Today's proof is taken from David Antin's translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.
Lemma 1: if an algebraic equation f(x) has a root α, then f(x) can be divided by x-α without a remainder and the degree of the result f'(x) is less than the degree of f(x).
Proof:
(1) Let f(x) = xn + a1xn-1 + ... + an-1x + an
(2) Let α be a root such that f(α) = 0
(3) Now, if we divide the polynomial by (x-α), we get the following (see here if proof needed):
f(x)/(x - α) = f1(x) + R/(x-α)
where R is a constant and f1(x) is a polynomial with order n-1.
(4) Multiplying both sides with x-α gives us:
f(x) = (x - α)f1(x) + R
(5) Now, if we substitute α for x we get:
f(α) = 0 which means that the constant in the equation is 0 so R = 0.
QED
Theorem: Fundamental Theorem of Algebra
For any polynomial equation of order n, there exist n roots ri such that:
xn + a1xn-1 + ... + an-1x + an = (x - r1)(x - r2)*...*(x - rn)
Proof:
(1) Let f(x) = xn + a1xn-1 + ... + an-1x + an
(2) We know that f(x) has at least one solution α1. [See here for proof]
(3) Using Lemma 1 above, we know that:
f(x)/(x - α1) = f'(x) where deg f'(x) = n-1.
So that we have:
f(x) = (x - α1)f'(x)
(4) But we know f'(x) has at least one solution (from here) so we can repeat steps #2 and #3 to get:
f'(x)/(x - α2) = f''(x) where deg f''(x) = n-2.
which combined with step #3 gives us:
f(x) = (x - α1)(x - α2)f''(x)
(5) Eventually we get to the point where the degree of fn(x) = 1.
In this case, fn(x) = x - αn.
(6) This establishes that there are n roots for a given equation f(x) where the degree is n.
(7) Putting this all together gives us:
f(x) = (x - α1)(x - α2)*...*(x - αn)
(8) Now, since f(x)=0 only when one of the values αi=x, we see that the n roots αi are the only solutions.
(9) So, we have proven that each equation is equal to n roots.
One important point to remember is that the n roots are not necessarily distinct. That is, it is possible that αi = αj where i ≠ j.
QED
References
- Heinrich Dorrie, 100 Great Problems of Elementary Mathematics
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