TITLE: Finding real roots of $ P(x)=x^8 - x^7 +x^2 -x +15$
QUESTION [11 upvotes]: Let $ P(x)=x^8 - x^7 +x^2 -x +15 $, Descartes' Rule of Signs tells us that the polynomial has 4 positive real roots , but if we group the terms as $$ P(x)= x(x-1)(x^6+1) +15 $$ we find that $ P(x) $ does not have any positive roots.
Where did I err ?

REPLY [5 votes]: Just for "fun" :
$\displaystyle  x^8-x^7+x^2-x+15 = \frac{(128x^4-64x^3-16x^2-8x-9)^2+4(16x^2-11x+121)^2+60(x-49)^2+43055}{2^{14}}>0\;\forall x\in \mathbb{R}$
Another Solution::
If $x<0,$ note that $x^8+(-x^7)+x^2+(-x)>0,$ so the polynomial cannot have any negative roots.
If $x\geq 0,$ then note that from AM-GM inequality we have:
$\left\{\begin{aligned}& \frac 78 x^8+\frac 18\geq x^7\\& x^2+\frac 14\geq x\end{aligned}\right\} ;$
Thus $\displaystyle\frac 78x^8-x^7+x^2-x+\frac 38>0;$