Using the Euclidean algorithm, deduce that $\gcd(x^3+2x^2+x +4;x^2+1)=1$

Question

So, I've tried it but I keep getting to $$\frac{x^2+1} 2$$ and don't know how to proceed.

Question is Using the Euclidean algorithm, deduce that $$\gcd(x^3+2 x^2+x +4,\;x^2+1)=1.$$

Answer 1

\begin{array}{c|cccc} & x^3+2x^2+x+4 & 1 & 0 \\ -x-2 & x^2+1 & 0 & 1 \\ \hline -\frac 12x^2 & 2 & 1 & -x-2 \\ & 1 & -\frac 12x^2 &\frac 12x^3+x^2+1 \end{array}

$$-\frac 12x^2\color{red}{(x^3+2x^2+x+4)}+ \left(\frac 12x^3+x^2+1\right)\color{red}{(x^2+1)} = 1$$

Answer 2

$$ \left( x^{3} + 2 x^{2} + x + 4 \right) $$

$$ \left( x^{2} + 1 \right) $$

$$ \left( x^{3} + 2 x^{2} + x + 4 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x + 2 \right) } + \left( 2 \right) $$ $$ \left( x^{2} + 1 \right) = \left( 2 \right) \cdot \color{magenta}{ \left( \frac{ x^{2} + 1 }{ 2 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x + 2 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x + 2 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ x^{2} + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{3} + 2 x^{2} + x + 4 }{ 2 } \right) }{ \left( \frac{ x^{2} + 1 }{ 2 } \right) } $$ $$ \left( x^{3} + 2 x^{2} + x + 4 \right) \left( \frac{ 1}{2 } \right) - \left( x^{2} + 1 \right) \left( \frac{ x + 2 }{ 2 } \right) = \left( 1 \right) $$

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