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the question says $P(x)=x^{32} -x^{25} +x^{18} -x^{11} +x^4 -x^3 +1$.how many possible imaginery and real roots does $p(x)=0$ has. how to determine the nature of roots for such equations of higher roots? note:By nature I mean no of possible real or imaginery roots.

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  • $\begingroup$ What do you mean by nature? $\endgroup$ – Prince M Jul 29 '16 at 0:57
  • $\begingroup$ Are you counting roots with multiplicity? $\endgroup$ – Gregory Grant Jul 29 '16 at 0:58
  • $\begingroup$ by nature I mean how many roots are real or imaginery $\endgroup$ – danny Jul 29 '16 at 0:59
  • $\begingroup$ I don't know anything about multiplicity :( $\endgroup$ – danny Jul 29 '16 at 1:02
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    $\begingroup$ If $f(x) = (x-1)^2$, then $x=1$ is a root but has multiplicity 2. Multiplicities are "repeated" roots, so to say. $\endgroup$ – Prince M Jul 29 '16 at 1:07
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Regardless of the degree, you can figure out how many roots, at maximum, are positive or negative. How? Decartes' rule of signs.

Here's how to apply it: The maximum number of positive roots of a polynomial $P(x)$ is equal to the number of sign changes in the coefficients of $P(x)$. The maximum number of negative roots is counted similarly in $P(-x)$.

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$P(x) \ge 1$ for $x \le 0$ and for $x \ge 1$.

For $x \in [0,1]$, we have $P(x) \approx x^4−x^3+1$, which has no real zeros.

Therefore, it seems that $P$ has no real zeros.

This is not a rigorous argument though.

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