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Let $p(x)$ be a polynomial with all roots real and distinct such none of its roots is equal to zero. Prove that the polynomial $x^2p''(x)+3xp'(x)+p(x)$ also has all roots real and distinct.

Unable to create this polynomial. Need some starters. Thanks.

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Taking $Q(x)=xp(x)$, we have all roots of $Q(x)$ are real and distinct. Using Rolle theorem all roots of $Q'(x)$ are real and distinct.

Taking $H(x) = xQ'(x)$, we also have $H'(x)$ has all roots are real and distinct.

And $H'(x) = (xQ'(x))' = (x^2p'(x) + xp(x))' = x^2p''(x) + 3xp'(x) + p(x)$.

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    $\begingroup$ One should mention that $0$ is not a root of $Q'$ (which is true because $Q'(0)=p(0)$), otherwise $xQ'$ might have a double root. $\endgroup$ – Hagen von Eitzen Jul 14 '16 at 16:42
  • $\begingroup$ Yep, there is an assumption "none of roots is equal to zero". I don't mention it but, you are right. :) $\endgroup$ – GAVD Jul 14 '16 at 16:44
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    $\begingroup$ The fun part is that the operator "multiply by $x$ and differentiate" takes polynomials with distinct non-zero real-only roots into polynomials with distinct non-zero real-only roots. The problem statement applies the operator twice, but one might carry on and on ... $\endgroup$ – Hagen von Eitzen Jul 14 '16 at 16:48
  • $\begingroup$ Yes, it is a tricky. If someone is familiar with the derivative, one can see where the proving polynomial comes from, $\endgroup$ – GAVD Jul 14 '16 at 16:58

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