What is the coefficient of the $x^3$ term in the expansion of $(x^2+x-5)^7$ (See details)? 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?
 A: This is pretty much the same as joriki's approach, but the emphasis here is on a slightly different idea. Write
$$
[x^2+x+(-5)]^7=\sum_{k=0}^7{7\choose k}(x^2+x)^k(-5)^{7-k}\,.
$$
Now note that for $k\ge 4$, the power $(x^2+x)^k$ starts with $x^4$ or higher power, and for $k<2$, it ends with $x^2$ or lower power. Thus only 2 terms ($k=2$ and $k=3$) have $x^3$ in them. Looking at what the corresponding coefficients are, we get
$$
{7\choose 3}(-5)^4+{7\choose 2}\cdot 2\cdot (-5)^5
$$
The rest of the computation is the same as in joriki's answer.
A: Given that you're looking for the coefficient on a small monomial and the exponent is small, we can compute directly. 
$$(x^2 + x - 5)^2 = \cdots + 2 x^3 - 9 x^2 - 10 x + 25$$
Rather than go through the usual method of multiplying two polynomials, you could do it term-wise. The only way to produce an $x^3$ is as $x^2 \cdot x$ or $x \cdot x^2$, so that coefficient is $1 \cdot 1 + 1 \cdot 1$, and so forth. Continuing:
$$(x^2 + x - 5)^3 = \cdots - 29 x^3 + 60 x^2 + 75 x - 125 $$
Ah, let me introduce a trick for repeated exponents: rather than multiplying in copies of $(x^2 + x - 5)$ one at a time, we can skip a lot of steps by simply squaring an intermediate value. Squaring the previous equation gives
$$\begin{align*}(x^2 + x - 5)^6 &= \cdots + (2 \cdot (-29) \cdot (-125) + 2 \cdot 60 \cdot 75) x^3 
\\ & + (2 \cdot 60 \cdot (-125) + 75^2) x^2 + (2 \cdot 75 \cdot (-125)) x + 125^2
\\&= \cdots + 16250 x^3 - 9375 x^2 - 18750 x + 15625 \end{align*}$$
Finally, when computing the seventh power, we can only compute the coefficient we care about:
$$\begin{align*}(x^2 + x - 5)^7 &= \cdots + (16250 \cdot (-5) + (-9375) \cdot 1 + (-18750) \cdot 1) x^3 + \cdots \\ &= \cdots - 109375 x^3 + \cdots\end{align*}$$
A: 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.
A: Why not use Calculus? Calculate the third derivative, evaluate at $0$, and divide by $3!=6$. Since the third derivative of $g^7$ is $210g^4{g'}^3+126g^5g'g''+7g^6g'''$, and $g(0)=-5$, $g'(0)=1$, $g''(0)=2$, and $g'''=0$, the evaluation gives surenough $-109375$.
A: Using the binomial theorem write this as $\sum_{k=0}^7\binom{7}{k}x^{2k}(x-5)^{7-k}$. Now the non-zero coefficients of $x^3$ occurs in the terms for $k=0$ and $k=1$. For $k=0$ we want the coefficient of $x^3$ in $(x-5)^7$ which is $\binom{7}{3}(-5)^4$; for $k=1$ we want the coefficient of $x$ in $(x-5)^6$ which is $6(-5)^5$. Thus the answer is 
$$\binom{7}{3}(-5)^4+7\cdot 6\cdot (-5)^5=-109375$$
