Always factorise polynomials It's always fun (however irresponsible) to make random conjectures, if you don't have to bear the burden of proving them. And thus, I conjecture -
Every polynomial expression (of the form-)
$$a_n\cdot x^n + a_{n-1}\cdot x^{n-1} + ... +a_{2}\cdot x^{2} + a_{1}\cdot x + a_0$$
[where $a_i \ \epsilon \ \Bbb{R}$]
is factorisable (with a remainder)/configurable, into the following form -
$$a_n\cdot(x+ k_{n}) \cdot (x + k_{n-1}) \cdot (...) \cdot (x + k_{2})\cdot (x + k_{1}) + r$$
[where all $k_i,r \ \epsilon \ \Bbb{R}$]
Is this true? Are there mathematical proofs to it? Have there been investigations on this ever before? Or is it just an obvious corollary that follows from some theorem/property?
If true, then is the configuration, viz. the set $\{a_n, k_i, r | i = 1(1)n\}$, unique for each unique polynomial $p(x)$? (Perhaps only if not unique, then -) Does there always exist a configuration where $r = 0$?
If not, then lastly, if $\{a_n, k_i, r | i = 1(1)n\} \ \epsilon \ \Bbb{C}$, then does there exist such a configuration that $r = 0$? In other words, is it possible to vanish out the remainder upon extending the set of the configurational constants?
 A: Indeed it is fun to conjecture things! Mathematics is good fun! :)
We can actually construct a nice disproof to the conjecture. Consider ${x^4 + x^2}$. Now do the remainder trick
$${x^4 + x^2 = (x^4 + x^2 - r) + r}$$
Now it remains to factor ${x^4 + x^2 - r}$, and we want to know if it's ever factor-able over ${\mathbb{R}}$ as linear factors for some ${r \in \mathbb{R}}$. Well, ${r=0}$ doesn't work (since ${x^4 + x^2 = 0}$ has solutions that are not within ${\mathbb{R}}$). Otherwise, let ${u=x^2}$. Then
$${\Leftrightarrow u^2 + u - r =0}$$
This will have two, non-zero roots (clearly ${u=0}$ is not a root):
$${u = \frac{-1\pm \sqrt{1 + 4r}}{2}}$$
We now have two possibilities - either:
(1) Both roots are real
(2) Both roots are not real (they contain non-zero imaginary part)
In case (1), clearly one of the roots will be negative. And this means ${x = \pm\sqrt{u}}$ will be imaginary, hence ${x^4 + x^2 - r}$ is not factor-able over ${\mathbb{R}}$ in this case. In case (2), clearly ${x=\pm \sqrt{u}}$ will also have to have non-zero imaginary part (since a real number multiplied by a real number will always give back a real number - that is, a number with imaginary part of $0$). Hence in any case, we have shown that
$${x^4 + x^2 - r, r \in \mathbb{R}}$$
must always have at least one root that does not belong to ${\mathbb{R}}$, hence is not factor-able over ${\mathbb{R}}$. It doesn't matter what "remainder" you choose - the polynomial you are left with to factor in this case will always have some non-real root. As required
