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How do you prove that a polynomial $P(x) = \sum_{i=0}^n a_i x^i$ where the first coefficient ($a_n$) is 1 can be written as

$$ P(x) = \prod_{i=1}^n (x-r_i) $$

where the $r_i$ are the roots of $P(x)$? I tried to apply the polynomial identity theorem (if P(x) and G(x) are polynomials of degree $< n$ that agree at $n$ points, then $P(x) = G(x)$) because these two representations agree at the $n$ roots, but I would still need one more point of agreement. Is there another way?

alexvas
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  • Are we assuming each of the $r_i$ are distinct? – muzzlator Apr 04 '13 at 20:30
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    Are we assuming an algebraically closed field? In the field $\mathbb Q$ of rationals, $x^3-x^2-2x+2$ has only one root $x=1$, yet $x^3-x^2-2x+2\ne x-1$. – Hagen von Eitzen Apr 04 '13 at 20:33
  • I am not assuming that the $r_i$ are distinct, so I allow $x^2 + 2x + 1 = (x+1)(x+1)$. Also, I allow my roots to be complex, so $x^2 + 1 = (x-i)(x+i)$ works. I believe the identity theorem (see question and accepted solution) still holds if the values at which the polynomials agree are complex. – alexvas Apr 04 '13 at 21:35

2 Answers2

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The difference of the two polynomials is of degree at most $n-1$ because the leading terms cancel. To show that it equals the zero polynomial, $n$ coincidences suffice.

Hagen von Eitzen
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Let us call $S=\prod(x-r_i)$, using the division algorithm for polynomials you can write $P=QS+R$ with $\deg R<\deg S$. It is easy to see that $R(r_i)=0$ for all $i$ (you need to have special care to deal with the case where there are multiple roots but a precise statement can be written down). Since $\deg R<n$ and $R$ coincides with the polynomial $0$ in $n$ points, we must have $R=0$ and then $S$ divides $P$ but both polynomials have the same degree so that they are scalar multiples. Now, you look at the first coefficient, which is $1$ in both polynomials, to conclude they are the same.

Quimey
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