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!

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?
