Question 199809
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If the polynomial is of degree 3, then there are guaranteed to be three zeros, at least one of which is a real number zero.  You are given one real number zero and one complex number zero.  But the thing we also know is that complex zeros <i>always</i> come in pairs, and those pairs are always conjugates.  That means that if *[tex \Large a + bi] is a zero, then *[tex \Large a - bi], the conjugate, is also a zero.


You are given:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 3] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 1 + 4i], therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 1 - 4i] must also be a zero.


Knowing the zeros of a polynomial allows us to determine the 1st degree factors of the polynomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 3 \ \ \Rightarrow\ \ x - 3 = 0], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 3] is a factor.


Likewise:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - (1 + 4i)] and *[tex \LARGE x - (1 - 4i)]


are factors, so the polynomial can be determined from the product of the three factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 3)(x - (1 + 4i))(x - (1 - 4i))]


When you multiply these, do the complex factors first.  Hints:


1. Treat *[tex \Large 1 + 4i] and *[tex \Large 1 - 4i] as single numbers.


2. *[tex \Large 1 + 4i] times *[tex \Large 1 - 4i] results in the difference of two squares


3. Remember *[tex \Large i^2 = -1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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