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Always factorise polynomials

Mathematics Asked by BeBlunt on November 29, 2021

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-)

$$a_ncdot x^n + a_{n-1}cdot x^{n-1} + … +a_{2}cdot x^{2} + a_{1}cdot x + a_0$$

[where $a_i epsilon Bbb{R}$]

is factorisable (with a remainder)/configurable, into the following form –

$$a_ncdot(x+ k_{n}) cdot (x + k_{n-1}) cdot (…) cdot (x + k_{2})cdot (x + k_{1}) + r$$

[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?

One Answer

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):

$${u = frac{-1pm sqrt{1 + 4r}}{2}}$$

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 ${x = pmsqrt{u}}$ 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

Answered by Riemann'sPointyNose on November 29, 2021

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