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I'm working with two polynomials:

$$ \begin{eqnarray} f(x) &=& 2t^{3} - 3t^{2} + 6 \\ g(x) &=& 6t^{2} - 6t \end{eqnarray} $$ and I'm doing the long division to find the $\operatorname{gcd}(f,g)$ using the Euclidean algorithm. I know from my notes that $\operatorname{gcd}(f,g)=1$, and after doing the process of long division I find that $2t^3-3t^2+6 = (t/3 - 1/6)(6t^2 - 6t) + (-t+6)$. I can't figure out how repeating the long division would imply I have a $\operatorname{gcd}$ of $1$. Any help with this would be much appreciated

Xander Henderson
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Sinnner
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  • $f(t)$ is coprime to $6t(t-1)\iff f(0)\neq 0\neq f(1),,$ by the Factor Theorem (over any domain where $6$ is invertible, e.g. working over $,\Bbb Q).\ \ $ – Bill Dubuque Jan 20 '20 at 23:08
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    You didn't finish the extended Euclidaen algorithm. The last step does $\ 6t^2!-6t\bmod -t!+!6.,$ Then you'll get $,a f + b g = 180,$ for $,a,b\in\Bbb Z[x],$ – Bill Dubuque Jan 20 '20 at 23:22
  • Finally, divide throughout by $180$ to get $1$ on the RHS. You'll get $,\dfrac{t+5}{30}, f - \dfrac{2t^2+9t-6}{180}\ g = 1\ \ \ $ – Bill Dubuque Jan 21 '20 at 00:51

1 Answers1

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When I talk about polynomials below, you can assume that the coefficients are complex numbers, but everything I'm going to say is going to work just as well if the coefficients are elements of a field.

First, we have the following fact: for any polynomials $f(t)$ and $g(t)$, if $g(t)\ne0$, then there exist unique polynomials, $q(t)$ and $r(t)$ such that $$f(t)=g(t)\cdot q(t)+r(t)$$

and either $r(t)=0$ or $\deg r(t)<\deg g(t)$. We find $q(t)$ and $r(t)$ by doing long division. If we divide $f(t)$ by $g(t)$, then $q(t)$ will be the quotient, and $r(t)$ will be the remainder.

The idea behind the Euclidean Algorithm is this: $GCD(f(t),g(t))=GCD(g(t),r(t))$.

You mentioned that you were given that

$$f(t)=2t^3-3t^2+6$$ $$g(t)=6t^2-6t$$

and that you divided and got that

$$2t^3-3t^2+6=(6t^2-6t)\cdot\left(\frac{1}{3}t-\frac{1}{6}\right)+(-t+6).$$

Then $q(t)=\frac{1}{3}t-\frac{1}{6}$, and $r(t)=-t+6$. So the Euclidean Algorithm tells us that

$$GCD(2t^3-3t^2+6,6t^2-6t)=GCD(6t^2-6t,-t+6).$$

The GCD on the right will be easier to find since we're dealing with smaller degree polynomials. If we divide again, we can show that this GCD will be equal to a GCD that is even easier to find.

Dividing $6t^2-6t$ by $-t+6$, we get that

$$6t^2-6t=(-t+6)\cdot(-6t-30)+180.$$

Now we have that

$$GCD(6t^2-6t,-t+6)=GCD(-t+6,180).$$

Hence we have that

$$\begin{align*} GCD(2t^3-3t^2+6,6t^2-6t)&=GCD(6t^2-6t,-t+6)\\ &=GCD(-t+6,180)\\ &=1\end{align*}$$

The fact that $GCD(-t+6,180)=1$ is because of the way GCD is defined. If $a(t)$ and $b(t)$ are polynomials that are not both zero, then $GCD(a(t),b(t)$ is the monic polynomial of largest degree that divides both $a(t)$ and $b(t)$. A polynomials is monic iff its leading coefficient is $1$. The only monic polynomial that divides $-t+6$ and $180$ is $1$. Hence $GCD(-t+6,180)=1$.

user729424
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  • @OP See here for more on unit normalization of gcds. – Bill Dubuque Jan 21 '20 at 02:38
  • @BillDubuque Thank you for your comment. I'm still a bit new to this website. When you write the "at" symbol followed by "OP", does this mean your comment is directed to the orginal poster of the question? I hope this isn't a silly a question. – user729424 Jan 21 '20 at 02:42
  • Yes, OP means "Original Post(er)". (SE software won't recognize that though, i.e. it won't show an inbox notification for the OP - for that you need to use the username, except in contexts when there are notifications by default, as in this comment). – Bill Dubuque Jan 21 '20 at 02:49
  • @BillDubuque Thanks for answering my question! – user729424 Jan 21 '20 at 02:54