By using long division, show that $2x^2+1$ is a factor of $2x^3-4x^2+x-2$. Express $\dfrac{-5x^2+11x-11}{2x^3-4x^2+x-2}$ in partial fractions.

1. Perform polynomial long division of $2x^3-4x^2+x-2$ by $2x^2+1$.

2. Conclude that the divisor is a factor based on the zero remainder.

Since the remainder is $0$, $2x^2+1$ is a factor of $2x^3-4x^2+x-2$.

1. Factorise the denominator using the result from part (a).

From part (a), $2x^2+1$ is a factor of $2x^3-4x^2+x-2$.
By inspection or division, $2x^3-4x^2+x-2 = (2x^2+1)(x-2)$.

2. Set up the partial fraction decomposition.

Let $\dfrac{-5x^2+11x-11}{(2x^2+1)(x-2)} = \dfrac{Ax+B}{2x^2+1} + \dfrac{C}{x-2}$.

3. Form the basic equation.

Multiplying throughout by $(2x^2+1)(x-2)$:
$-5x^2+11x-11 = (Ax+B)(x-2) + C(2x^2+1)$

4. Solve for the constants $A$, $B$, and $C$.

Substitute $x = 2$:
$-5(2)^2 + 11(2) - 11 = C(2(2)^2 + 1)$
$-20 + 22 - 11 = 9C$
$-9 = 9C \implies C = -1$

Substitute $x = 0$:
$-11 = B(-2) + C(1)$
$-11 = -2B - 1 \implies -2B = -10 \implies B = 5$

Compare coefficients of $x^2$:
$-5 = A + 2C$
$-5 = A + 2(-1) \implies A = -3$

5. Write the final partial fraction expression.

$\dfrac{-5x^2+11x-11}{2x^3-4x^2+x-2} = \dfrac{-3x+5}{2x^2+1} - \dfrac{1}{x-2}$