Question 303728
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If a polynomial function has a zero *[tex \Large \alpha], then the polynomial has a factor *[tex \Large x\ -\ \alpha].  Also, irrational zeros of polynomial functions with rational coefficients ALWAYS come in conjugate pairs.  That means that if *[tex \Large \sqrt{3}] is a root, then *[tex \Large -\sqrt{3}] is also a root.


So now you know your four zeroes:  2, 0, *[tex \Large \sqrt{3}] and *[tex \Large -\sqrt{3}], so you can write the four factors of the polynomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ x\left(x\ -\ 2\right)\left(x\ -\ \sqrt{3}\right)\left(x\ +\ \sqrt{3}\right)]


The last two give you the difference of two squares, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ x\left(x\ -\ 2\right)\left(x^2\ -\ 3\right)]


A little FOIL:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ x\left(x^3\ -\ 2x^2\ -\ 3x\ +\ 6\right)]


And then distribute the remaining factor


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ x^4\ -\ 2x^3\ -\ 3x^2\ +\ 6x]





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