Factoring: how to prove a polynomial can be written as product of x minus its zeros? For a lower degree polynomial function, we can simply expand the product of factors using distributive properties and compare that to the original polynomial. But how to prove that an arbitary degree single variable polynomail indeed equals to product of x minus all its zeros (complex numbers), in another word, validatity of factoring?
 A: First we prove (Equality 1), for all $x$
$$f(x)=(x-a)\cdot g(x)+b$$
With $a,b$ any number, $f(x)$ a polinomial function of degree $n$ and $g(x)$ polynomial function of degree $k\leq n$.
The proof will be done by complete induction on $n$.
(1) When $n=1$,
$$px+q=px-pa+q+pa=(x-a)p+(q+pa)$$ 
for $g(x)=p\land b=q+pa$
$$=(x-a)\cdot g(x)+b$$
It's true for $n=1$
(2) The equality is assumed to be true for $k\leq n$.
For $n=k+1$
$$f(x)=a_{k+1}x^{k+1}+\ldots +a_{1}x+a_0$$
So that 
$$h(x)=f(x)-a_{k+1}(x-a)$$
$h(x)$ has a degree $\leq k\leq n$ so, by our assumption,
$$h(x)=f(x)-a_{k+1}(x-a)=(x-a)g(x)+b$$
$$f(x)=(x-a)(g(x)+a_{k+1})+b$$
(remember that $f(x)$ has a degree of $k+1$ so we proved that if its true for $k$ it´s true for $k+1$.)
$\square$


Equality 1 gives us $f(x)=(x-a)\cdot g(x)+b$, but since $g(x)$ is also a polinomial,
$$f(x)=(x-a)\big((x-c)j(x)+r\big)+b=(x-a)(x-c)j(x)+(x-a)r+b$$
The same goes for $j(x)$ and so on.
$$f(x)=(x-a)(x-c)(x-d)\ldots +\text{a bunch of stuff (rigorous AF)}$$
When $x=$ one of the zeroes:
$$0=f(x)=0+\text{a bunch of stuff}\Rightarrow \text{a bunch of stuff}=0$$
So, when $x=$ one of the zeroes
$$f(x)=(x-a)(x-c)(x-d)\ldots$$ $\square$
