A specific question about a factor theorem proof I know there are tons of answers to questions about the factor theorem, however, I couldn't find what I was looking for so apologies if someone has already answered this before.
Since
$\frac{x^k - y^k}{x-y} = x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1}$
${x^k - y^k} = (x-y)(x^{k-1}y^0 + x^{k-2}y^1 + x^{k-3}y^2 + ... + x^1y^{k-2} + y^{k-1})$
hence ${x^k - y^k} = (x-y)q_k(x)$ 
As usual, we write our polynomial $ p$ as 
$p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 $
$p(x)-p(y) = $
$a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 $
$ -(a_ny^n + a_{n-1}y^{n-1} + ... + a_1y + a_0)$
$= a_n(x^n-y^n) + a_{n-1}(x^{n-1}-y{n-1}^) + ... + a_1(x-y)$
The terms have the form $a_k(x^k-y^k)$. But $x^k-y^k = (x-y)q_k(x)$, and substitute this in:
$p(x)-p(y)=$
$a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)$
$(x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1)$
$= (x-y)q(x)$ where $q(x) = a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1$
If $p(a) = 0$ and $y = a$ then
$p(x)-p(a) = (x-a)q(x) $ Q.E.D
My question is: why we don't write $p(x) - p(y)$ as
$a_n(x-y)q_n(x) + a_{n-1}(x-y)q_{n-1}(x) + ... + a_1(x-y)q_1(x)$
$(x-y)(a_nq_n(x) + a_{n-1}q_{n-1}(x) + ... a_1q_1(x))$.
NOTE! That I'm asking whether or not $a_1q_1(x) = a_1$
Can anyone please explain this to me thanks! 

It's exactly what's happening. 
Observe that $q_1$ is just the constant $1$ polynomial. 
