$Q[x]$ is the ring and $I=<x^{3}+6x+3>$. I want to find $(2x-5+I)^{-1}$. The problem, I dont know the technique, how to reduce the degree of the elements in the multiply polynomial, which is greater my creator's degree. I'll show you what I mean:
$$(2x-5+I)(ax^{2}+bx+c+I)=1+I$$ I want to find $a,b,c$. Soafter multypling, I get: $$2ax^{3}+(2b-5a)x^{2}+(2c-5a)x-5c+I=1+I$$ Now I want to exchange x^{3} for something that fits the form of dergee 2 and less(using the theorem that there is only one polynomial for each degree lesser then the degree of the ideal). And I don't remember the technique. Can someboyd help me from here showing and explaining how do I find the polynomial' which is $X^{3}$ but with degree lesser then $3$?
You have to write a Bézout's relation between $2x-5$ and $x^3+6x+3$, which is rather straightforward with Euclidean division: $$x^3+6x+3=\frac18(4x^2+10x+49)(2x-5)+\frac{269}8,$$ whence \begin{align*}\frac{269}8&=-\frac18(4x^2+10x+49)(2x-5)+x^3+6x+3\\&\equiv-\frac18(4x^2+10x+49)(2x-5)\pmod{x^3+6x+3}\end{align*}so that $$(2x-5)^{-1}=-\frac1{269}(4x^2+10x+49).$$