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I got stuck on this question: Find the monic gcd of $f(x)=x^5-6x^4+13x^3-11x^2+x+5$ and $g(x)=x^2-3x+2$.

I worked through the Euclidean algorithm, first multiplying $g(x)$ with $x^3$ but then the remainder term has a larger power than $g(x)$. I have attached the working.

I know I am probably making a stupid mistake but can someone let me know how to get around this problem?

enter image description here

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    $\begingroup$ I don't see an attachment. Regardless: the algorithm calls for you to replace $f(x)$, not $g(x)$, by $f(x)-x^3g(x)$. So the degree of the first factor does go down. Then you can use a multiple of $g(x)$ to reduce the first factor's degree again ... you always replace the "larger" polynomial with the new difference, just like in the Euclidean algorithm for integers. $\endgroup$ – Greg Martin May 22 '14 at 11:51
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    $\begingroup$ Is there a missing ^2 and the polynomial is $f(x)=x^5-6x^4+13x^3-11x²+x+5$? $\endgroup$ – mathmax May 22 '14 at 11:53
  • $\begingroup$ Hi I amended my post.. The problem is I get f=qg+r and r is bigger than g?! $\endgroup$ – user136069 May 22 '14 at 12:24
  • $\begingroup$ When you do the polynomial long division, $x^3$ isn't the quotient --- it's just the first term of the quotient --- you have to keep going, get an $x^2$ term and an $x$-term and a constant in the quotient. $\endgroup$ – Gerry Myerson May 22 '14 at 13:06
  • $\begingroup$ Ohh ok. I just looked at the first term and didn't even use long division, thanks I should be ok now :) and Monica means the leading coefficient is 1 right? So I can always scale my answer if needs be $\endgroup$ – user136069 May 22 '14 at 14:13
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Consider factoring $g(x)$. By inspection, $$g(x) = x^2 - 3x + 2 = (x -1)(x-2)$$ Now check if either $(x-1)$ or $(x-2)$ is a factor of $f(x)$. Clearly, $x - 2$ cannot be a factor of $f(x)$. Why not?

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  • $\begingroup$ Hi: yep sorry it is -11x^2 let me add my attachment $\endgroup$ – user136069 May 22 '14 at 12:21
  • $\begingroup$ So an alternative to the Euclidean algorithm is to realise 1 and 2 are not roots of f hence the gcd of f and g is 1? $\endgroup$ – user136069 May 22 '14 at 12:28
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    $\begingroup$ That will do it! $\endgroup$ – Namaste May 22 '14 at 12:30

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