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Let $a$ be a root of $X^4 + 2X + 1 = 0$. How to express $(a+1)/(a^2 - 2a + 2)$ as a polynomial in $a$ with rational coefficients? I know $X^4 + 2X + 1 = (X+1)(X^3 - X^2 + X + 1)$ but I don't know how to proceed from here.

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    You should format your math text using MathJax. Here is a reference that may be useful: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-02-21
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    @Homaniac I am a bit confused with "express as a polynomial in $a$". Do you mean to express $(a + 1)(a^2 - 2a + 2)^{-1}$ as $c_0 + c_1 a + \dots + c_n a^n$ for some $c_i \in \mathbb{Q}$? – 2017-02-21
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    Yes I think so~ ^^ – 2017-02-21
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    @Homaniac I am going to hospital right now. Hint: $x^3 - x^2 + x + 1$ is irreducible over $\mathbb{Q}$. Hence $(a^2 - 2a + 2)^{-1}$ can be written as $c_0 + c_1a + c_2a^2$. Good luck! – 2017-02-21
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    Since your polynomial is not irreducible, there will be two correct answers, depending on whether $a$ is a root of $x+1$ or of $x^3-x^2+x+1$. – 2017-02-21
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    How can 1/(a2−2a+2) can be written as c0+c1a+c2a^2? – 2017-02-21
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    @GerryMyerson I forget to assume $a \ne 1$ in my previous comment. If $a \ne 1$, then $(a^2 - 2a + 2)^{-1}$ can be written as $c_0 + c_1 a + c_2 a^2$. If $a = 1$, then $\frac{1 + 1}{1^2 - 2(1) + 2} = 2$. – 2017-02-21
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    @Homaniac The reason is that assuming $a \ne 1$. We have $a$ being a root of $x^3 - x^2 + x + 1$, i.e. $a^3 - a^2 + a + 1 = 0 \iff a^3 = a^2 - a -1$. So if we can write $(a^2 - 2a + 2)^{-1}$ as $c_0 + c_1 a + \dots + c_n a^n$, it can also be written as $c_0 + c_1 a + c_2 a^2$. – 2017-02-21
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    @Alex, I think you mean $a\ne-1$. – 2017-02-21
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    @GerryMyerson Thanks! Indeed, it should be $a \ne -1$. So if $a \ne -1$, then the previous comment still work. If $a = -1$, we have $\frac{-1 + 1}{(-1)^2 - 2(-1) + 2} = 0$. – 2017-02-22
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    So, Homaniac, anything to say about the answers you have received? – 2017-02-23
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    Thank you friends, I'm immensely grateful for all the insights man! Particularly yours @GerryMyerson – 2017-02-23

2 Answers 2

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$${a+1\over a^2-2a+2}=Qa^2+Ra+S$$ where $Q,R,S$ are the unknown rational numbers we have to find.

We'll assume $a$ is a root of the irreducible polynomial $x^3-x^2+x+1$. It follows that $a^3=a^2-a-1$, and $a^4=a^3-a^2-a=(a^2-a-1)-a^2-a=-2a-1$.

$$a+1=(a^2-2a+2)(Qa^2+Ra+S)$$ Now multiply out the right side, and reduce to degree two by using $a^3=a^2-a-1$ and $a^4=-2a-1$. Then equate the coefficients of $1$, of $a$, and of $a^2$ on the two sides of the equation. That will give you a system of three equations in the three unknowns $Q,R,S$. Solve the system, and you win.

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Let $M = \begin{pmatrix} 0 & 0 & -1\\ 1 & 0 & -1 \\0 & 1 & 1 \end{pmatrix}$ be the companion matrix of the polynomial $p(x) = x^3 -x^2 +x +1$. It has the property $p(M) = 0$. Let $A$ be the algebra generated by $M$. It is three dimensional and a basis is $1, M, M^2$. Note that the coefficients of an element of $A$ w.r.t. this basis can be found in its first column (try out $M^0, M$ and $M^2$). Then $(M+I)/(M^2-2M+2)$ gives $\begin{pmatrix} -1 & -2 & -1\\ -1 & -3 & -3 \\2 & 1 & -2 \end{pmatrix}$ so the result (reading from the first column) is $-1 -a + 2a^2$.