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If I had some polynomial $q$ and knew it had a repeated root at $x = a$ for some real $a$, I can write down $q(x) = A(x-a)^2 p(x)$ for some polynomial $p$, and some real $A$. Is this any more or less general than $q(x) = $D$(x-a)^2p(x)$ + $C$ for real $C,D$, what I mean by this is that are all the possible polynomials I get by varying $A$ in the first example the same as the possible polynomials I get by varying $C$ and $D$? Thank you so much for reading.

Nav Bhatthal
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  • For $C=0$ you get the same polynomials, for $C\neq 0$ you get polynomials that do not have a root at $a$. I don't get your question. – Raskolnikov Jul 19 '23 at 15:55

2 Answers2

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The two forms are not equivalent.

$q(x) = A(x-a)^2 p(x)$ for some polynomial $p$, and some real $A$

This describes a polynomial $q$ with root $a$ of multiplicity $\ge 2$.

$q(x) = $D$(x-a)^2p(x)$ + $C$ for real $C,D$

This describes a polynomial $q$ whose derivative $q'$ has $a$ as a root. $a$ is also a root of $q$ iff $C=0$.

dxiv
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  • You are a great teacher, also who always helps when I comment to help you to fix the answer, because you are not helping to get answer/comment (up)votes. I hope you will return to the site soon as a great teacher . – lone student Oct 17 '23 at 06:35
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    Thank you for the kind words, @lonestudent. Right, I have been less active lately, but I have not left the site, and I have not closed the door on "returning" at some point. – dxiv Oct 18 '23 at 03:33
  • Due to homelessness, I cannot be on the site much. I'm overstepping my bounds by bothering you again, but if you have a few minutes here, I'd love for you to guide me. As you know, irreducibility and similar things are not at the school level. Generally, I always avoid using such terminologies. I wrote an answer, but that is beyond my knowledge. Actually, I wouldn't want to embarrass myself by sending this answer to MSE. In this sense, can you give me a little help from your wealth of knowledge to understand how consistent this answer is ?... – lone student Nov 01 '23 at 06:23
  • @lonestudent I hope you stay safe, wherever you are, and find all the support to get through. Kudos for sticking to math despite the other trials. It's always been a safe space where we are all equal, and right vs. wrong is not a matter of sway. – dxiv Nov 02 '23 at 05:01
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    Thank you very much for your nice and kind words, for not rejecting my request for help, which was not the right request to write in a comment, and for not ignoring my request when I was really stuck in my answer while I was writing an answer to learn it myself. I wish you health, wellness and well-being . Thank you so much, I learned a lot from you . – lone student Nov 08 '23 at 03:02
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$C$ must be $0$ in your second expression in order to have a root at $x=a$ so there is just one constant. In fact there are none, since you can absorb the original $A$ in the quotient polynomial $p(x)$.

Ethan Bolker
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