Question 635770
<font face="Times New Roman" size="+2">

"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
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>