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#