2.6 Zeros of Polynomials and Hornerβs Method 1 Zeros of Polynomials β’ Definition: Degree of a Polynomial A polynomial of degree π has the form π π₯ = ππ π₯ π + ππβ1 π₯ πβ1 + β― + π1 π₯ + π0 , ππ β 0 Where the ππβ² π are constants (either real or complex) called the coefficients of π π₯ β’ Fundamental Theorem of Algebra If π π₯ is a polynomial of degree π β₯ 1, with real or complex coefficients, π π₯ = 0 has at least one root. β’ Corollary There exists unique constants π₯1 , π₯2 , β¦ , π₯π and unique positive integers π1 , π2 , β¦ , ππ such that ππ=1 ππ = π and π π₯ = ππ (π₯ β π₯1 )π1 (π₯ β π₯2 )π2 β¦ (π₯ β π₯π )ππ 2 Remark: 1. Collection of zeros is unique 2. ππ are multiplicities of the individual zeros 3. A polynomial of degree π has exactly π zeros, counting multiplicity. β’ Corollary Let π(π₯) and π(π₯) be polynomials of degree at most π. If π₯1 , π₯2 , β¦ , π₯π with π > π are distinct numbers with π π₯π = π(π₯π ) for all π = 1,2, β¦ , π, then π π₯ = π(π₯) for all values of π₯. Remark: If two polynomials of degree π agree at at least (n+1) points, then they must be the same. 3 Hornerβs Method β’ Hornerβs method is a technique to evaluate polynomials quickly. Need π multiplications and π additions to evaluate π π₯0 . β’ Assume π π₯ = ππ π₯ π + ππβ1 π₯ πβ1 + β― + π1 π₯ + π0 . Evaluate π π₯0 . Let ππ = ππ , ππ = ππ + ππ+1 π₯0 , for π = π β 1 , π β 2 , β¦ , 1,0 Then π0 = π π₯0 . β’ At the same time, we also computed the decomposition: π π₯ = π₯ β π₯0 π π₯ + π0 , Where π π₯ = ππ π₯ πβ1 + ππβ1 π₯ πβ2 + β― + π2 π₯ + π1 4 Remark: 1. πβ² π₯0 = π(π₯0 ), which can be computed by using Hornerβs method in (n-1) multiplications and (n-1) additions. 2. Hornerβs method is nested arithmetic 5 β’ Example. Use Hornerβs method to evaluate π π₯ = 2π₯ 4 β 3π₯ 2 + 3π₯ β 4 at π₯0 = β2. Solution: Step 1: π4 = 2, π4 = 2 Step 2: π3 = 0, π3 = π3 + π4 π₯0 = 0 + 2 β2 = β4 Step 3: π2 = β3, π2 = π2 + π3 π₯0 = β3 + β4 β2 = 5 Step 4: π1 = 3, π1 = π1 + π2 π₯0 = 3 + 5 β2 = β7 Step 5: π0 = β4, π0 = π0 + π1 π₯0 = β4 + β7 β2 = 10 Thus π β2 = π0 = 10 Remark: π4 π₯ 4 + π3 π₯ 3 + π2 π₯ 2 + π1 π₯ + π0 = π4 π₯ + π3 π₯ + π2 π₯ + π1 π₯ + π0 6
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