Question 674302
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Ok, Scruffy, go get your leash and collar and we'll take a walk.


Put your trinomial in standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ -\ 25t\ +\ 144]


We are looking for two factors of the form *[tex \LARGE t\ +\ p] and *[tex \LARGE t\ +\ q] such that *[tex \LARGE pq\ =\ 144], *[tex \LARGE \ \ p\ +\ q\ =\ -25], and *[tex \LARGE p,\,q\ \in\ \mathbb{Z}].  We get to use coefficients of 1 on the *[tex \LARGE t] terms because the lead coefficient in the given trinomial is 1.


In order for the sum to be negative and the product to be positive, both of the *[tex \LARGE p] and *[tex \LARGE q] values must be negative.


So what are the possible integer factors of 144?
<pre>

   -1    -144  :  -1 X -144 = 144, but -1 + -144 = -145 &#8800; -25
   -2     -72  :  -2 X  -72 = 144, but -2 + -72 = -74 &#8800; -25
   -3     -48  :
   -4     -36  :
   -6     -24  :
   -8     -18  :
   -9     -16  :
  -12     -12  :

</pre>


You fill in the rest of the table above and determine whether an appropriate pair of integers exists. If it does, that is your solution, if not, the polynomial is prime. 


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