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Is the following reasoning correct?

According to the complex conjugate root theorem, the number of complex roots of a polynomial is always equal to its degree.

Since odd degree polynomials have a maximum of 2 turning points, they can have a maximum of 3 real roots. And since even degree polynomials have a maximum of 1 turning point, they can have a maximum of 3 real roots.

Therefore, the maximum number of a real roots a polynomial of any degree can have is 3, all other roots are non-real.

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I don't know where you got the information that an odd degree polynomial has at most two turning points:

enter image description here

Actually the maximum number of turning points of a degree $n$ polynomial is $n-1$ and the one in the picture indeed has four of them, and five real roots.

Generalizing, if $n$ is any positive integer, the polynomial $$ (x-1)(x-2)\dots(x-n) $$ has $n$ distinct roots.

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  • $\begingroup$ What is that nice application? $\endgroup$ – Lee Mosher Jul 9 '17 at 14:02
  • $\begingroup$ @LeeMosher Grapher, which comes along with Mac OS X $\endgroup$ – egreg Jul 9 '17 at 14:07
  • $\begingroup$ Hah, seems I overlooked that since... forever. $\endgroup$ – Lee Mosher Jul 9 '17 at 14:08

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