Question: Let $a,b>0.$ Can the polynomial $$x^{10} − x^7 + 2x^5 + ax^3 − bx + 1$$
have exactly three (counting multiplicity) positive zeroes? Can it have three simple positive zeroes together with one positive zero of multiplicity two? Explain.
I understand Decartes' Rule of Signs. There are four sign changes therefore there are a maximum of four positive zeroes that are possible. What I'm not sure is if a zero with a multiplicity of two were to count as two zeroes. For example, $(x-1)(x-2)(x-3)(x-4)^2$, are there five or four zeroes present?
Thank you!