I'm a relatively new student to this topic, please give comprehensive explanations
This is what I have so far:
We can use Euclidean algorithm to solve the following and we have: $$\gcd(x^3-2x^2+2, x^2-x-2)=\gcd(x^3-2x^2+2-(x^2-x-2)(x-1),x^2-x-2)=\gcd(x, x^2-x-2)$$
And then:
$$\gcd(x, x^2-x-2-(x)(x-1))=\gcd(x,2)$$
Finally, I have,
$$2=g(x)-(x-1)(f(x)-(x-1)g(x))$$
So
$$1=g(x)(1+3x^2-x)+f(x)(3-3x)$$
But I'm not sure how to proceed, or might have made a mistake somewhere.
Please help and thank you
Well, why not. This is the Extended Euclidean Algorithm, written as a continued fraction rather than using back substitution. As usual, this requires fractions. The piece of luck is that there is no factor of $5$ denominator. This means that you can just replace any factor $\frac{1}{2}$ by the factor $3.$ at the end. Note also that you could follow this exact method but get rid of the fractions earlier, in the line $ \left( x \right) = \left( -2 \right) \cdot \color{magenta}{ \left( \frac{ - x }{ 2 } \right) } + \left( 0 \right) $
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ \left( x^{3} - 2 x^{2} + 2 \right) $$
$$ \left( x^{2} - x - 2 \right) $$
$$ \left( x^{3} - 2 x^{2} + 2 \right) = \left( x^{2} - x - 2 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( x \right) $$ $$ \left( x^{2} - x - 2 \right) = \left( x \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( -2 \right) $$ $$ \left( x \right) = \left( -2 \right) \cdot \color{magenta}{ \left( \frac{ - x }{ 2 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x - 1 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{2} - 2 x + 2 \right) }{ \left( x - 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - x }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x^{3} + 2 x^{2} - 2 }{ 2 } \right) }{ \left( \frac{ - x^{2} + x + 2 }{ 2 } \right) } $$ $$ \left( x^{3} - 2 x^{2} + 2 \right) \left( \frac{ x - 1 }{ 2 } \right) - \left( x^{2} - x - 2 \right) \left( \frac{ x^{2} - 2 x + 2 }{ 2 } \right) = \left( 1 \right) $$