# Finding the real roots of an octic polynomial (degree eight). [duplicate]

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.

• 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. Commented Jun 21 at 17:47
• might as well substitute $x = t - 1$ and see how the coefficients change; Commented Jun 21 at 18:00
• on the other hand, suspecting no real roots, writing as a sum of squares would confirm that. For contests they call this SOS Commented Jun 21 at 18:01
• 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).
– J.D
Commented Jun 21 at 18:04

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.

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 $$8!f(x)$$.
• The relationship between $f$ and $f'$ becomes even more apparent if one writes $f(x) = 8! \sum_{k=0}^8 \frac{x^k}{k!}$. Commented Jun 22 at 17:16