Question: I am attempting to find the coefficient of $x^5$ in the expression $(2 + x - x^2)^5$ by using the multinomial theorem. I was capable of performing this task only when the number of elements is 2 (using the binomial theorem), however, I am stuck on this one.

Answer: 
The trinomial expansion formula:
$$(a+b+c)^n=\sum\_{i+j+k=n} {{n} \choose {i,j,k}} a^ib^jc^k = \sum\_{i+j+k=n} \frac{n!}{i!j!k!} a^ib^jc^k.$$
So:
$$(2+x+(-x^2))^5=\sum\_{i+j+k=5} {5 \choose {i,j,k}} 2^ix^j(-x^2)^k= \\ \cdots+\frac{5!}{0!5!0!}2^0x^5(-x^2)^0+\frac{5!}{1!3!1!}2^1x^3(-x^2)^1+\frac{5!}{2!1!2!}2^2x^1(-x^2)^2=\cdots+81x^5.$$
Note that $x^5=(1)^2(x)^1(x^2)^2=(1)^1(x)^3(x^2)^1=(1)^0(x)^5(x^2)^0$.

The coefficient is $\displaystyle \binom{5}{2}(2)^2\binom{3}{1}(1)^1(-1)^2+\binom{5}{1}(2)^1\binom{4}{3}(1)^3(-1)^1+\binom{5}{0}(2)^0\binom{5}{5}(1)^5(-1)^0=81$