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I have learnt so far that a polynomial of degree $n$ has $n$ roots. To find out the number of real roots it has, we have to take its derivative, equate it to $0$ and then find the roots. Those roots are the extremes in the graph of the original polynomial. Using those extremes, we can find out how many times the graph of the original polynomial intersects the $X-axis$, so we know the number of real roots. Now, consider a polynomial of degree $3$. Suppose its derivative has no real roots. Does this $necessarily$ mean that the original polynomial has only one real root? Moreover, consider a polynomial of degree $4$. If its second derivative has no real roots, then (taking into account that the first derivative has one real root) will the original polynomial have two real roots? And does this continue for higher degrees?

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If a polynomial has a non-real ('complex') root, say $x=a+ib$ then the complex conjugate, then $\bar{x}=x-ib$ must be a root as well (that's not very hard to prove). Does that help you perhaps? – postmortes Nov 29 '13 at 9:12
up vote 2 down vote accepted

Yes, a polynomial of degree$~3$ (or of any odd degree) whose derivative has no real roots defines a strictly monotonic function $\Bbb R\to\Bbb R$, which being bijective has exactly one zero. But no, for a polynomial of degree$~4$ (or of any even degree) all the information you can give about its derivatives will not allow telling whether it has any roots: the answer to that question can be altered by adding a constant to the polynomial, which does not change any of its derivatives.

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