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In school we spend several hours factorizing polynomials. But now as I've started gainnning some knowledge on polynomial rings, it suddenly occurred to me that none of the books I practiced then, suggested the factorized form of the polynomial $x^2-4$ as $$(cx+2c)\left(\frac{1}{c}x-\frac{2}{c}\right)$$ for arbitrarily chosen $c\ne0$ even though both of $(cx+2c),\left(\frac{1}{c}x-\frac{2}{c}\right)\in\mathbb R[x].$ They always took the form for $c=1$ as the answer. Why is it so? When I was said to factorize a polynomial in school what did I actually told to perform?

Added: Due to Wikipedia, "the aim of factoring is usually to reduce something to “basic building blocks”, such as numbers to prime numbers, or polynomials to irreducible polynomials." Well here $(cx+2c),\left(\frac{1}{c}x-\frac{2}{c}\right)\in\mathbb R[x]$ are irreducible. However for $6,-3$ is not a prime factor.

Sriti Mallick
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3 Answers3

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Factorisation is usually only considered up to units. As $c\neq 0$ is a unit in $\mathbb{R}[x]$, $(x+2)(x-2)$ is in this sense the same factorisation as $(cx+2c)\left(\frac{1}{c}x-\frac{2}{c}\right)$.

This is very much analogous to the difference (or lack thereof) between the factorisations $(x+2)(x-2)$ and $(x-2)(x+2)$.

Abel
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It's the same reason that when asked to factor the integer $6$, you never see people give $(-2) \cdot (-3)$ as the answer.

In many settings where factorization makes sense, there are a variety of reasonable choices for the factors, but there is a standard way to normalize the choice. When factoring integers, we usually choose the positive representative of each prime (i.e. $2$ rather than $-2$). When factoring polynomials over a field, we usually choose the factors to be monic. When factoring polynomials over the integers, we usually choose the leading coefficient of each factor to be positive.

e.g. the factorization of $2 - 8x^2$ over the rationals would usually be written as $(-8)(x-1/2)(x+1/2)$, but its factorization over the integers would usually be written as $(-1)2(2x-1)(2x+1)$. (actually, we would probably write $-2$ rather than $(-1) 2$, but I wanted to emphasize that $-2$ here is meant to be the product of a sign and the factor $2$, rather to be the number $-2$)

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You was supposed to find the form \begin{equation} p(x) = c \prod_{k=1}^n (x-x_k) \end{equation} where $x_k$ are the roots of $p$ and $c$ is a unique constant. This kind of representation is generally true to complex polynomials. It is called the fundamental theorem of algebra. In your case it was applicable. In the case of real polynomials the form isn't necessarily applicable. Then $p$ has complex conjugate root pairs, that can be seen by assuming that if $z$ is a root of $p$ then the conjugated equation tells that $\overline{z}$ is a root of $p$.

Juha-Matti Vihtanen
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