Why all such polynomials have $-1$ as a root?

Question

Why all polynomials of this form have $-1$ as a root?

$ x^5+x^4+x^3+x^2+x+1 $ and similar polynomials like
$ x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$

Answer 1

Hint $\ $ Evaluated at $\,x = -1\,$ each odd term $\,x^{2n+1}\,$ has value $-1$ and each even term $\,x^{2n}\,$ has value $1,$ and there are an equal number of odd and even terms, so they sum to $0\,\,$ i.e. $\,f(-1) = 0.$

Your cases are sums of terms $\,x^{2n}+x^{2n+1} = (1+x)x^{2n}\,$ so you can explicitly factor out $\,x+1\,$ which shows that $\,x = -1\,$ is a root.

Generally, if $\,f(x) = a_0+a_1 x+\,\cdots\,+ a_n x^n\,$ then $\,f(-1) = a_0-a_1+a_2-a_3+\,\cdots+ (-1)^n a_n$ is the alternating sum of the coefficients, which, e.g. $= 0\,$ if $\,n\,$ is odd and $\,a_{2k} = a_{2k+1},\, $ or, more precisely, iff $\,a_0 + a_2 + \cdots = a_1 + a_3 +\,\cdots,\,$ i.e. the odd and even parts agree at $\,x = 1,$ i.e.

$$ f(-1)\,=\, 0 \iff \overbrace{\frac{f(1)+f(-1)}2}^{\large {\rm even\ part\ at\ } x = 1}\,=\, \overbrace{\frac{f(1)-f(-1)}2}^{\large {\rm odd\ part\ at\ } x = 1}$$

Answer 2

Because when plugging $-1$, all the terms with even degrees will sum up to $n/2+1$, and all the terms with odd degrees will sum up to $-n/2-1$, adding $1$ you get $0$, and thus $-1$ will be a root. ($n$ being the number of terms of those polynomials)

$$\underbrace{x^n+x^{n-2}+\cdots+x^2+1}_{\displaystyle \color{white}{\overset{}{\color{black}{\dfrac n2+1}}}}+\underbrace{x^{n+1}+x^{n-1}+\cdots +x}_{\displaystyle\overset{}{-\dfrac n2-1}}$$

I hope this helps.
Best wishes, $\mathcal H$akim.

Answer 3

Those polynomial functions can be compressed using geometric sums, $$ x^{2n-1}+x^{2n-2}+...+x+1=\frac{x^{2n}-1}{x-1} $$ The numerator allows another factorization $$ x^{2n}-1=(x^2-1)(x^{2(n-1)}+x^{2(n-2)}+...+x^2+1) $$ so that the original polynomial is equal to $$ (x+1)(x^{2(n-1)}+x^{2(n-2)}+...+x^2+1) $$ This could of course also be seen directly by grouping the even number of terms in the original polynomial.

Answer 4

Let $$P(X)=\sum\limits_{k=0}^{N}X^k.$$ Then,

$$P(-1)=\sum\limits_{k=0}^{N}(-1)^k=1+\sum\limits_{k=1}^{N}(-1)^k.$$