Question 635770
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"Prime" in this context is a relative term.  Yes, this quadratic binomial is prime over the real numbers, but it is not prime over the complex numbers.


A second degree polynomial has two linear factors of the form *[tex \LARGE x\ -\ a] and *[tex \LARGE x\ -\ b] where *[tex \LARGE a] and *[tex \LARGE b] are the zeros of the polynomial.


Set your polynomial equal to zero and solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x^2\ +\ 25\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x^2\ =\ -25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ -\frac{25}{9}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\sqrt{-\frac{25}{9}}]


However, there is no real number that is the square root of a negative number, hence the solution set of the above equation <b><i>over the real numbers</i></b> is the null set.  Since there are no real number zeros of the equation, there are no real number factors of the polynomial.


As an aside, if you define *[tex \LARGE i] as the imaginary number such that *[tex \LARGE i^2\ =\ -1], then the equation has the solutions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\frac{5i}{3}]


and the factors of the polynomial, <b><i>over the complex numbers</i></b> are *[tex \LARGE x\ -\ \frac{5i}{3}] and *[tex \LARGE x\ +\ \frac{5i}{3}]



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