Horner s Method

Zeros of Polynomials
A polynomial of degree n?1 has the form:
P(x)=a_n x^n+a_(n-1) x^(n-1)+?+a_1 x+a_0,
where the a_i s are constants and a_n?0.
Theorem 1: (Fundamental Theorem of Algebra)
If P(x) is a polynomial of degree n?1 with real or complex coefficients, then P(x)=0 has at least one (possibly complex) root. ?

Example 1: Determine all the zeros of the polynomial P(x)=x^3-5x^2+17x-13.
Solution:
P(1)=0
So x=1 is a zero of P. (x-1) is a factor of P. Dividing P(x) by (x-1) gives:
P(x)=(x-1)(x^2-4x+13)
The zeros of (x^2-4x+13) is obtained by using the quadratic formula:
(-b±?(b^2-4ac))/2a
(-(-4) ?((-4)^2-4(1)(13) ))/2(1) =(4±?(-36))/2=2±3i
Zeros of P(x) are x_1=1, x_2=2+3i and x_3=2-3i. ?
Corollary 1: If P(x) is a polynomial of degree n?1 with real or complex coefficients, then there exist unique constants x_1,x_2,…,x_k , possibly complex, and unique positive integers m_1,m_2,…,m_k , such that ?_(i=1)^k??m_i=n? and
P(x)=a_n (x-x_1 )^(m_1 ) (x-x_2 )^(m_2 )… (x-x_k )^(m_k ) ?