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This is for a modern algebra course.

Find the greatest common divisor of each of the following pairs of $p(x)$ an $q(x)$ of polynomials. If $d(x)=gcd(p(x),q(x))$, find two polynomials $a(x)$ and $b(x)$ such that $a(x)p(x)+b(x)q(x)=d(x)$.

a. $p(x)=7x^3+6x^2-8x+4$ and $q(x)=x^3+x-2$ where $p(x),q(x)\in\mathbb{Q}[x]$

1st attempt:

I used an online factoring calculator to try to factor $p(x)=7x^3+6x^2-8x+4$ and it told me it's not factorable (I am not really sure how to factor polynomials with cubic terms easily by hand). and since $p(X)$ is not a multiple of $7q(x)=7x^3+7x-14$ I think maybe $gcd((p(x),q(x))=1$ in this case?

$$d(x)=a(x)p(x)+b(x)q(x)$$ $$1=a(x)(7x^3+6x^2-8x+4)+b(x)(x^3+x-2)$$ $$1=7x^3a(x)+6x^2a(x)-8xa(x)+4a(x)+x^3b(x)+x-2b(x)$$ $$1=x^3(7a(x)+b(x))+6x^2a(x)-8xa(x)+(4a(x)-2b(x))$$

Kind of got stuck here...

b. $p(x)=x^3+x^2-x+1$ and $q(x)=x^3+x-1$ where $p(x),q(x)\in\mathbb{Z}_2[x]$

1st attempt: I also can't seem to factor either of these...

I'm so confused..

Any help would be greatly appreciated!

Math Major
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1 Answers1

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In the general case, to obtain $a(x)$ and $b(x)$ ,you have to use the Extended Euclidean algorithm, exactly as in $\mathbf Z$.

In $\mathbf Q[x]$, this leads to polynomials having coefficients with a large number of digits. So there exists a normalised form of the algorithm, in which the successive remainders are monic polynomials,which makes divisions simpler.

The second example, in $\mathbf Z_2[x]$, is easier: $p(x)=x^3+x^2+x+1$, $q(x)=x^3+x+1$. The successive divisions are: \begin{align*} x^3+x^2+x+1&=1\times(x^3+x+1)+x^2\\ x^3+x+1&=x\cdot x^2+x+1\\ x^2&=(x+1)(x+1)+1, \end{align*} hence $\,\gcd(p(x),q(x)=1$, and the Extended Euclidean algorithm displays as follows: $$\begin{array}{lccc} r_k(x)& u_k(x)&v_k(x)&Q(x)\\ \hline p(x)& 1&0& \\ q(x)&0&1&1\\\hline x^2&1&1&x\\ x+1&x&x+1&x+1\\ 1&x^2+x+1&x^2\end{array}$$ Thus we obtain: $$1=(x^2+x+1)(x^3+x^2+x+1)+x^2(x^3+x) $$

Bernard
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  • Thanks for your help but I'm afraid your q(x) term is missing a -1 – Math Major May 01 '15 at 13:23
  • You mean a $-2$, I suppose? We're in $\mathbf Z_2$. – Bernard May 01 '15 at 14:10
  • Oh so maybe I am just confused. in the equation given in the book $q(x)=x^3+x-1$ but in your answer the -1 term is dropped from the original equation, so we can just do that in $Z_2$? – Math Major May 01 '15 at 14:20
  • I took $-2$ from your question, so there was ver little job to do. I'll change to the actual value of your book in a moment, it will be more instructive. Note that in $\mathbf Z_2$, $-1=1$. – Bernard May 01 '15 at 14:31
  • @Math Major: Sorry, I've just founf actually I read $-2$ from $q(x)$ in your first example! :o( – Bernard May 01 '15 at 14:35
  • no worries! I really appreciate your explanation! – Math Major May 01 '15 at 17:13