As complicated as 321

Lemma: The polynomial $x^{2n} + ... + x + 1$ has no real roots.

Proof: Consider the equation $x^{2n + 1} - 1 = 0$ . This polynomial has only one root, namely $x = 1$ , but it also factors into $(x-1)(x^{2n} + x^{2n-1} + ... + x +1)$ . Because the factor of $x - 1$ is present, it accounts for the lone root of $x = 1$ , meaning the other factor $(x^{2n} + x^{2n-1} + ... + x +1)$ has no real roots.

Proceeding the with problem at hand now, we see that the given polynomial above splits additively into

$(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) + (x^4 + x^3 + x^2 + x + 1) +(x^2 + x + 1).$

Each of these polynomials are always greater than zero, so their sum is also always greater than zero. Hence, there are no real roots.

It was pretty interesting learn about the Descartes´ Rule of Signs. A very useful tool to predict the existence of real roots of a polynomial.

However, in this particular case, reading the comments on facebook, i do not found a conclusive argument to the inexistence of real roots.

In the better case, the equation does not have positive roots or there are 6, 4, 2 or none negative real roots. What helps, but not concludes the question.

I propose the following, for all x, positives, negatives and for x = 0, for all x real:

$\small x^6 + x^5 + 2x^4 + 2x^3 + 3x^2 + 3x + 3 = (x^2+x).((x^2+1)^2+2) + 3$

As,

$\small -1/4 \leqslant (x^2+x) \Rightarrow -1 < -1/4 \leqslant (x^2+x) \Rightarrow -1 < (x^2+x)$

And,

$\small 3 \leqslant ((x^2+1)^2+2) \Rightarrow 0 < 3 \leqslant ((x^2+1)^2+2) \Rightarrow 0 < ((x^2+1)^2+2)$

Thus,

$\ -1 < (x^2+x).((x^2+1)^2+2)$

Therefore,

$\small -1 + 3 < (x^2+x).((x^2+1)^2+2) + 3 = x^6 + x^5 + 2x^4 + 2x^3 + 3x^2 + 3x + 3$

Finally, for all x real,

$\large x^6 + x^5 + 2x^4 + 2x^3 + 3x^2 + 3x + 3 > 2$

$\large \Longleftrightarrow$

$\large x^6 + x^5 + 2x^4 + 2x^3 + 3x^2 + 3x + 3 > 0$

What means that there is no real roots for the polynomial

$\large x^6 + x^5 + 2x^4 + 2x^3 + 3x^2 + 3x + 3$

I hope this helps. And I would be very happy to have any reply.

I've made this answer before reading Rohit Kumar. So, in this very opportunity, i also wanna greet him for the Brilliant answer that he made.

In this case, I reasoned that the potential root must be negative or zero, as otherwise P(r)=\=0(or P(r) is not equal to 0, or just that it won't work). -1 cancles out all the x terms but the +3 isn't helpful, making the result 3. 0 doesn't work for the same reason. And without trying, I know -2 won't work because while -2^6=64, -2^5=32, meaning they will not cancel out well enough.-3,-4,-5 and so on will only get worse, as this yields bigger differences between powers. From this, it makes it pretty clear that such a root is impossible.