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Does anyone know of a relationship between the number of zeros (complex and real) and the degree of a polynomial? Specifically, if a polynomial has a double root, is it compensated with complex zeros?

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    $\begingroup$ The reason a double root is called a double root is because it counts as two roots; if you count them that way then an nth-degree polynomial has exactly $n$ roots over the complex numbers. Otherwise all you can say is that an nth-degree polynomial has at most $n$ roots over the complex numbers. $\endgroup$ – MJD Jul 14 '14 at 15:37
  • $\begingroup$ "at most n different roots" $\endgroup$ – Timbuc Jul 14 '14 at 15:44
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    $\begingroup$ Look into the Fundamental theorem of algebra. $\endgroup$ – Théophile Jul 14 '14 at 15:53
  • $\begingroup$ Yes, the property holds for complex roots too. $\endgroup$ – Yves Daoust Jul 14 '14 at 16:18
  • $\begingroup$ When a polynomial has real coefficients, you can always factor it as the product of first degree binomials for the real roots and second degree trinomials with real coefficients for the complex pairs. $$x^5-3x^4+4x^3-4x^2+3x-1=(x-1)(x-1)(x-1)(x^2+1),$$ $$x^4+8x^3+26x^2+40x+25=(x^2+4x+5)(x^2+4x+5).$$ $\endgroup$ – Yves Daoust Jul 14 '14 at 16:23
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The fundamental theorem of algebra states that an $n^{th}$ degree polynomial has exactly $n$ roots, provided you count them with their multiplicity.

This is equivalent to saying that any polynomial has a root.

Because if you divide a polynomial of the $n^{th}$ degree by the monomial $x-r$ (where $r$ is a root), you get an $(n-1)^{th}$ polynomial, which has a root, and you can continue until $n=1$ (this is called deflating the polynomial).

Said differently, any polynomial can be factored as the product of $n$ binomials. $$x^3-3x^2+2x=(x-0)(x-1)(x-2),$$ $$x^5-3x^4+4x^3-4x^2+3x-1=(x-1)(x-1)(x-1)(x-i)(x+i),$$ $$x^4+8x^3+26x^2+40x+25=(x+2+i)(x+2+i)(x+2-i)(x+2-i).$$

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