This one is extremely...tedious...to solve. You really need some luck and clever ideas on how to proceed.
First, I apply the Rational Roots Theorem to come up with a list of all possible rational roots for f(x) = 0. I list all of the factors of the constant term p and the leading coefficient term q, and then consider all reduced form rational terms of the form +-(p " factor")/(q " factor"):
p " factors": 1, 3, 5, 9, 15, 25, 45, 75, 125, 225, 375, 1125
#q " factors": 1, 3
Possible rational roots:
+- (1, 1/3, 3, 5, 5/3, 9, 15, 25, 25/3, 45, 75, 125, 125/3, 225, 375, 1125)
This is fairly huge. There are 32 possible rational roots. Since this is a degree 7 polynomial, there will be 7 roots (although some may be complex or irrational in nature).
From here, it's trial and error time. Synthetic division can help; test a possible root using synthetic division and look for a final value of 0, which would indicate you've located a zero. I'll omit all the trials that did not succeed for me to save space.
Try x = -3
-3__|color(white)("a")3color(white)("aa")-32color(white)("aa")28color(white)("aaa")591color(white)("aa")-1181color(white)("aa")-2810color(white)("aaa")5550color(white)("a")-1125
underline(color(white)("aaaaaaaaa")-9color(white)("a")123color(white)("")-453color(white)("aaa")-414color(white)("aaaaa")4785color(white)("")-5925color(white)("aaaa")1125)
color(white)("aaaa")3color(white)("aa")-41color(white)("a")151color(white)("aaa")138color(white)("aa")-1595color(white)("aaaa")1975color(white)("aa")-375color(white)("aaaaa")color(red)(0)
This result tells us that x = -3 is a zero. From now on, we can use the reduced final row for our synthetic testing.
Try x= 5
5__|color(white)("aaaa")3color(white)("aa")-41color(white)("aaa")151color(white)("aaa")138color(white)("aa")-1595color(white)("aaaa")1975color(white)("aa")-375
underline(color(white)("aaaaaaaaaaa")15color(white)
("a")-130color(white)("aaa")105color(white)("aaaa")1215color(white)("aa")-1900color(white)("aaaa")375)
color(white)("aaaaaa")3color(white)("aa")-26color(white)("aaaa")21color(white)("aaa")243color(white)("aa")-380color(white)("aaaaaaa")75color(white)("aaaaa")color(red)(0)
This result tell us that x = 5 is a zero. Ignoring the final 0 result, we are now down to a 5th degree polynomial.
The next step is a trick. We already know x = -3 is a zero. It could be a zero of multiplicity; in other words, it could occur as more than one factor.
Try x = -3 (again)
-3__|color(white)("aaaa")3color(white)("aa")-26color(white)("aaaa")21color(white)("aaaa")243color(white)("aa")-380color(white)("aaaaaa")75
underline(color(white)("aaaaaaaaaaaa")-9color(white)
("aaaa")105color(white)("a")-378color(white)("aaaa")405color(white)("aaaa")-75)
color(white)("aaaaaaaa")3color(white)("aa")-35color(white)("aaa")126color(white)("a")-135color(white)("aaaaa")25color(white)("aaaaaaa")color(red)(0)
This result tells us that x = -3 is a zero a 2nd time. An additional try of -3 (not shown here) shows that -3 is not a zero a 3rd time. (An application of Descarte's Rule of Signs would demonstrate that -3 could not be a zero three times.)
Try x = 5 (again)
5__|color(white)("aaaaa")3color(white)("aaa")-35color(white)("aaaa")126color(white)("aaa")-135color(white)("aaaaa")25
underline(color(white)("aaaaaaaaaaaaa")15color(white)("aa")-100color(white)("aaaaa")130color(white)("aaa")-25)
color(white)("aaaaaaa")3color(white)("aaa")-20color(white)("aaaaa")26color(white)("aaaa")-5color(white)("aaaaaa")color(red)(0)
This result tells us that x = 5 is a zero a 2nd time. Could it be a zero a third time?
Try x = 5 (again x 2)
5__|color(white)("aaaa")3color(white)("aaa")-20color(white)("aaaaa")26color(white)("aaaa")-5
underline(color(white)("aaaaaaaaaaaa")15color(white)("aaa")-25color(white)("aaaaaa")5)
color(white)("aaaaaa")3color(white)("aaa")-5color(white)("aaaaaa")1color(white)("aaaaaaa")color(red)(0)
This result tells us that x = 5 is a zero a 3rd time.
At this point we are now left with a quadratic polynomial of 3x^2-5x+1. which we can complete using the Quadratic Formula:
x = (-b+-sqrt(b^2-4ac))/(2a)
x = (5+-sqrt((=5)^2-4(3)(1)))/(2*3)
x = (5+-sqrt(25-12))/6 =(5+-sqrt(13))/6
Our final collection of all 7 roots is:
x = -3, -3, 5, 5, 5, (5+sqrt(13))/6, (5-sqrt(13))/6