20.4. Factorization of Polynomials

So, once we have seen polynomial division over the fields and GCD of polynomials we can define what we call as factorization. So, a trivial factorization that is possible for any polynomial is the following form. So, you are given the polynomial f(x), a trivial factorization will be the following: you take any constant from the field and you take that constant $\alpha$ multiplied with the multiplicative inverse of $\alpha$ of course $\alpha$ is not 0 here. Otherwise, the inverse is not well defined; it does not exist. So, if I take any non 0 $\alpha$ from the field and multiply $\alpha$ and $\alpha^{-1}$ with that polynomial $f(x)$ I will get back the original polynomial $f(x)$ itself in that sense I can always say that there is a trivial factorization of $f(x)$ namely $\alpha$ and $\alpha^{-1}$ are trivial factors for any $f(x)$.

In this chunk, we learn about the concept of factorization in the context of polynomials. A trivial factorization refers to the situation where any polynomial can be expressed as a product involving a constant and its multiplicative inverse. For example, if we have a polynomial $f(x)$, we can take a non-zero constant $\alpha$ and multiply it by its inverse $\alpha^{-1}$. This will yield back the polynomial $f(x)$. Therefore, if we think of it this way, every polynomial has trivial factors represented by any constant from the field along with its inverse.

Imagine a recipe for a cake that requires certain ingredients. If you had a special ingredient that you could multiply (say, flour) with another ingredient that cancels it out (like water, which you only require in a specific amount), you can see how they can balance each other, much like $\alpha$ and $\alpha^{-1}$ help regain the original recipe (polynomial).

Now, what we want to define is what we call irreducible polynomial. Namely polynomials which cannot be factored into products of lower degree polynomials. That is a rough idea of what we call as irreducible polynomial. So, let us now formally define, so intuitively it is a non-constant polynomial which cannot be factored into product of two non-constant polynomials. And why we are taking the case that it cannot be factored into product of two non-constant polynomials. Because of this trivial factorization, because if you give me any polynomial $f(x)$ I can always factorize it.

Irreducible polynomials are defined as those that cannot be factored into lower degree polynomials, except for trivial factorizations that involve constants. This means if you have a polynomial $f(x)$ that is non-constant, it cannot be expressed as the product of two smaller polynomials, both of which are also non-constant. This property is crucial because it helps to identify 'building block' polynomials that cannot be simplified further, which are foundational in polynomial algebra.

Let's think of an irreducible polynomial as a prime number in the world of integers, like the number 7. Just as prime numbers cannot be divided evenly by any number other than 1 and themselves, irreducible polynomials cannot be broken down into simpler polynomial factors. For example, if $x^2 + 1$ cannot be factored into simpler polynomial forms, it’s like saying it’s a prime number the way 7 is in the number line.

So, the next thing that we want to define is the factor theorem for polynomials over fields. So, the factor theorem states the following. If you take any polynomial over the field then the polynomial $(x – \alpha)$ will be considered as a factor of your polynomial $f(x)$ if and only if the polynomial $f(x)$ when evaluated at $x = \alpha$ gives you the element 0, where 0 is the additive identity.

The factor theorem provides a convenient way to check if a polynomial has a specific linear factor. It states that a polynomial $(x – \alpha)$ is a factor of polynomial $f(x)$ if substituting $\alpha$ into $f(x)$ results in zero. This theorem directly connects the factors of polynomials with their roots, making it easier to factor polynomials by finding their roots.

Think of the factor theorem like having a key that fits into a lock. If you find the right key (or value, $\alpha$), it opens up the lock (the polynomial evaluates to zero). Just as not every key fits every lock, not every value will satisfy this condition, but when you find one that does, you know that the corresponding polynomial factor exists.

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