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With polynomials in general, is there more than one factorisation combination (i.e. is it similar to numbers, such as with '56' which can be both 28x2 and 7x8)?

For instance, with $x^{2}-2x+1$, are there any other factorisations other than $\left(x-1\right)^{2}$ ?

All help is appreciated.

qwerty
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    Well...$x^4=x^2\times x^2=x^3\times x = x\times x\times x\times x$ if you want to use non-irreducible factors. – lulu Sep 06 '22 at 11:31
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    With $56$, there is a unique way if you break it into irreducible factors as $2\times 2\times 2\times 7$. (Note you have a typo, it is not $23\times 2$ but rather $28\times 2$). One of the big punchlines is the fundamental theorem of arithmetic which proves that the integers form a unique factorization domain. It is a standard result that if $R$ is a unique factorization domain, then so too is $R[X]$... the ring of polynomials with coefficients in $R$. – JMoravitz Sep 06 '22 at 11:50
  • So, in other words both the scenario of factoring $56$ or another number and factoring $x^2-2x+1$ or another polynomial with integer coefficients there is a unique way to factor these into irreducible "prime" factors, unique up to order and sign. – JMoravitz Sep 06 '22 at 11:53

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The complete factorization of both numbers and polynomials is unique. The keyword here is "complete".

$56$ can be factorized into irreducible prime factors $2\times 2 \times 2 \times 7$ the same way as $p(x)=x^3+3x^2+3x+1$ can be factorized into the product of irreducible simple roots $(x+1)^3$. These factorizations are unique (up to ordering and signs of the factors).

Expressing $56$ both as $8\times 7$ or $2\times 28$ is the same as expressing $p(x)$ both as $(x+1)^3$ and $(x^2+2x+1)(x+1)$. It is just a non-unique way of multiplying prime factors with each other.

I hope this clears up any misconceptions.

Andreas Tsevas
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