$ is a factor of $x^3 - 12x + 16$. Express $\frac{2 + x^2}{x^3 - 12x + 16}$ as the sum of three partial fractions.

1. Let $P(x)$ be the polynomial and evaluate $P(2)$

Let $P(x) = x^3 - 12x + 16$.
Substitute $x = 2$ into $P(x)$:
$P(2) = (2)^3 - 12(2) + 16$

2. Calculate the value and conclude using the Factor Theorem

$P(2) = 8 - 24 + 16 = 0$
Since $P(2) = 0$, by the Factor Theorem, $x - 2$ is a factor of $x^3 - 12x + 16$.

1. Factorise the denominator completely using the result from part (i)

From (i), $x - 2$ is a factor of $x^3 - 12x + 16$.
By long division or comparing coefficients:
$x^3 - 12x + 16 = (x - 2)(x^2 + 2x - 8)$
Factorising the quadratic factor:
$x^2 + 2x - 8 = (x - 2)(x + 4)$
So, $x^3 - 12x + 16 = (x - 2)^2(x + 4)$

2. Set up the partial fraction decomposition

$\frac{x^2 + 2}{(x - 2)^2(x + 4)} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} + \frac{C}{x + 4}$

3. Form the basic equation to solve for constants

Multiplying throughout by $(x - 2)^2(x + 4)$:
$x^2 + 2 = A(x - 2)(x + 4) + B(x + 4) + C(x - 2)^2$

4. Solve for the constants by substitution and comparing coefficients

Let $x = 2$:
$2^2 + 2 = B(2 + 4)$
$6 = 6B \implies B = 1$

Let $x = -4$:
$(-4)^2 + 2 = C(-4 - 2)^2$
$18 = 36C \implies C = \frac{1}{2}$

Comparing coefficients of $x^2$:
$1 = A + C$
$1 = A + \frac{1}{2} \implies A = \frac{1}{2}$

5. Write the final partial fraction expression

$\frac{2 + x^2}{x^3 - 12x + 16} = \frac{1}{2(x - 2)} + \frac{1}{(x - 2)^2} + \frac{1}{2(x + 4)}$