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Use Euclid’s Algorithm to find the monic gcd of (i) X^3 + 3X^2 + 4X + 2 and 2X^2 + 7X + 5

I ended up with 13/4X+13/4 = 45/13(169/180X +169/180), but these are not monic and I don't know where I went wrong as I just followed Euclid's algorithm.

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You haven't done anything wrong. $\frac{13}4x + \frac{13}4$ is indeed a greatest common divisor of those two polynomials. To make it monomial, just divide it by the leading coefficient to get: $$ \frac{\frac{13}4x + \frac{13}4}{\frac{13}4} = x + 1 $$ We can do this because dividing our gcd by a non-zero constant doesn't change which other polynomials it divides (at least as long as you permit rational / real coefficients, which you seem to do; if we were limited to integer polynomials, then that would be a different story, the Euclidean algorithm wouldn't even work in the first place).

Arthur
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  • thank you! is it ok if I ask you one more question? –  Nov 13 '19 at 18:19
  • If it's very related to this one. Otherwise it's better to ask it as a separate post. – Arthur Nov 13 '19 at 18:21
  • it is related, would the gcd of the polynomials X^4+X^3+X^2+X+1 and X^4+1 be 1 as they are irreducible ? –  Nov 13 '19 at 18:23
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    @MorenaDragomir Whether they are irreducible depends on your field: Over the integers or the rationals, sure, they are irreducible. Over the real numbers, any polynomial can be factored to quadratic and linear terms, for instance: $x^4 = (x^2+\sqrt 2 x + 1)(x^2 - \sqrt2x+1)$. At any rate, the $\gcd$ of two irreducible polynomials is either equal to one of the polynomials (if one is a constant multiple of the other), or it is $1$. Did you really need me to tell you that? Irreducible polynomials only have themselves and $1$ as divisors so there are only those two options for what the gcd can be. – Arthur Nov 13 '19 at 18:28