Question 1181514
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Both complex and irrational zeros come in conjugate pairs.  This means that if *[tex \Large a\ +\ bi] is a zero, then so is *[tex \Large a\ -\ bi].  And if *[tex \Large a\ +\ b] where *[tex \Large a\ \in\ \mathbb{Q}] and *[tex \Large b\ \not\in\ \mathbb{Q}] then *[tex \Large a\ -\ b] is also a zero.


Since the degree of a polynomial function is equal to the number of zeros, and since you are given one complex zero and one irrational zero there must, at a minimum, be four zeros, you are looking for a fourth-degree polynomial.


We also know that if *[tex \Large p] is a zero of a polynomial function, then *[tex \Large x\ -\ p] must be a factor of the polynomial.


You were given *[tex \Large 5\ -\ i] as a zero, so *[tex \Large x\ -\ (5\ -\ i)] is a factor of the desired polynomial.  Likewise, *[tex \Large x\ -\ (5\ +\ i)] is a factor.  Also, you were given *[tex \Large \sqrt{11}] as a zero so *[tex \Large x\ -\ \sqrt{11}] is a factor of the desired polynomial and likewise *[tex \Large x\ -\ (-\sqrt{11})] is a factor.


Putting it all together, your polynomial function, in factored form, is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ \(x\,-\,5\,+\,i\)\(x\,-\,5\,-\,i\)\(x\,-\,\sqrt{11}\)(x\,+\,\sqrt{11}\)]


The only thing left for you to do is to multiply the four binomials and collect like terms.  Hints: The product of two conjugates is the difference of two squares and *[tex \Large i^2\ =\ -1].

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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