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So I was bored in class and decided to graph polynomials in geogebra, I noticed that $x^{4}+x^{3}+x^{2}+x+1$ and $x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$, are all above the x-axis.

Now I am wondering if it is possible to prove that these polynomials (or maybe the first one) are above x-axis without finding the stationary points. (I ask this since I can't solve cubic equations without newtons method).

I am also wondering that if every polynomial that has the format above, which starts with an even number power will be above the x-axis.

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6 Answers 6

up vote 22 down vote accepted

For $n$ odd:

  • if $x\neq 1$: $$ 1+x+\cdots+x^{n-1} = \frac{x^n-1}{x-1}\neq 0 $$
  • if $x=1$: $$1+x+\cdots+x^{n-1}=n> 0$$

As it is a continuous function the intermediate value theorem concludes that it is $>0$.

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it is also important that $n$ is odd :) –  mookid Mar 13 '14 at 14:02
I would replace the last $\ne 0$ to $> 0$ –  leonbloy Mar 13 '14 at 14:04
I like your suggestion! –  mookid Mar 13 '14 at 14:05
My goodness how did I not realise that there was a geometric series.... I guess my brain has gotten lazy, through the holidays. Anyways thank you! –  user135199 Mar 13 '14 at 14:18

It turns out that any polynomial or rational function that is always positive can be written as a sum of squares.


$$ x^4 + x^3 + x^2 + x + 1 = \left(\frac{x^2 + x}{\sqrt{2}}\right)^2 + \left(\frac{x + 1}{\sqrt{2}}\right)^2 + \left(\frac{x^2}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 $$

Alas, I don't know of any systematic way to figure out how to come up with such a representation, although this one is easily extended to the particular family of polynomials you are interested in.

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Apparently, this is Hilbert's seventeenth problem. –  Ian Mateus Mar 22 '14 at 22:21

The polynomial equation $x^n-1=0$ has n complex roots, one of which is definitely $1$, and, if n is even, then so is $-1$. But if you divide the polynomial with $x-1$ for odd n, that root which is $1$ just disappears. And if you were already taught complex numbers in class, then you already know by now that the complex solutions to $x^n=a$ form a regular n-sided polygon of radius $\sqrt[n]a$, centered in the origin. Now, for $n=3$, we have an equilateral triangle with one tip in $x=1$, and the other two tips obviously not on the real axis. The same for $n=5$, (where we have a regular pentagon), and for all other odd values of n as well. So all other roots, except for $x=1$, which we've eliminated through polynomial division, are non-real. See n-th root and/or cyclotomic polynomial for more details.

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For the first one, a more elementary approach would be:

$$\begin{align}x^4 + x^3 + x^2 + x + 1 &= x^2\left(x^2 + x + \frac{1}{2}\right) + \frac{1}{2}x^2 + x + 1\\ &\ge\frac{1}{2}\left((x+1)^2 + 1\right)\\ &\ge \frac{1}{2}\\ &> 0 \end{align}$$

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Now for real $y,$ $$y^2-y+1=\left(y-\frac12\right)^2+\frac34\ge\frac34$$

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By pairing monomials differing by one degree it is possible to obtain a solution to the problem.

We show that any polynomial of the following form:

$$x^{2n}+x^{2n-1}+...x+1$$ where $n \in \mathbb {N}$, is strictly positive for all $x \in \mathbb{R}$

If $x<-1$, then $x+1<0$, implying that $x^{2i-1}(x+1)>0$, and hence:

$$x^{2n}+x^{2n-1}+...x+1= \sum_{i=1}^n (x^{2i} + x^{2i-1}) +1=\sum_{i=1}^n x^{2i-1}(x+1) +1> 0$$

If $ -1\leq x <0$, then $x+1\geq0$, hence $x^{2i-2}(x+1) \geq 0$. This implies that

$$x^{2n}+x^{2n-1}+...x+1= x^{2n}+ \sum_{i=1}^n (x^{2i-1} + x^{2i-2}) =x^{2n}+\sum_{i=1}^n x^{2i-2}(x+1) \geq x^{2n}+ 0>0$$

Finally if $x\geq 0$ it is clear that $x^{2n}+x^{2n-1}+...x+1\geq1>0$

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