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I'm not familiar with the terminology "differential operator", so when I encounter it in this exercise I'm quite stumped.

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Am I right to assume that, given a polynomial $p(x)$ of real coefficients, what $D$ does is to carry out this operation? $$D(p(x))= p_2(x)\frac{d^2p}{dx^2} + p_1(x)\frac{dp}{dx} + p_0(x)p$$

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    Almost. The last last term ought to be $p_0(x)\times p(x)$. – lulu Aug 25 '18 at 11:40
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    If possible, you could check those prestigious linear algebra textbooks. Operators would be explained there. Or, might not be the best, check wiki. – xbh Aug 25 '18 at 11:43
  • Thanks. I've fixed the error. Also I think this is a shorthand, so even though I know what a operator is, I still have to guess how it's related to what's written in the question. –  Aug 25 '18 at 11:47

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Essentially. What $D$ does can be worked out explicitly as

$$ \begin{align} D(p(x))=& \left( p_2(x)\frac{d^2}{dx^2}+p_1(x)\frac{d}{dx}+p_0(x) \right) p(x) &&\\ =& p_2(x)\frac{d^2}{dx^2}(p(x))+p_1(x)\frac{d}{dx}(p(x))+p_0(x)p(x) &&\\ =& p_2(x)\frac{d^2p}{dx^2}+p_1(x)\frac{dp}{dx}+p_0(x)p(x), && \end{align} $$

where the direction of multiplication matters. I suggest writing this out in full your first few times dealing with such operators so that you're less likely to make silly mistakes.

For the question in the exercise, think about how differentiation (first-order, second-order, etc.) affects the degree of a given polynomial. In particular, consider what the $k^{th}$ derivative does to the degree.

Stuck
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