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Sorry if the title isn't very descriptive. It's just that I'm really not sure how to phrase the question.

Anyways here is the question. Let $a$ and $b$ be positive roots of polynomials $(x+1)^2-2$ and $(x+2)^2-5$. Let $c = a+b$. Part one says find a polynomial $p(x) \in Q[x]$ such that $p(c) = 0$. This is extremely simple. However, I'm completely stuck on part two, which says to find a couple of polynomials $q(x), r(x) \in Q[x]$ such that $\frac{q(c)}{r(c)} = a$. No idea how to start. Any help/hint would be much appreciated.

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I really need help on this. Can anyone give me a hint? – Aden Dong Dec 18 '12 at 1:17
Here's what we know: $a^2=1-2a$ and $b^2=1-4b$, so any time we consider polynomials in $c$ we get some polynomial which we can always reduce to something with just $a$, $b$, and $ab$ (no higher powers). – process91 Dec 18 '12 at 1:34
up vote 1 down vote accepted

First, note that $a^2=1-2a$ and $b^2=1-4b$, so any polynomial in $c$ will reduce to some linear combination of $1$, $a$, $b$, and $ab$ (since all higher powers can be reduced by the relations above).

Now let's see what we get (I had no particular strategy here, just trying to get $a$ to come out): $$c^2=a^2+2ab+b^2=1-2a+2ab+1-4b\\ \tfrac 1 2 (c^2+4c-2)=ab+a$$

Now we can work with this result:

$$\begin{align} (ab+a)^2&=a^2b^2+a^2+2a^2b\\ &=(1-2a)(1-4b)+1-2a+2(1-2a)b\\ &=2-4a-2b+4ab\\ \Rightarrow\tfrac 1 2 [(ab+a)^2-2+4c]&=b+2ab \end{align}$$

This is actually quite useful, as it gets rid of the mixed product from $c^2$, that is we now have that $$c^2=2-2a-5b+(b+2ab)$$ and now you can work backwards through this to find a polynomial equal to $a$. Surprisingly, we do not even need $r(x)$ in order to do this. Let me know if you need me to expand out the details of this last bit.

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Yep. After I saw your hint (which was most helpful), I was able to arrive at more or less the same answer as yours several minutes ago. – Aden Dong Dec 18 '12 at 2:31
@AdenDong Glad to hear it! Did yours make use of the $r(x)$ polynomial? I was surprised that we didn't need it, and thought perhaps I had made a mistake. – process91 Dec 18 '12 at 2:35
Well after I worked it out, it seemed that there was no need for the $r(x)$. Basically I just found a polynomial $q(x)$ such that $q(c)=a$ and let $r(x) = p(x)+1$, so that $r(c) = p(c)+1=1$. – Aden Dong Dec 18 '12 at 2:36
@AdenDong Good to know. You could also just let $r(x)=1$. – process91 Dec 18 '12 at 2:39
Ahh yes of course. I think I was too caught up in the thought that there has to be come sort of relation between the first part and the second part of the questions. But anyways, thanks a lot for your hint! – Aden Dong Dec 18 '12 at 2:40

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