351 3.3 Zeros of Polynomial Functions For example, a polynomial function of degree 3 has at most three distinct zeros but can have as few as one zero. Consider the following polynomial. ƒ1x2 = x3 + 3x2 + 3x + 1 ƒ1x2 = 1x + 123 Factored form of ƒ1x2 The function ƒ is of degree 3 but has one distinct zero, -1. Actually, the zero -1 occurs three times because there are three factors of x + 1. The number of times a zero occurs is referred to as the multiplicity of the zero. EXAMPLE 4 Finding a Polynomial Function That Satisfies Given Conditions (Real Zeros) Find a polynomial function ƒ1x2 of degree 3 with real coefficients that satisfies the given conditions. (a) Zeros of -1, 2, and 4; ƒ112 = 3 (b) -2 is a zero of multiplicity 3; ƒ1-12 = 4 SOLUTION (a) These three zeros give x - 1-12 = x + 1, x - 2, and x - 4 as factors of ƒ1x2. Because ƒ1x2 is to be of degree 3, these are the only possible factors by the number of zeros theorem. Therefore, ƒ1x2 has the form ƒ1x2 = a1x + 121x - 221x - 42, for some real number a. To find a, use the fact that ƒ11 = 3. ƒ112 = a11 + 1211 - 2211 - 42 Let x = 1. 3 = a1221-121-3 ƒ112 = 3 3 = 6a Multiply. a = 1 2 Divide by 6. Thus, ƒ1x2 = 1 2 1x + 121x - 221x - 42, Let a = 1 2. or, ƒ1x2 = 1 2 x3 - 5 2 x2 + x + 4. Multiply. (b) The polynomial function ƒ1x2 has the following form. ƒ1x2 = a1x + 221x + 221x + 22 Factor theorem ƒ1x2 = a1x + 223 1x + 22 is a factor three times. To find a, use the fact that ƒ1-12 = 4. ƒ1-12 = a1-1 + 223 Let x = -1. 4 = a1123 ƒ1-12 = 4 a = 4 Solve for a. Thus, ƒ1x2 = 41x + 223, or, ƒ1x2 = 4x3 + 24x2 + 48x + 32. Multiply. S Now Try Exercises 53 and 57. Remember: 1x + 223 ≠x3 + 23
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