Roots of the Derivative of a Polynomial

**joeyjoejoe** · January 9th 2010, 03:20 PM

Just took a qualifying exam (and did pretty well on it I think), but got stumped by one of the questions.

Show that if a polynomial has all real roots then it's derivative has only real roots as well.

I couldn't construct a counterexample, so it seems true to me.

If the roots are all distinct you can use Rolle's Theorem on each interval between the roots. But what would you do if you had multiple real roots? Why are you guaranteed real roots of the derivative in this case? Thanks in advance for any help.

**Drexel28** · January 9th 2010, 04:14 PM

Originally Posted by joeyjoejoe

Just took a qualifying exam (and did pretty well on it I think), but got stumped by one of the questions.

Show that if a polynomial has all real roots then it's derivative has only real roots as well.

I couldn't construct a counterexample, so it seems true to me.

If the roots are all distinct you can use Rolle's Theorem on each interval between the roots. But what would you do if you had multiple real roots? Why are you guaranteed real roots of the derivative in this case? Thanks in advance for any help.

Descarte's rule of signs.

**Jose27** · January 12th 2010, 07:12 PM

Or you could use the theorem that says: if $P(x)$ is a complex polynomial then the roots of $P'(x)$ are contained in the convex hull of the roots of $P$

**CaptainBlack** · January 12th 2010, 09:01 PM

Originally Posted by joeyjoejoe

Just took a qualifying exam (and did pretty well on it I think), but got stumped by one of the questions.

Show that if a polynomial has all real roots then it's derivative has only real roots as well.

I couldn't construct a counterexample, so it seems true to me.

If the roots are all distinct you can use Rolle's Theorem on each interval between the roots. But what would you do if you had multiple real roots? Why are you guaranteed real roots of the derivative in this case? Thanks in advance for any help.

If the polynomial $P(x)$ of order $N$ has a root of multiplicity $n$ at $x=a$ then we may write:

$P(x)=(x-a)^n Q(x)$

where $Q(x)$ is a polynomial for which $Q(a) \ne 0$ . Then:

$P'(x)=n(x-a)^{n-1}Q(x)+(x-a)^n Q'(x)=(x-a)^{n-1}[nQ(x)+(x-a)Q'(x)]$

So as the expression in the square brackets is a polynomial which is non-zero at $x=a$ we conclude that $P'(x)$ is a polynomial with a root of multiplicity $n-1$ at $x=a$ .

So if a polynomial $P(x)$ of degree $N$ has all real roots $a_1, a_2, ..., a_n$ of multiplicities $k_1, k_2, ..., k_n$ then:

$\sum k_i=N$

and $a_1, a_2, ..., a_k$ are roots of $P'(x)$ of multiplicities $k_1-1, k_2-1, ..., k_n-1$ and in addition between each adjacent pair of roots of $P(x)$ there is at least one additional root of $P'(x)$ we have a total of at least :

$(n-1)+\sum (k_i-1) =N-1$

real roots counting multiplicities for $P'(x)$ , but this is the order of $P'(x)$ so we have exactly $N-1$ real roots counting multiplicities for $P'(x)$

Note: there are a few gaps in the above that need filling, but you get the picture?

CB

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