Question 645882
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I've never heard of a polynomial of any degree that had a zero of a lower case "o", so I suppose I shall presume that you meant 0 (which is to say zero).  You are given three of the 4 zeros that you know must exist because the polynomial is of degree 4.  Fortunately, one of your zeros is a complex number.  Complex zeros of polynomials with real coefficients always appear in conjugate pairs.  If you have a complex number of the form *[tex \LARGE a\ +\ bi] then its conjugate is *[tex \LARGE a\ -\ bi].  Furthermore, if *[tex \LARGE \alpha] is a zero of a polynomial, then *[tex \LARGE x\ -\ \alpha] is a factor of the polynomial.  Now that you know the four zeros, you can construct the four factors.  Simply multiply the four factors together, collect like terms, and you will have your desired polynomial.


Hint for multiplying the complex factors: The product of two conjugates is the difference of two squares, but remember that *[tex \LARGE i^2\ =\ -1], so the product of two <b><i>complex</i></b> conjugates is the SUM of two squares.


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