Greatest Common Divisor of two Polynomials. Find the $\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$ over $\mathbb{Q}[x]$.  Then find two polynomials $a(x),b(x) \in \mathbb{Q}[x]$ such that, $$a(x)(x^3-6x^2+14x-15) + b(x)(x^3-8x^2+21x-18)=\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$$  
I have managed to find,
$$x^3-6x^2+14x-15=(x-3)(x^2-3x+5)$$
$$x^3-8x^2+21x-18=(x-3)(x-3)(x-2)$$
Now since $x^2-3x+5$ is irreducible over $\mathbb{Q}[x]$ and so the greatest common divisor is $(x-3)$.  Now to find $a(x)$ and $b(x)$ I have no clue how to do that.  I have looked online and it seems there is extended euclidean algorithm for polynomials but I haven't formally learned it in my class yet, so I was wondering if there is another efficient way to find these polynomials. Any help is appreciated, thanks!
 A: This is just the Extended Euclidean Algorithm. Instead of back-substitution, I have always preferred to write the construction steps in the style of continued fractions. Furthermore, I have always depended on the kindness of strangers.
$$  \left(   x^{3}  - 6 x^{2}  + 14 x  - 15 \right)  $$ 
$$  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  $$ 
$$  \left(   x^{3}  - 6 x^{2}  + 14 x  - 15 \right)  =  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  \cdot \color{magenta}{  \left( 1  \right) } +  \left(  2 x^{2}  - 7 x  + 3 \right)  $$ 
 $$  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  =  \left(  2 x^{2}  - 7 x  + 3 \right)  \cdot \color{magenta}{  \left(   \frac{ 2 x  - 9 }{ 4 }  \right) } +  \left(   \frac{ 15 x  - 45 }{ 4 }  \right)  $$ 
 $$  \left(  2 x^{2}  - 7 x  + 3 \right)  =  \left(   \frac{ 15 x  - 45 }{ 4 }  \right)  \cdot \color{magenta}{  \left(   \frac{ 8 x  - 4 }{ 15 }  \right) } +  \left( 0 \right)  $$ 
 $$ \frac{ 0}{1} $$ 
 $$ \frac{ 1}{0} $$ 
 $$ \color{magenta}{  \left( 1  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left( 1  \right) }{ \left( 1  \right) } $$ 
 $$ \color{magenta}{  \left(   \frac{ 2 x  - 9 }{ 4 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{ 2 x  - 5 }{ 4 }  \right) }{ \left(   \frac{ 2 x  - 9 }{ 4 }  \right) } $$ 
 $$ \color{magenta}{  \left(   \frac{ 8 x  - 4 }{ 15 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{ 4 x^{2}  - 12 x  + 20 }{ 15 }  \right) }{ \left(   \frac{ 4 x^{2}  - 20 x  + 24 }{ 15 }  \right) } $$ 
 $$  \left(   x^{2}  - 3 x  + 5 \right)  \left(   \frac{ 2 x  - 9 }{ 15 }  \right)  -  \left(   x^{2}  - 5 x  + 6 \right)  \left(   \frac{ 2 x  - 5 }{ 15 }  \right)  =  \left( -1  \right)  $$ 
 $$  \left(   x^{3}  - 6 x^{2}  + 14 x  - 15 \right)  =  \left(   x^{2}  - 3 x  + 5 \right)  \cdot \color{magenta}{  \left(   x  - 3 \right) } +  \left( 0 \right)  $$ 
 $$  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  =  \left(   x^{2}  - 5 x  + 6 \right)  \cdot \color{magenta}{  \left(   x  - 3 \right) } +  \left( 0 \right)  $$ 
 $$  \mbox{GCD} =   \color{magenta}{  \left(   x  - 3 \right) }   $$ 
 $$  \left(   x^{3}  - 6 x^{2}  + 14 x  - 15 \right)  \left(   \frac{ 2 x  - 9 }{ 15 }  \right)  -  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  \left(   \frac{ 2 x  - 5 }{ 15 }  \right)  =  \left(   -  x  + 3 \right)  $$ 
............
A: If I interpreted your question correctly and you want to do something like this, where in:
$a(x)(p)+b(x)(q)=gcd(p,q)$, $a=1$, and $b=1$
You can restate your expression as:
$$a(x)(x-3)(x^2-3x+5)+b(x)(x-3)(x-3)(x-2)$$
$$(x-3)\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)$$
Since this has to equal the $gcd$, which is $x-3$:
$$(x-3)\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)=x-3$$
$$\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)=1$$
Note: There are infinitely many solutions for $a(x)$ and $b(x)$ (but might not satisfy the definition of polynomials), since $0\le a(x)\le 1$ and $0\le b(x)\le 1$
Here, $a(x)(x^2-3x+5)$ and $b(x)(x-3)(x-2)$ sum up to $1$, so one instance is that they can individually sum up to $\dfrac 12$, since $\dfrac 12+\dfrac 12=1$.
So:
$$a(x)(x^2-3x+5)=\dfrac 12$$
$$b(x)(x-3)(x-2)=\dfrac 12$$
Can you figure this out?
A: You have the complete extended Euclid algorithm (which IMHO is the best way) given already in Will Jaggy's post.  
Given that you already extracted $x-3$ as GCD, here is a perhaps quicker way (which may not generalise much). We simplify a bit by noting we seek linear $a(x), b(x)$, s.t.:  
$a(x)\color{blue}{(x^2-3x+5)} + b(x) \color{blue}{(x-3)(x-2)}=1$.  Setting $x=2, 3$ we get $a(x) = \frac1{15}(-2x+9)$.  
Thus we are left with 
$\tfrac1{15}(-2x+9)\color{blue}{(x^2-3x+5)} + (px+q)\color{blue}{ (x-3)(x-2)}=1$
Comparing coeffs of the cubic term and the constant, $\implies p=\frac2{15},\; 3+6q = 1\implies q = -\frac13$.
So
$$\tfrac1{15}(-2x+9)\color{blue}{(x^2-3x+5)} + \tfrac1{15}(2x-5)\color{blue}{(x-3)(x-2)}=1$$
Multiply this by $\color{blue}{x-3}$ if you want to see the equation in form $a(x) p_1(x) + b(x) p_2(x) = \gcd(p_1, p_2)$.
