How to use "results from partial fractions"? Let ${a_n}$ be a sequence whose corresponding power series $A(x)=\sum_{i\geq 0}a_ix^i$ satisfies
$$A(x)=\frac{6-x+5x^2}{1-3x^2-2x^3}$$
The denominator can be factored into $(1-2x)(1+x)^2$. Using results from partial fractions, it can be shown that there exists constants $C_1,C_2,C_3$ such that
$$A(x)=\frac{C_1}{1-2x}+\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$$
Determine these constants and find $a_5$ using this new expression.
What exactly are "results from partial fractions" and how should I use them?

We describe what we do once we have the coefficients $C_1,C_2,C_3$. It all comes from the expansion
$$\frac{1}{1-t}=1+t+t^2+t^3+\cdots.\tag{$1$}$$
Putting $t=2x$, we find that $\frac{C_1}{1-2x}$ has expansion 
$$C_1+2C_1 x+4C_1x^2+8C_1x^3+\cdots.$$
Putting $t=-x$, we can in a similar way get the power series expansion of $\frac{C_2}{1+x}$. 
For $\frac{C_3}{(1+x)^2}$ we need an additional idea. Look at Equation $(1)$, and differentiate both sides. We get
$$\frac{1}{(1-t)^2}=1+2t+3t^2+4t^3+\cdots.$$
Put $t=-x$, and multiply by $C_3$.
Now that we have the three power series expansion, we can read off the coefficient of $x^n$ in their sum. 
Details: Almost all the work in finding the $C_i$ has been done by DonAntonio. Bringing $\frac{C_1}{1-2x}=\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$ to the common denominator $(1-2x)(1+x)^2$, we find that the numerator is $C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)$. Thus, identically, we must have
$$C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)=6-x+5x^2.$$
 There are various ways to find the constants. Put $x=-1$. We get $C_3(3)=12$, so $C_3=4$. Put $x=1/2$. We get $C_1(9/4)=27/4$, so $C_1=3$. Finally, the constant term on the left is $C_1+C_2+C_3$, while on the right it is $6$. Since $C_1+C_3=7$, we get $C_2=-1$. So our original function is equal to
$$\frac{3}{1-2x}-\frac{1}{1+x}+\frac{4}{(1+x)^2}.\tag{$2$}$$
Finally, we compute the coefficient $a_5$ of $x^5$. By the discussion in the main answer, the coefficient of $x^5$ in the expansion of $\frac{1}{1-2x}$ is $2^5$.
The coefficient of $x^5$ in the expansion of $\frac{1}{1+x}$ is $(-1)^5$.
The expansion of $\frac{1}{(1-t)^2}$ was obtained by differentiating $1+t+t^2+\cdots$. so the $t^5$ term is $6t^5$. Putting $t=-x$ we get that the coefficient of $x$ is $6(-1)^5$. Putting things together, and remembering our $C_i$, we get
$$a_5=(3)(2^5)+(-1)(-1)^5+(4)(6)(-1)^5.$$  
