I fail to see a simple way to answer this.
As such, this is my long winded approach:
Using the multinomial theorem,
$$(x_1 + x_2 + \cdots + x_m)^n
= \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m}
\prod_{1\le t\le m}x_{t}^{k_{t}},$$
we have the specific parameters $m=3$, $n=7$, $x_1=x^2$, $x_2=x$, and $x_3=-5$.
Via the theorem,
$$(x^2+x-5)^7=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}\prod_{1 \le t \le 3}x_{t}^{k_t}=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}x^{2k_1}x^{k_2}(-5)^{k_3}.$$
The coefficient of the $x^3$ term is the summation of the multinomial coefficient multiplied by the $(-5)^{k_3}$ factor evaluated at all the solutions of the equation $2k_1+k_2=3$ where $0\le k_1\le 7$ and $0 \le k_2 \le 7$.
Those values are $(k_1,k_2)=\{(1,1),(0,3)\}.$ Given that $k_1+k_2+k_3=7$, $k_3$ are respectively $5$ and $4$.
Hence, the coefficient of the $x^3$ is
$${7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4.$$
Given that the definition of the multinomial coefficient is
$${n \choose k_1,k_2,\ldots ,k_m}=\frac{n!}{k_1!k_2!\cdots k_m!},$$
$$
\begin{align}
{7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4&=\frac{7!(-5)^5}{1!1!5!}+\frac{7!(-5)^4}{0!3!4!}\\
&=7\cdot 6(-5)^5+\frac{7\cdot 6\cdot 5(-5)^4}{3!}\\
&=-109375
\end{align}$$
This could be atrociously wrong. Either way, I am desperate for a much simpler process. This is ridiculous to do in a timed testing environment without the formulas given.
I would like to see a very simple but also very general way of arriving at the correct answer (preferably without college methods, but I am open to any methods).
What says you, Math.SE?
Best Answer
Well, in full generality, the approach you took is the right one (though you made a minor mistake in the end), but if you want to do without all the machinery, it's probably easier to do this calculation armed only with a bit of common sense.
You want three factors of $x$ – where are you going to get them from? Either from three individual $x$s, or from one $x^2$ and one $x$. In either case, the remaining factors have to be $-5$. In the first case, you can choose $3$ out of $7$ factors to be $x$, so this yields $\binom73(-5)^4$. In the second case, you can select one out of $7$ factors to be $x^2$ and then one of the $6$ remaining factors to be $x$, so this yields $7\cdot6(-5)^5$. The sum is $35(-5)^4+42(-5)^5=(35-210)(-5)^4=-175\cdot625=-109375$, which is also what your approach would have yielded if you hadn't accidentally increased the exponent of $-5$ by one in the second term.