Question 234892
<font face="Garamond" size="+2">


Both signs are +, so both of the signs in the factors must be plus.  The constant term is 1, so the only possible integer factors are 1 and 1.  So, so far we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (\ \ +\ 1)(\ \ +\ 1)]


The only way to make a *[tex \Large 5a^2] is with a *[tex \Large 5a] and an *[tex \Large a], so if this trinomial is factorable over the integers, the factors must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (5a\ +\ 1)(a\ +\ 1)]


Verification that this is the proper factorization is left as an exercise for the student.


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