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I have been spending some time on a question which I encountered in the PYQs of a small maths competition from $2$ years ago. It is from the topic quadratic equations and polynomials, it however deals with a polynomial of degree $8$ and it is very complicated and convoluted. I present the question below:

Find the real roots of the polynomial $x^8 + 8x^7 + 56x^6 + 336x^5 + 1680x^4 + 6720x^3 + 20160x^2 + 40320x + 40320$.

Checking from the trend of other similar questions, my guess is that there is a method to factorise this polynomial which can help us to find its roots. I tried multiple methods like regrouping, such as writing $x^6(x^2 + 8x + 56) + x^3(336x^2 + 1680x + 6720) + 20160x^2 + 40320x + 40320$ and tried to factorise each of the three quadratics, but that has not been helpful.

I even tried to use rational root theorem and tried to find the rational roots of this polynomial by hit and trial and put the values of $\pm 1$, $\pm 2$, $\pm 3$ and other factors of $40320$ to find such a factor where it attains value $0$. I have not managed to find any.

I even thought of completing the square but that is also not useful and I have not been able to get anything fruitful..

I think this question has a different strategy compared to the other questions where factorisation/hit and trial/completing the square was sufficient. I will appreciate any help! Thanks.

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    $\begingroup$ WolframAlpha shows that the polynomial has no real root. I'm guessing you could try to check the increasing/decreasing nature of the polynomial through its derivative if all else fails. $\endgroup$
    – Sahaj
    Commented Jun 21 at 17:47
  • $\begingroup$ might as well substitute $x = t - 1$ and see how the coefficients change; $\endgroup$
    – Will Jagy
    Commented Jun 21 at 18:00
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    $\begingroup$ on the other hand, suspecting no real roots, writing as a sum of squares would confirm that. For contests they call this SOS $\endgroup$
    – Will Jagy
    Commented Jun 21 at 18:01
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    $\begingroup$ I mean I don't know if this is helpful , but when you take the derivative it is almost the same as the original function (maybe something to do with problem). $\endgroup$
    – J.D
    Commented Jun 21 at 18:04

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You're right: the polynomial initially resists simple attempts at finding its real roots through elementary methods like RRT (as a side note; you really only needed to try out the negative factors of the constant term) and factorisation.

But if we define $f(x) := x^8+\color{blue}{8x^7+56x^6+336x^5+1680x^4+6720x^3+20160x^2+40320x+40320}$ and find its derivative $f'(x) = \color{blue}{8x^7+56x^6+336x^5+1680x^4+6720x^3+20160x^2+40320x+40320}$ we realize a useful relationship:

$$\forall x, \ f(x) = x^8 + f'(x)$$

Since $f$ is an even degree polynomial with a positive leading coefficient it must attain a minimum value at some $x=x_0$ which must also be a local minimum; so we must have $f'(x_0)=0$. That means $f(x) \geq f(x_0) = x_0^8 \geq 0$. If $x_0 = 0$, then $f$ has a root at $0$, contradiction. Thus we must have $f(x_0) = x_0^8 >0$ thereby giving $$f(x) > 0 \ \ \forall x \in \mathbb R$$ thereby showing the polynomial has no real roots.


ADDENDUM: Generalization.

After some pondering, I've generalized this result as follows:

Theorem: $$f(x):=\sum_{r=0}^{2n} \frac{x^r}{r!}$$ has no real roots for $n \in \mathbb Z$.

The proof is similar to what I did for the specific case for $n=4$ for $8!f(x)$.

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    $\begingroup$ didn't expect calculus to be so useful here. thanks for your solution! $\endgroup$
    – Sarvagya
    Commented Jun 21 at 18:30
  • $\begingroup$ @Sarvagya You're welcome. You can click on the checkmark to indicate that the question has been resolved. $\endgroup$
    – Sahaj
    Commented Jun 22 at 10:12
  • $\begingroup$ The relationship between $f$ and $f'$ becomes even more apparent if one writes $f(x) = 8! \sum_{k=0}^8 \frac{x^k}{k!}$. $\endgroup$
    – Martin R
    Commented Jun 22 at 17:16
  • $\begingroup$ You're right. I was editing in a general result exactly at this moment right now. $\endgroup$
    – Sahaj
    Commented Jun 22 at 17:17
  • $\begingroup$ The general result has been proven here: math.stackexchange.com/a/3601257/42969 (and also here math.stackexchange.com/a/1397473/42969 with a different method). $\endgroup$
    – Martin R
    Commented Jun 22 at 17:22

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