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The polynomial is $2x^4-7x^3+7x^2-14x+6$. I have found that it has following zeros: $3, \frac{1}{2},i\sqrt{2}, -i\sqrt{2}$ , so it can be written as:
$2(x-3)(x-\frac{1}{2})(x-i\sqrt{2})(x+i\sqrt{2})$.

The question is: if we need to write it as a product of prime factors in $\mathbb R$, will we only write: $2(x-3)(x-\frac{1}{2})(x^2+2)$, and if we need factorization in $\mathbb C$, then: $2(x-3)(x-\frac{1}{2})(x-i\sqrt{2})(x+i\sqrt{2})$?

user121
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Yes, that's right. A (full / "prime") factorisation of a polynomial over $\Bbb R$ will have a combination of linear and quadratic factors, where the quadratic factors have no real roots. And assuming the factors are all monic, there is only one such factorisation for any given polynomial. Over $\Bbb C$, any (full) factoring will only have linear factors, and again, assuming they are all monic, there is only one such factorisation for any given polynomial.

Arthur
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  • Thanks. And why do we have to assume that they are all monic? – user121 Oct 22 '19 at 06:23
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    @user121 Because then it's not unique any more. You get things like $$2(x-3)(2x-1)\left(\frac12x^2+1\right)$$just throwing constant factors around to make "new" factorisations which are really only rewritings of the factorisation you already have. Also, of course, as in the fundamental theorem of arithmetic, you can change the order of the factors, so in that respect you can also construct "new" factorisations. – Arthur Oct 22 '19 at 06:29