353 3.3 Zeros of Polynomial Functions EXAMPLE 5 Finding a Polynomial Function That Satisfies Given Conditions (Complex Zeros) Find a polynomial function ƒ1x2 of least degree having only real coefficients and zeros 3 and 2 + i. SOLUTION The complex number 2 - i must also be a zero, so the polynomial has at least three zeros: 3, 2 + i, and 2 - i. For the polynomial to be of least degree, these must be the only zeros. By the factor theorem there must be three factors: x - 3, x - 12 + i2, and x - 12 - i2. ƒ1x2 = 1x - 323x - 12 + i243x - 12 - i24 Factor theorem ƒ1x2 = 1x - 321x - 2 - i21x - 2 + i2 Distribute negative signs. ƒ1x2 = 1x - 321x2 - 4x + 52 Multiply and combine like terms; i2 = -1. ƒ1x2 = x3 - 7x2 + 17x - 15 Multiply again. The multiplication steps are not shown here. Any nonzero multiple of x3 - 7x2 + 17x - 15 also satisfies the given conditions on zeros. The information on zeros given in the problem is not sufficient to give a specific value for the leading coefficient. S Now Try Exercise 69. Zeros of a Polynomial Function The theorem on conjugate zeros helps predict the number of real zeros of polynomial functions with real coefficients. • A polynomial function with real coefficients of odd degree n, where n Ú 1, must have at least one real zero (because zeros of the form a + bi, where b≠0, occur in conjugate pairs). • A polynomial function with real coefficients of even degree n may have no real zeros. EXAMPLE 6 Finding All Zeros Given One Zero Find all zeros of ƒ1x2 = x4 - 7x3 + 18x2 - 22x + 12, given that 1 - i is a zero. SOLUTION Because the polynomial function has only real coefficients and 1 - i is a zero, by the conjugate zeros theorem 1 + i is also a zero. To find the remaining zeros, first use synthetic division to divide the original polynomial by x - 11 - i2. 1 - i )1 -7 18 -22 12 1 - i -7 + 5i 16 - 6i -12 1 -6 - i 11 + 5i -6 - 6i 0 By the factor theorem, because x = 1 - i is a zero of ƒ1x2, x - 11 - i2 is a factor, and ƒ1x2 can be written as follows. ƒ1x2 = 3x - 11 - i243x3 + 1-6 - i2x2 + 111 + 5i2x + 1-6 - 6i24 We know that x = 1 + i is also a zero of ƒ1x2. Continue to use synthetic division and divide the quotient polynomial above by x - 11 + i2. 1 + i )1 -6 - i 11 + 5i -6 - 6i 1 + i -5 - 5i 6 + 6i 1 -5 6 0 11 - i 21-6 - i 2 = -6 - i + 6i + i2 = -7 + 5i
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