math111_logo Appendix: Polynomial

2. Representing polynomial by roots

The real polynomial 1 + t2 does not have any real root. In order for it to have roots, we must add a special number i = √-1 to the real numbers R. This gives rise to the complex numbers C. It turns out that this extension is sufficient to enable any polynomial (not just the special 1 + t2) of positive degree to have roots. This striking result is the fundamental theorem of algebra.

Now consider a complex polynomial p(t) of degree n. By the fundamental theorem, p has a root r1. This implies p(t) = (t - r1)p1(t) for another polynomial p1(t), which must have degree n - 1. Then by aplying the fundamental theorem to p1, we have p(t) = (t - r1)p1(t) = (t - r1)(t - r2)p2(t) for another polynomial p2(t) of degree n - 2. Continuing like this, we eventually prove the following.

Any complex polynomial of degree n can be decomposed as follows

p(t) = a(t - r1)(t - r2)...(t - rn), a ≠ 0.

Assume the highest order term in p(t) is tn. Then a = an = 1. Expanding the decomposition, we get

p(t) = (t - r1)(t - r2)...(t - rn) = tn - σ1tn-1 + σ2tn-2 + ... + (-1)n-1σn-1t + (-1)nσn,

where

σ1 = ∑i ri = r1 + r2 + r3 + ... + rn,
σ2 = ∑i<j rirj = r1r2 + r1r3 + r2r3 + ... + rn-1rn,
σ3 = ∑i<j<k rirjrk = r1r2r3 + r1r3r4 + r2r3r4 + ... + rn-2rn-1rn,
... ...
σn = ∏i ri = r1r2r3 ... rn,

are the symmetric polynomials of the roots r1, r2, ..., rn. This leads to the so called Viete Theorem.

Let r1, r2, ..., rn be all the roots of p(t) = a0 + a1t + a2t2 + ... + an-1tn-1 + tn. Then

ai = (-1)n-iσn(r1, r2, ..., rn).

In the special case p(t) = a0 + a1t + t2, p(t) has two roots r1, r2, and we have

a0 = r1r2, a1 = - (r1 + r2).


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