Finding real roots of $ P(x)=x^8 - x^7 +x^2 -x +15$ 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 ?

we have to calculate no. of ral roots of $x^8-x^7+x^2-x+15=0$
$\bf{Solution::}$ Let $f(x) = x^8-x^7+x^2-x+15$, we check real solution in $\bf{x\in \mathbb{R}}$
$\bullet$ If $x\leq 0$, Then $f(x) = x^8-x^7+x^2-x+15>0$
$\bullet$ If $0<x<1$, Then $f(x) = x^8+x^2(1-x^5)+(1-x)+14>0$
$\bullet$ If $x\geq 1$, Then $f(x)= x^7(x-1)+x(x-1)+15>0$
So we observe that $f(x)= x^8-x^7+x^2-x+15>0\;\forall x\in \mathbb{R}$
So $f(x)=x^8-x^7+x^2-x+1=0$ has no real roots for all $x\in \mathbb{R}$
