Factoring (C)

In most cases, however, we do not have even a disguised quadratic equation, and our only recourse is to factor the polynomial expression. We have the following useful result:

Theorem 2685

If p(z) is a polynomial then p(a) = 0 if and only if there is a polynomial q(z) so that p(z) = (z-a)q(z).

This is easy to prove. If we can write p(z) = (z-a)q(z) then p(a) = (a-a)q(a) = 0. On the other hand, suppose that

$$p(z) = A_Nz^N + A_{N-1}Z^{N-1} + \cdots + A_1z + A_0$$

and p(a) = 0. Then

$$p(z) = p(z)-p(a) = A_N(z^N-a^N) + A_{N-1}(z^{N-1}-a^{N-1}) + \cdots + A_1(z-a)$$

and for every positive integer M we have

$$z^M-a^M = (z-a)(z^{M-1} + z^{M-2}a + z^{n-3}a^2 + \cdots + za^{M-2} + a^{M-1})$$

so we can see that z-a divides p(z). (Another use of geometric progressions!) QED

Once one has found one zero,say z=a, one finds (by long division), the polynomial q so that p(z) = (z-a)q(z) and then looks for the zeros of q, since every zero of q is a zero of p. Now our problem is to find an intelligent way to find some of the zeros.