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**Unformatted text preview: **= an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 . If z is a zero of f , then f (z ) = 0, which means
an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 = 0. Next, we consider f (z ) and apply Theorem 3.12 below.
8 You really should do this once in your life to convince yourself that all of the theory actually does work! 224 Polynomial Functions f (z ) = an (z )n + an−1 (z )n−1 + . . . + a2 (z )2 + a1 z + a0
= an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 since (z )n = z n = an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 since the coeﬃcients are real = an zn + an−1 z n−1 + . . . + a2 z + a1 z + a0 since z w = zw = an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 since z + w = z + w 2 = f (z )
=0
=0
This shows that z is a zero of f . So, if f is a polynomial function with real number coeﬃcients,
Theorem 3.15 tells us if a + bi is a nonreal zero of f , then so is a − bi. In other words, nonreal
zeros of f come in conjugate pairs. The Factor Theorem kicks in to give us both (x − [a + bi]) and
(x − [a − bi]) as factors of f (x) which means (x − [a + bi])(x − [a − bi]) = x2 + 2ax + a2 +...

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