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Let $$f(x)=(x-a)(x-b)^3(x-c)^5(x-d)^7 $$ where $a,b,c,d$ are real numbers with $a < b < c < d$ . Thus $ f ( x )$ has $16$ real roots counting multiplicities and among them $4$ are distinct from each other. Consider $f ' ( x )$, i.e. the derivative of $f ( x )$. Find the following:
$(i)$ the number of real roots of $f ' ( x )$, counting multiplicities,
$(ii)$ the number of distinct real roots of $f ' ( x )$.


This is a polynomial of degree $16$ hence the derivative will be of degree $15$ and hence it will have $15$ roots. But are they real ?
How to find distinct real roots ? Rolle's theorem tells only about existence of root.

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  • $\begingroup$ How about finding $f'(x)$ explicitly? $\endgroup$ – Brian Cheung Apr 24 '16 at 13:06
  • $\begingroup$ Hint: $12$ of the roots are obvious (every multiple root of $f(x)$ is a zero of $f'(x)$. $\endgroup$ – lulu Apr 24 '16 at 13:07
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If $p(x) = (x-\lambda)^n q(x)$ then from the product rule:
$p'(x) = n(x-\lambda)^{n-1}q(x)+(x-\lambda)^nq'(x)=(x-\lambda)^{n-1}\left[nq(x)+(x-\lambda)q'(x)\right]$
This shows that if $\lambda$ is a root of $p(x)$ of multiplicity $n>1$, then $\lambda$ is also a root of $p'(x)$ with multiplicity $n-1$.
This gets you 2 + 4 + 6 = 12 real roots from b, c, and d, and the other 3 come from Rolle's theorem.
To summarize: 15 real roots, 6 of which are distinct.

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You may want to consider the following: A root of some polynomial $P$ with multiplicity $n$ is a root of its derivative $P'$ with multiplicity $n-1$. Hence here, $b$, $c$ and $d$ wil be distinct real roots with respective multiplicities $2$,$4$,$6$, accounting for $12$ of the $15$ roots of $f'$. dividing $f'$ by $(x-b)^{2}(x-c)^{4}(x-d)^{6}$ will give you a degree $3$ polynomial, and ascertaining the v

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  • $\begingroup$ ascertaining the nature of its roots can be done by calculating the disriminant $\endgroup$ – Noe Blassel Apr 24 '16 at 13:30

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