0
$\begingroup$

It is obvious the GCD of those two is 1.

The thing that makes me post this question is that why when using Euclidean Algorithm the GCD seems to be $ \frac{5}{4}$ ?

I want furthermore to find the inverse of the polynomial and i know i can do that when the GCD = 1 by using Extended Euclidean Algorithm. Yet as i've said i have 5/4.

$\endgroup$
4
  • 2
    $\begingroup$ $5/4$ is a unit (invertible) in the ring of polynomials with inverse $4/5$. It is like if you find that $-1$ is the gcd between two (rational) integers. People don't count an invertible factor. $\endgroup$
    – OR.
    Jun 18, 2017 at 10:58
  • 1
    $\begingroup$ Divide your whole equation by $5/4$ and you get an equation with 1. $\endgroup$
    – OR.
    Jun 18, 2017 at 11:19
  • $\begingroup$ OH RIGHT!! Thank you very much !! $\endgroup$
    – Eduard6421
    Jun 18, 2017 at 11:26
  • 1
    $\begingroup$ That's the reason why people don't count units as interesting factors, you can divide them out as you need. $\endgroup$
    – OR.
    Jun 18, 2017 at 11:29

1 Answer 1

1
$\begingroup$

So what? The conclusion (assuming that the computations are correct) is that $\frac54$ is a GCD of your polynomials. This is the same thing as saying that they are relatively prime.

$\endgroup$
3
  • $\begingroup$ The thing is that furthermore i want to find the inverse of the polynomial. If from Euclid i didn't get 1 i don't know how am i supposed to continue\ $\endgroup$
    – Eduard6421
    Jun 18, 2017 at 11:16
  • $\begingroup$ @Eduard6421 What's the problem? Start with $\frac54$ and express it as $\alpha(x)p(x)+\beta(x)q(x)$, where $p(x)$ and $q(x)$ are your polynomials. Then $1=\frac45\alpha(x)p(x)+\frac45\beta(x)q(x)$. $\endgroup$ Jun 18, 2017 at 11:18
  • $\begingroup$ I don't know how did i let that slip, that s completely obvious.. Thank you a lot! $\endgroup$
    – Eduard6421
    Jun 18, 2017 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.