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I found this proof for the Descartes' Rule of Signs.

Towards the end the author writes this:

Now return to our original polynomial, $$f(x) = x_n + a_{n-1}x^{n-1} + ... + a_1x + a_0.$$ We can express f(x) in factored form as
$$f(x) =N(x)(x - p_1)(x - p_2) ... (x-p_m)$$
where $N(x)$ has no positive roots (but may, of course, nevertheless have variations in sign), and the $p_i$ are the $m$ positive roots of $f(x)$, listed repeatedly, if necessary, according to their multiplicity.

I wanted to know how $N(x)$ could have variations in signs if it has no positive roots.

What I assumed is $N(x)$ will have negative roots which will be of the form $(x+k)$ or they will have complex roots, which will be of the form $(x^{2n}+k)$, where n is a positive integer and k is a positive real number, in the factorized polynomial. In this case, I see that all of the signs of coefficients in the expanded polynomial format will be positive and there would be no sign variations.

How is my assumption wrong? Is there an example of a polynomial without positive roots but with sign variations?

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You have failed to consider complex roots that aren't wholly imaginary, whose corresponding quadratic factors in $N(x)$ might have negative linear terms. For example, if $f(x)=x^2-x+1$ then $N(x)=f(x)$.

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  • $\begingroup$ Oh yeah yes! Since the proof is based on the number of sign changes, can we also say that the number of sign changes in N(x) is an even number? $\endgroup$ – Tanmay Gujar Apr 20 '20 at 5:14
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    $\begingroup$ @TanmayGujar Yes. $\endgroup$ – Parcly Taxel Apr 20 '20 at 5:15
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How about $x^2-x+1$? Its roots are complex (with positive real part), but it has no positive real roots.

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