Answer
$6$ or $4$ or $2$ or $0$ positive roots
$0$ negative roots
$6$ or $4$ or $2$ or $0$ real roots.
Work Step by Step
Descartes' Rule of Signs states that the possible number of the positive roots of a polynomial is equal to the number of sign changes in the coefficients or less than the number of sign changes by a multiple of $2$ AND the possible number of negative roots of a polynomial is equal to the number of sign changes or less than the total number of sign changes by a multiple of $2$ after substituting $−x$ for $x$. The substitution has the effect of negating all of the odd-power terms in the polynomial.
Therefore,
$P(x)=x^8-x^5+x^4-x^3+x^2-x+1$, has $6$ sign changes, between $x^8$ and $-x^5$, and $-x^5$ and $x^4$, and $x^4$ and $-x^3$, and $-x^3$ and $x^2$, and $x^2$ and $-x$, and $-x$ and $+1$. Therefore, it has $6$ or $4$ or $2$ or $0$ positive roots.
$P(-x)=x^8+x^5+x^4+x^3+x^2+x+1$, has no sign change, Therefore, it has no negative roots.
Therefore, It has a total of $6$ or $4$ or $2$ or $0$ real roots.
