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$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$

I'm trying to understand why do all polynomials with real roots are factorizable. The explanation relies on the fact that all polynomials which are divided by a first degree polynomial (x-b) where b is a root will have

$(x-b)(a_{n-1}x^{n-1}+\cdots+a_0)+R$ R is a real number

I dont understand why this is so. Why is there no possibility of remainders which are not real numbers?

On a separate note, does this mean that polynomials with no real roots are unfactorizable?

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    No to the second question, e.g., $(x^2+1)^2$ is factorizable.2012-10-03
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    @GerryMyerson, but not into linear binomials2012-10-03
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    When you do long division of the polynomial $f$ (which I assume has all real coefficients, though you didn't explicitly say this) by $x-b$ (which also has all real coefficients), all the computations involve real numbers only. So there's no possibility of getting a non-real remainder.2012-10-03
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    Hmm, but that doesn't explain why you don't get a real remainder that can't be reduced to a polynomial.2012-10-03

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