Question 1004232
.
from https://answers.yahoo.com/question/index?qid=20110301123054AAIHayZ (5 years ago):

<pre>
Use the rational root theorem to find the possible rational roots. 
The rational roots theorem says that possible rational roots are +/- factors the constant term (36 here) 
divided by factors of the leading coefficient (1 here). Possible rational roots are 

+/- 1, 2, 3, 4, 9, 12, 18, 36 

Test each zero using the rational root test. To do this, use synthetic division to test the roots. 
I won't show the work here, but the roots that work are -2 and -3. As factors, this is x+2 and x+3. 

From the synthetic division, we have x^2-4x+6 left over, which is irreducible. 

In factored form: 
f(x) = (x+2)(x+3)(x^-4x+6) 
</pre>