sum of the roots of x^2001 + (1/2 - x)^2001

[malick]

The 2001 AIME question #3 states: Find the sum of the roots of the polynomial x^2001 + (½ - x)^2001.

The answer states:Expanding, coeff x^2001 = 0, coeff x^2000 = 2001/2, coeff x^1999 = -250·2001. Hence sum = 500.

I understood everything until the hence sum = 500 part.

Am I missing something?

 

[pka]

There is probably a typo in the solution. The word should be QUOTIENT not SUM.

It is the quotient of the coefficients of x<SUP>1999</SUP> & x<SUP>2000</SUP>, that is \(\displaystyle \frac{{\frac{{\left( {2001} \right)\left( {2000} \right)}}{8}}}{{\frac{{2001}}{2}}} = 500.\)

The reason is that the sum of roots principle applies to monic polynomials.
The sum of the roots of \(\displaystyle 3x^6 + 4x^5 - 5x^4 - x^2 + 1\) is \(\displaystyle \frac{{ - 4}}{3}\).
We had to divide through by the leading coefficient 3 to make it a monic polynomial.

 

[Matt]

Find the sum of the roots of the polynomial x^2001 + (½ - x)^2001.

The answer states:Expanding, coeff x^2001 = 0, coeff x^2000 = 2001/2, coeff x^1999 = -250·2001. Hence sum = 500.

I understood everything until the hence sum = 500 part.

Am I missing something?

\(\displaystyle \L\text{sum of roots}=-\frac{\text{coeff of }x^{1999}}{\text{coeff of }x^{2000}}\)

More information about the sum/product of the roots of a polynomial here.

 

[malick]

oh yeah, like -b/a in quadratics, I just didn't know that it applied to every polynomial. Thanks.