Resolve the following into partial fractions (a) (x-11)/(x2-x-2) (b) (3-x)/((x2+3)(x+3)) (c) (x3-6x+9)/(x2+x-2)

Since the denominator has distinct linear factors, we write:

Let $x = 2$:

Let $x = -1$:

The partial fraction decomposition is $\boxed{\text{}\frac{4}{x+1} - \frac{3}{x-2}\text{}}$

Q1 b)

We need to resolve $\frac{3 - x}{(x^2 + 3)(x + 3)}$ into partial fractions.

The denominator has an irreducible quadratic factor ($x^2 + 3$) and a linear factor ($x+3$).

Let $x = -3$:

Expand the equation and group terms by powers of $x$:

Compare coefficients:
Coefficient of $x^2$: $0 = A+C$
Since $C = \frac{1}{2}$, then $A = -\frac{1}{2}$.
Coefficient of $x$: $-1 = 3A+B$
Substitute $A = -\frac{1}{2}$:

(Check constant term: $3 = 3B+3C = 3(\frac{1}{2}) + 3(\frac{1}{2}) = \frac{3}{2} + \frac{3}{2} = \frac{6}{2} = 3$. This is consistent.)

The partial fraction decomposition is $\boxed{\text{}\frac{1-x}{2(x^2 + 3)} + \frac{1}{2(x+3)}\text{}}$

Q1 c)

We need to resolve $\frac{x^3 - 6x + 9}{x^2 + x - 2}$ into partial fractions.

So, $\frac{x^3 - 6x + 9}{x^2 + x - 2} = x - 1 + \frac{-3x + 7}{x^2 + x - 2}$.

Let $x = 1$:

Let $x = -2$:

The partial fraction decomposition is $\boxed{\text{}x - 1 - \frac{13}{3(x+2)} + \frac{4}{3(x-1)}\text{}}$