# Using binomial coefficients to find sum of roots of a polynomial.

Find the sum of the roots, real and non-real, of the equation $$x^{2001}+\left(\frac 12-x\right)^{2001}=0$$, given that there are no multiple roots.

While trying to solve the above problem (AIME 2001, Problem 3) which was asked here on MSE, I came across three solutions on AoPS. ( The MSE solution uses Vieta's formula which I am clear about )

The first solution (on AoPS) involves the use of Vieta's formula's and is quite clear.

The third solution states the following :

Note that if $$r$$ is a root, then $$\frac{1}{2}-r$$ is a root and they sum up to $$\frac{1}{2}.$$

We make the substitution $$y=x-\frac{1}{4}$$ so $$(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.$$

Expanding gives $$2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots$$ so by Vieta, the sum of the roots of $$y$$ is 0.

Since $$x$$ has a degree of 2000, then $$x$$ has 2000 roots so the sum of the roots is $$2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.$$

I do not understand two things in the above solution :

1. "Note that if $$r$$ is a root, then $$\frac{1}{2}-r$$ is a root and they sum up to $$\frac{1}{2}.$$"

a) Here what is being referred to as "they"? (Shouldn't $$\frac{1}{2}-r$$ be a factor and not a root). Answered by the third comment

b)Why is the sum 1/2? Also answered by the third comment

c)Why is $$\frac{1}{2}-r$$ a root? Answered by the sixth comment

1. How is the final expression arrived upon to find the sum of all 2000 roots? (Answered by @YvesDaoust)

The second solution is more mystifying (possibly because it is similar to the one above):

We find that the given equation has a $$2000^{\text{th}}$$ degree polynomial. Note that there are no multiple roots. Thus, if $$\frac{1}{2} - x$$ is a root, $$x$$ is also a root. Thus, we pair up $$1000$$ pairs of roots that sum to $$\frac{1}{2}$$ to get a sum of $$\boxed{500}$$.

1. Again, why is $$\frac{1}{2} - x$$ a root. By "$$x$$ is also a root" does it mean $$x$$ representing the set of all roots? Answered by the sixth comment

2. Why does the pairing up occur? Why is the sum of each pair 1/2? (Answered by @YvesDaoust)

I wonder if it could be solved as follows :

Let the roots be $$P_1,P_2,...P_{2000}$$. The polynomial can be expressed as a product of factors as follows : $$(2001/2)($$x$$-P_1)($$x$$-P_2)....($$x$$-P_{2000}) = 0$$. The above expression is the same as $$x^{2001}+\left(\frac 12-x\right)^{2001}=0$$.

Thus, $$x^{2001}+\left(\frac 12-x\right)^{2001}$$ = $$(2001/2)($$x$$-P_1)($$x$$-P_2)....($$x$$-P_{2000})$$

Here the coefficient of $$x^{1999}$$ on the $$RHS$$ should represent $$\sum\limits_{i=1}^{2000}P_i\times(-1/2)$$

On the $$LHS$$ the corresponding term would be the term with $$x^{1999}$$ and thus the coefficient of this term on the $$LHS$$ should also be the required sum.

On the LHS the coefficient of the $$x^{1999}$$ term is -$${2001}\choose{2}$$*$$(1/2)^2$$ which represent the sum of the roots.

[Picked up this approach here, but I don't see how this would work ] (https://youtu.be/S6FRtmDUl-s?t=2806)

In this solution I find some errors(?) :

1. Are there any inconsistencies in the reasoning?Wouldn't the sum of roots differ from the binomial coefficient since the expression involves unique values of $$P_i$$ (no multiple roots).
2. The answers do not match, which seems to suggest so.
3. Is there a way of arriving at the answer without using Vieta's formula and by expressing the polynomial as a product of factors and then using binomial coefficients as attempted above?
• That's seven questions! – Lord Shark the Unknown Oct 13 '18 at 5:00
• The first solution you give (which you call the third solution) is essentially the same as the second solution you give (which mercifully you call the second solution). – Lord Shark the Unknown Oct 13 '18 at 5:01
• In your question 1, are you asking why $r+\left(\frac12-r\right)=\frac12$? – Lord Shark the Unknown Oct 13 '18 at 5:03
• $\dfrac12-r$ is a root whenever $r$ is. This is because $$P(\dfrac12-r)=(\dfrac12-r)^{2001}-(\dfrac12-(\dfrac12-r))^{2001}= (\dfrac12-r)^{2001}-r^{2001}=-P(r).$$ So if $P(r)=0$ then $P(\dfrac12-r)=0$ also. – Jyrki Lahtonen Oct 13 '18 at 7:26
• You're right. I made a sign error. Anyway, the argument, as per your modification, survives with the correct sign as well. – Jyrki Lahtonen Oct 20 '18 at 20:28

First solution:

By symmetry of $$x^n+(\frac12-x)^n$$, if $$x$$ is a root, so is $$\frac12-x$$. Now if you take the roots in pairs ($$1000$$ pairs), the sum of the individual pairs is $$x+\frac12-x=\frac12$$. Hence in total $$1000\cdot\dfrac12$$.

Second solution:

There is no difference with the first.

Extra solution:

By Vieta, the sum of the roots is the negative ratio of the two coefficients of the highest degree. Then using the Binomial theorem,

• degree $$2001$$: $$1-1=0$$,

• degree $$2000$$: $$\dfrac12\dfrac{2001}2\left(-\dfrac12\right)$$,

• degree $$1999$$: $$\dfrac{2001}2\left(\dfrac{2001-1}2\right)\left(-\dfrac12\right)^2$$.

The requested ratio is indeed $$500$$.

• Yes, that answers question 2 and 4! I think the extra solution is the same as the first solution on \$[AoPS](artofproblemsolving.com/wiki/… whose working I did not add. @YvesDaoust – JC2000 Oct 13 '18 at 9:47

In answer to questions 5,6,7 (the attempted solution) :

1. It was explained to me elsewhere that the leading coefficient on the $$RHS$$ should not be $$(1/2)$$ but in fact be $$(2001/2)$$ since the leading coefficient for the polynomial of order 2000 would be $${2001}\choose{2000}\times(1/2)$$.

2. Thus on solving for the sum of the roots the answer would be $$500$$.

3. It was also pointed out to me that this would essentially be the same as using Vieta's.

(Apologies for cross posting, I just saw a meta question regarding this and realized my error. I intend to follow the guidelines henceforth).