GCD of polynomials by using Euclid's algorithm

Question

Let $g = x^2 +6x -7$ and $f = x^4 - 1$. Find the GCD of $f$ and $g$.

So I started by evaluating $f/g$ and the result is $q = x^2-6x+43, r = -300x+300$. I tried to follow the algorithm one step further but I can't see how the result ends up to be $x-1$.

Answer 1

As you say, euclidean division yields \begin{align*} (x^4-1) &= (x^2 - 6x + 43)(x^2 + 6x - 7) + (-300x+300)\\ &= q(x)\cdot(x^2 + 6x - 7) + r(x) \end{align*} Then $\gcd(x^4-1,\,x^2+6x-7)=\gcd(x^2+6x-7, -300x+300)$. You can apply again euclidean division to conclude

$$\gcd(x^4-1,\,x^2+6x-7) = -300x+300$$

But you already knew this. What you may have missed is that the greatest common divisor is unique up to units. In particular $$-300x+300 = -300\cdot(x-1)$$ Now, since 300 is a unit, we can say:

$$\gcd(x^4-1,\,x^2+6x-7) = x-1$$

The point is that the constant term is not important. You can divide by $-300$ and you still have a greatest common divisor.