MHB Show that the polynomial has no real roots

[Saitama]

Problem:
Show that the polynomial $x^8-x^7+x^2-x+15$ has no real root.

Attempt:
I am not sure what should be the best way to approach the problem.

I thought of defining $f(x)=x^8-x^7+x^2-x$ because $f(x)+15$ is basically a shifted version of $f(x)$ along the y-axis. So if $15$ is greater than the minimum value of $f(x)$, I can conclude that the polynomial in the question has no real roots.

$f(x)$ can be written as $x(x-1)(x^6+1)$ i.e $f(x)$ has $1$ and $0$ as its real roots. $f'(x)=8x^7-7x^6+2x-1$ and this has no nice real roots. I can't find the minimum value of $f(x)$ if I can't solve $f'(x)=0$. I am stuck here.

Any help is appreciated. Thanks!

 

[Evgeny.Makarov]

So $f(x)<0\iff 0<x<1$. We have $f(x)=x(x−1)(x^6+1)\ge-|x(x−1)(x^6+1)|$ and
\[
|x(x−1)(x^6+1)|\le|x|\cdot|x-1|\cdot|x^6+1|<1\cdot1\cdot2=2
\]
for $0<x<1$. Therefore, $f(x)>-2$.

 

[Opalg]

See http://mathhelpboards.com/challenge-questions-puzzles-28/prove-equation-has-no-real-solution-8986.html?highlight=real+roots.

 

[Saitama]

So $f(x)<0\iff 0<x<1$. We have $f(x)=x(x−1)(x^6+1)\ge-|x(x−1)(x^6+1)|$ and
\[
|x(x−1)(x^6+1)|\le|x|\cdot|x-1|\cdot|x^6+1|<1\cdot1\cdot2=2
\]
for $0<x<1$. Therefore, $f(x)>-2$.

Thanks Evgeny.Makarov! :)

See http://mathhelpboards.com/challenge-questions-puzzles-28/prove-equation-has-no-real-solution-8986.html?highlight=real+roots.

I did not know that the very same problem had been already discussed here, thanks Opalg, nice to know I was along the right lines. :)