Show that $x - 2$ is a factor of $x^{3} - 2x^{2} + 3x - 6$. Express $\dfrac{11 - 2x}{x^{3} - 2x^{2} + 3x - 6}$ as partial fractions.

1. Evaluate the polynomial at $x = 2$

Let $P(x) = x^3 - 2x^2 + 3x - 6$.
$P(2) = (2)^3 - 2(2)^2 + 3(2) - 6$
$P(2) = 8 - 8 + 6 - 6 = 0$

2. State the conclusion using the Factor Theorem

Since $P(2) = 0$, by the Factor Theorem, $x - 2$ is a factor of $x^3 - 2x^2 + 3x - 6$.

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

From part (i), $x - 2$ is a factor.
$x^3 - 2x^2 + 3x - 6 = x^2(x - 2) + 3(x - 2) = (x - 2)(x^2 + 3)$

2. Set up the partial fraction decomposition

Let $\dfrac{11 - 2x}{(x - 2)(x^2 + 3)} = \dfrac{A}{x - 2} + \dfrac{Bx + C}{x^2 + 3}$

3. Form the basic equation by clearing the denominator

$11 - 2x = A(x^2 + 3) + (Bx + C)(x - 2)$

4. Find the value of A by substituting $x = 2$

Substitute $x = 2$:
$11 - 2(2) = A(2^2 + 3) + 0$
$7 = 7A \implies A = 1$

5. Find the values of B and C by comparing coefficients and substituting values

Compare coefficients of $x^2$:
$0 = A + B \implies 0 = 1 + B \implies B = -1$

Substitute $x = 0$:
$11 = A(3) + C(-2)$
$11 = 1(3) - 2C$
$2C = -8 \implies C = -4$

6. Write the final partial fraction expression

$\dfrac{11 - 2x}{x^3 - 2x^2 + 3x - 6} = \dfrac{1}{x - 2} + \dfrac{-x - 4}{x^2 + 3} = \dfrac{1}{x - 2} - \dfrac{x + 4}{x^2 + 3}$