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Stu(0)P = X^4 + pX^2 + qx + r

Stu(1)P = 4x^3 + 2px + q

Stu(2)P = -[2px^2 + 3qx + 4r]/4

Should anyone know how to get from the 3 lines above to Stu(3)P shown here on next line:-

Stu^3(P) = -[(2p^3 - 8pr + 9q^2)x + (p^2)q + 12qr)]/(p^2)

I've read Wikipedia and the relevant webpages emphasising the importance of the minus sign in the remainder, and found the general notation of how one line leads on to the next with appropriate quotients factored in etc

Can anyone please show me the answer by demonstration no matter how simple/obvious/cumbersome the process?

Above taken from here.


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Well, polynomial long division of $\mathrm{Stu}^1(P)$ by $\mathrm{Stu}^2(P)$ gives $$ 4x^3 + 2px + q = \left(-\frac{2px^2 + 3qx + 4r}{4}\right) \left( \frac{-8px + 12 q}{p^2} \right) + \frac{(2p^3 - 8pr + 9q^2)x + p^2 q + 12qr}{p^2}. $$ The last term is the remainder, $-\mathrm{Stu}^3(P)$.

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Thanks Hans, Your very helpful answer was what I needed and which no one else that I asked help of seemed to understand. It was the term 'long division' that you used that actually cracked it for me. I am greatful for the time and effort that you took to reply Have a good healthy life. Paddy – user9883 Apr 25 '11 at 0:54
You're welcome. :) – Hans Lundmark Apr 25 '11 at 6:46

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