How to find the coefficient of $x^8$ in $\prod\limits_{i=1}^{10}{\left(x-i\right)}$? How to find the coefficient of $x^8$ in $(x-1) (x-2)  . . .(x-10)$.
Is there any general formula to solve this kind of problems?
 A: The roots of the polynomial $(x - 1)(x-2)\cdots(x-10) = 0$ are $x = 1,2,\dots,10$.
Apply Vieta's Formulas to deduce that the coefficient of $x^8$ is the sum of all pairwise products of  these roots (a.k.a the second elementary symmetric function of the roots):
$$(1\cdot2 + 1\cdot 3 + \dots + 1\cdot 10) + (2\cdot 3 + 2\cdot 4 + \dots + 2\cdot 10) + \dots + (9\cdot 10)\\
= 1320$$
Hence, the coefficient of $x^8$ is $1320$. The Wolf verifies.
A: Let  $$(x−1)(x−2)...(x−10)=x^{10}+a_9x^9+a_8x^8+\cdots+a_1x+a_0$$
Using  Vieta's formulas, the coefficient
of $x^9$ is $1+2+\cdots+10$
of $x^8$ is $\displaystyle\frac{\left(\sum_{r=1}^{10}r\right)^2-\sum_{r=1}^{10}r^2}2$
A: It's important to understand this general fact about multiplying sums together.

When you have the product of several sums $(a_1+a_2+...)(b_1+b_2+...)...$, and apply distributivity to multiply everything out as much as possible, you end up with (this is a bit of a mouthful) the sum of every possible product that can be formed by multiplying together one term from each sum.

For example, $(a + b)(c + d + e)=ac+ad+ae+bc+cd+be$. The products are formed by forming every possible combination of one term from the first sum and one term from the second.
This is very hard to describe in words, but I'll try. Here, when you multiply everything out, which terms are going to have $x^8$? Well, there's ten sums in your product. You're looking for combinations of one term from each sum which take $8$ $x$'s, and since each of these combinations will have $10$ factors in total, each one will include exactly two non-$x$'s. When you factor the $x^8$, you'll end up with the sum of all of those products of two non-$x$'s, so the coefficient is the sum of every possible product of two of the non-$x$ terms in your sums, in other words, every possible product of two of $\{-1, -2, ... -10\}$.
Viete's formulas, mentioned in the other two answers, is basically just the technical name for a particular case of the above reasoning.
A: Your polynomial is (almost) a falling factorial power:
$$
x^{\underline{11}} = x (x - 1) \cdots (x - 10)
$$
It is known that the coefficents of those are given by Stirling numbers of the first kind:
$$
x^{\underline{r}} = \sum_k (-1)^{r - k} \genfrac{[}{]}{0pt}{}{r}{k} x^k
$$
You want:
$$
(-1)^{11 - 9} \genfrac{[}{]}{0pt}{}{11}{9} = 1320
$$
(computed by Casio)
A: To transform this into a standard problem, multiply by $x$ so that you are asking for the coefficient of $x^9$ in $x(x-1)(x-2)\ldots(x-10)$. That is (by definition of the latter) a falling factorial power $x^\underline{11}$ (the exponent says there are $11$ factors, the underline that they are decreasing by unit steps). The coefficients of the expansion of these as a polynomial in $x$ are called Stirling numbers of the first kind:
$$
  x^\underline n = \sum_{k\geq0}s(n,k)x^k = \sum_{k\geq0}(-1)^{n-k}\genfrac[]0{}nkx^k.
$$
So the value asked for in the question is $(-1)^2\genfrac[]0{}{11}9=1320$.
