solve gcd $n^3+3n^2-5$ and $n+2$ with bezout

Question

Hello I have to solve $\gcd (n^3+3n^2-5 , n+2)$ with bezout

Here’s how I did it:

bezout says $a$ and $b$ are co prime if and only if $au+bv = 1$.

Then I did : $(n+2)(n^2+n-2) -(n^3+3n^2-5)$

I found one so their $\gcd$ is equal to $1$.

Did I get the right method ?

Sorry for my bad english !!!

yes but also if it's good like have I solve the problem ? @YvesDaoust — achla95, May 17 '21 at 15:38
$(n+2)(n^2+n-2) -(n^3+3n^2-5)=1$ is what the extended Euclidean algorithm gives in $\mathbb Q[n]$ as a polynomial ring: WA — lhf, May 17 '21 at 20:18

score 1 · Accepted Answer · answered May 17 '21 at 19:59

$$A(n^3+3n^2-5) + B(n+2) = 1$$

This isn't a blueprint for a general method of finding a solution. I am going to take advantage of the fact that the second polynomial is $n+2$, a first degree polynomial.

Let $m = n+2$. Then

$$n^3+3n^2-5 = (m-2)^3+3(m-2)^2-5 = m^3 - 3m^2 -1$$

So we need to solve

$$A(m^3 - 3m^2 -1) + B(m) = 1$$

If we let $A=-1$, we get

\begin{align} A(m^3 - 3m^2 -1) + B(m) &= 1 \\ -1(m^3 - 3m^2 -1) + B(m) &= 1 \\ Bm &= m^3 -3m^2 \\ B &= m^2 - 3m \\ B &= (n+2)^2-3(n+2) \\ B &= n^2 + n - 2 \end{align}

It is easy to check that

$$A(n^2 + n - 2) + B(n+2) = (-1)(n^3+3n^2-5) + (n^2 + n - 2)(n+2) = 1 $$

Thank you for your explanations this helped me a lot despite having taken time to understand. — achla95, May 18 '21 at 14:43

solve gcd $n^3+3n^2-5$ and $n+2$ with bezout

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