What is the coefficient of $x^5$ in the expression $(2 + x - x^2)^5$

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 1

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: $$(2+x+(-x^2))^5=\sum_{i+j+k=5} {5 \choose {i,j,k}} 2^ix^j(-x^2)^k= \\ \cdots+\frac{5!}{0!5!0!}2^0x^5(-x^2)^0+\frac{5!}{1!3!1!}2^1x^3(-x^2)^1+\frac{5!}{2!1!2!}2^2x^1(-x^2)^2=\cdots+81x^5.$$

Answer 2

Hint :

$$2+x-x^2=(2-x)(x+1)$$

Now can you count the cases when the multiplication of the terms from each bracket will make up to $x^5?$

Answer 3

Hint: To get $x^5$, the $x^2$ can only be raised to the power $0, 1$ or $2$, so using the multinomial theorem, you only want to look at the multinomial coefficients $${{5}\choose{0,5,0}}, {{5}\choose{1,3,1}}, {{5}\choose{2,1,2}}.$$

Answer 4

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