What is $\gcd (x^{3}+6x^{2}+11x+6,x^{3}+1)$

Question

When applying straight-forward Euclid's algorithm the result have fractional coefficients, but by factoring linear terms you get $x+1$. Which answer is right?

Answer 1

Here is what I get. Note the check at the very end, in order to express the gcd as a linear combination of the initial terms, the coefficients do get denominators, your 91. Let me paste that here:

$$ \left( x^{3} + 6 x^{2} + 11 x + 6 \right) \left( \frac{ 6 x - 11 }{ 91 } \right) - \left( x^{3} + 1 \right) \left( \frac{ 6 x + 25 }{ 91 } \right) = \left( - x - 1 \right) $$

OR

$$ - \left( x^{3} + 6 x^{2} + 11 x + 6 \right) \left( \frac{ 6 x - 11 }{ 91 } \right) + \left( x^{3} + 1 \right) \left( \frac{ 6 x + 25 }{ 91 } \right) = \; \; x+1 $$

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$$ \left( x^{3} + 6 x^{2} + 11 x + 6 \right) $$

$$ \left( x^{3} + 1 \right) $$

$$ \left( x^{3} + 6 x^{2} + 11 x + 6 \right) = \left( x^{3} + 1 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 6 x^{2} + 11 x + 5 \right) $$ $$ \left( x^{3} + 1 \right) = \left( 6 x^{2} + 11 x + 5 \right) \cdot \color{magenta}{ \left( \frac{ 6 x - 11 }{ 36 } \right) } + \left( \frac{ 91 x + 91 }{ 36 } \right) $$ $$ \left( 6 x^{2} + 11 x + 5 \right) = \left( \frac{ 91 x + 91 }{ 36 } \right) \cdot \color{magenta}{ \left( \frac{ 216 x + 180 }{ 91 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 6 x - 11 }{ 36 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 6 x + 25 }{ 36 } \right) }{ \left( \frac{ 6 x - 11 }{ 36 } \right) } $$ $$ \color{magenta}{ \left( \frac{ 216 x + 180 }{ 91 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 36 x^{2} + 180 x + 216 }{ 91 } \right) }{ \left( \frac{ 36 x^{2} - 36 x + 36 }{ 91 } \right) } $$ $$ \left( x^{2} + 5 x + 6 \right) \left( \frac{ 6 x - 11 }{ 91 } \right) - \left( x^{2} - x + 1 \right) \left( \frac{ 6 x + 25 }{ 91 } \right) = \left( -1 \right) $$ $$ \left( x^{3} + 6 x^{2} + 11 x + 6 \right) = \left( x^{2} + 5 x + 6 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 0 \right) $$ $$ \left( x^{3} + 1 \right) = \left( x^{2} - x + 1 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x + 1 \right) } $$ $$ \left( x^{3} + 6 x^{2} + 11 x + 6 \right) \left( \frac{ 6 x - 11 }{ 91 } \right) - \left( x^{3} + 1 \right) \left( \frac{ 6 x + 25 }{ 91 } \right) = \left( - x - 1 \right) $$

Answer 2

$(x^3+6x^2+11x+6,x^3+1)=((x+1)(x^2-x+1),(x+1)(x+2)(x+3)) = (x+1)((x^2-x+1),(x+2)(x+3)) = (x+1)$

Answer 3

You can calculate the gcd in $\mathbb{Q}[x]$ following Jyrki Lahtonen's comments. So, just to see what happens, lets work in $\mathbb{Z}[x]$, also a unique factorization domain. Note that If $(a,c)=1$ then $(a,b)=(a,cb)$

$$(x^3+6x^2+11x+6, x^3+1)=(6x^2+11x+5, x^3+1)$$ $$=(6x^2+11x+5, 6x^3+6), \text{since} (11x+5,6)=1$$ $$=(6x^2+11x+5, 11x^2+5x-6)$$ $$=(66x^2+121x+55, 66x^2+30x-36)$$ $$=(91x+91, 66x^2+30x-36)$$ $$=(x+1, 11x^2+5x-6)$$ $$=(x+1, 6x+6)$$ $$=x+1$$

Answer 4

Both methods may agree (depending on the "fractional coefficients" you got). When it comes to factoring any constant times a polynomial $P(x)$ is the same as $P(x)$ because you have the same roots. For instance, there is no substantial difference between $x+1$ and $(36/91)(x+1)=(36/91)x+(36/91)$ because they both have a single, nondegenerate root at $x=-1$.