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So, that's my first time attempting a problem of this kind. I followed the steps of the Euclidean algorithm and got the following:

By long division for polynomials we have:

$x^6+x^4+x^3+x^2+x+1$ = $x(x^5+2x^3+x^2+x+1)+(-x^4+1)$

then we repeat the steps to get:

$x^5+2x^3+x^2+x+1$ = $-x(-x^4+1) + (2x^3+x^2+2x+1)$

I'm fine up to that point. But then when I moved on, I saw that my solution is different than the correct answer. Basically I divided $-x^4+1$ by $2x^3+x^2+2x+1$

but I got that: $-x^4+1$ = $-x/2(2x^3+x^2+2x+1) + x^3/2 + x/2 + 1$

whereas the solution says

$-x^4+1$ = $-(x/2 + 1/4) (2x^3+x^2+2x+1) + (3/4 x^2 + 3/4)$

How do they obtain that result? and where am I going wrong?

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    @Lana: You weren't finished. You're done when you get a zero remainder. – 2017-02-08
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    Guys, yea I know. I'm asking about a particular step though. – 2017-02-08
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    They did two steps at once (to save time, perhaps), Don't worry about it. Just check the final answer. No, the book is right, my mistake, You didn't finsh the division of _that_ step. – 2017-02-08
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    When you divided the quartic by the cubic, you weren't supposed to stop until the remainder has degree _less_ than 3. You stopped too soon. – 2017-02-08
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    difficult to read... there is no need to use fractions, one may always multiply through by a (constant) common denominator, or divide out by a common numerator. It would appear that the gcd, as a monic, is $x^2 + 1$ – 2017-02-08
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    @Will Jagy: Sure, but unless the book and/or the class does that, it's best to stay with the standard steps of the algorithm -- at least at first. – 2017-02-08

2 Answers 2

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As I was saying, we are allowed to multiply or divide by constants at any convenient point, therefore getting rid of fractions. $$ (x^2 + 1)(x^4 +x+1) = x^6 + x^4 + x^3 + x^2 + x + 1 $$ $$ (x^2 + 1)(x^3 +x+1) = x^5 + 2 x^3 + x^2 + x + 1 $$

If, as we think, the two extra factors are coprime, then the original gcd is $x^2 + 1.$ This is a worthwhile confirmation:

$$ (-2x^2+ x - 4)(x^4 + x + 1) + (2 x^3 - x^2 + 2x +1)(x^3 + x + 1) = -3 $$ Notice that the outcome is not $1,$ it is $-3.$ However, this is a nonzero constant in the field. Therefore this is a proof that $x^4 + x + 1, x^3 + x + 1$ are coprime.

Also $$ \scriptsize (-2x^2+ x - 4)(x^6 + x^4 + x^3 + x^2 + x + 1) + (2 x^3 - x^2 + 2x +1)(x^5 + 2 x^3 + x^2 + x + 1) = -3x^2 - 3 = -3(x^2 + 1)$$

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    yup, I think i was confused because they jumped ahead. When I continued with it I got the same result. – 2017-02-08
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Try rearranging some of the terms with the commutative law of addition and searching for common binomial factors.

Let $P(x) = x^6 + x^4 + x^3 + x^2 + x + 1$. The 6 terms here means that we can split this expression into 3 groups of binomial factors.

$\therefore P(x) = x^6 +x^3 + x^4 + x + x^2 + 1$

$ = x^3(x^3 + 1) + x(x^3 + 1) + (x^2 + 1)$

$= x(x^2+1)(x^3+1) + 1(x^2 + 1)$

$=(x^2 + 1)(x^4 + x + 1)$

Now let $G(x) = x^5 + 2x^3 + x^2 + x + 1$. Split $2x^3$ into $x^3 + x^3$ to form only monic coefficients and turn 5 terms into 6.

$\therefore G(x)= x^5 + x^3 + x^3 + x + x^2 + 1$

$= x^3(x^2 + 1) + x(x^2 + 1) + 1(x^2 + 1)$

$=(x^3 + x + 1)(x^2 + 1)$

The highest common factor should be clear here, but just to clarify we must see if $x^4 + x + 1$ is divisible by $x^3 + x + 1$ i.e. if either of them are prime.

$x^4 + x + 1 = x(x^3 + 1) + 1 = x(x+1)(x^2 - x + 1) + 1$ which is clearly prime.

Hence the greatest common divisor (highest common factor) of $P(x)$ and $G(x)$ is $x^2 + 1$.