TITLE: Why does factorizing of polynomials work
QUESTION [7 upvotes]: I'm very comfortable with GCSE Mathematics, and I wanted to expand and develop my understanding of mathematics. One such thing that I am doing is trying to understand why exactly many of the basic principles I have taken for granted work. Now this question may seem really obvious; of course $x^2 + 3x + 2 = (x+1)(x+2)$. 

Why can polynomials be factorized into forms like $(x-a)...$

REPLY [3 votes]: As I'm sure you're aware, you can represent a polynomial as either $p(x)=a_0+a_1x+a_2x^2+\ldots$ or as $p(x)=(x-r_1)(x-r_2)\ldots$ where each $r_k$ is a root of $p(x)$. In other words, as a sum of powers of $x$, or as a product of roots.
So, to my mind, there's a few interesting questions here, listed in order of difficulty.

Why are $r_k$ roots of $p(x)$?
Why can we write $p(x)$ as both a product and a sum, simultaneously?
Are the product and sum formulae equal?
Can we write every polynomial as a product of roots?


$r_k$ are roots of $p(x)$

Starting off with an easy one, each $r_k$ must be a root of $(x-r_1)(x-r_2)\ldots$ since we can substitute in $r_k$ and find that one of the bracketed terms will be $(r_k-r_k)=0$ so the whole expression must be $0$. Hence, each $r_k$ is a root of $p(x)$.

$p(x)$ can be expressed as both a product and as a sum

We demonstrate this algebraically by multiplying out all the factors. Notice that as we multiply everything out, we pick either $x$ or $r_k$ from each bracket. So one term will be a product of $x$, one term will be only $r_k$, and all other combinations in-between.
$$\begin{aligned}(x-r_1)(x-r_2)\ldots(x-r_n)
&=(\underbrace{x\times x\ldots}_n)+\ldots+(-1)^n(r_1r_2\ldots r_n)
\\&=x^n+b_{n-1}x^{n-1}+\ldots+b_1x+b_0\end{aligned}$$
Where, each $b_k$ is a constant, made of some combination of $r_k$. So now, we've got a representation of $p(x)$ as a sum of powers of $x$, as desired. In fact, the value of each coefficient, $b_k$, is given by Vietas's Formulas. You may ask why there's no coefficient, $b_k$ for $x^n$; it was just easier to read this way but you could multiply both sides by $b_k$.

The product and sum formulae are the same: $\sum a_kx^k=\prod(x-r_k)$
(Matching of polynomial coefficients)

We know that $p(x)=a_0+a_1x+\ldots$ but we've just shown that $p(x)=b_0+b_1x+\ldots$. For me, the next question I'd ask is whether $a_k$ and $b_k$ are the same coefficients. Sure, we've shown that the polynomial equals a sum of powers of $x$ but is it the only possible sum?
Well for our first representation, $p(0)=a_0$ and in our second representation, $p(0)=b_0$. So $p(0)=a_0=b_0$, so we can subtract $a_0$ or $b_0$ from each representation, then divide through by $x$ to see that $a_1+a_2x+\ldots=b_1+b_2x+\ldots$. So then, we can substitute $x=0$ again to see that $a_1=b_1$. We can repeat this method to see that each $a_k=b_k$, so the two representations are identical.
This means our product is the exact same thing as the sum $a_0+a_1x+\ldots$. So we could also have gone the other way around, and factorized $a_0+a_1x+\ldots$ into $(x-r_1)(x-r_2)\ldots$, which may answer your question. But this is all assuming those roots actually exist. How do we know they do?

Polynomials, of degree $n$, have exactly $n$ complex roots

This is the hardest part to prove and I imagine it's what you were most interested in. Why is it valid to say that $p(x)=(x-r_1)(x-r_2)\ldots$? Why should $p(x)$ have any roots, anyway? We assumed it did, but we didn't prove it.
As @flawr and @user357980 have pointed out, what we're now looking for is the Fundamental Theorem of Algebra. This theorem tells us that every $n$th degree polynomial, has $n$ roots.* 
What we have to watch out for is that the roots of the polynomial may be Complex Numbers. We could have a polynomial like $p(x)=x^2+1$, where the way to factorise it is as $p(x)=(x+i)(x-i)$. Nevertheless, though we may have to use complex numbers, this theorem gives us the knowledge that we can factorize polynomials.
Please let me know if you think I've made any errors.

* To be more precise, the polynomials need to be non=constant and with coefficients and roots in $\mathbb{C}$