What are all the possible rational zeros for #f(x)=3x^56x^4+2x^26x+12# and how do you find all zeros?
1 Answer
Assuming:
#f(x) = 3x^56x^4x^3+2x^26x+12#
Possible rational zeros:
#+1/3,+2/3,+1,+4/3,+2,+3,+4,+6,+12#
Actual zeros:
#2," "+sqrt(1/6(1+sqrt(73)))," "+sqrt(1/6(sqrt(73)1))i#
Explanation:
I think there is a missing term

The given quintic has no solutions expressible using radicals, let alone rational roots.

There is no
#x^3# term  so one might well have been missed. 
If the missing term is
#x^3# then the quintic is solvable using precalculus level methods and has one rational zero.
So, suppose:
#f(x) = 3x^56x^4x^3+2x^26x+12#
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+1/3, +2/3, +1, +4/3, +2, +3, +4, +6, +12#
We find:
#f(2) = 3(color(blue)(2))^56(color(blue)(2))^4(color(blue)(2))^3+2(color(blue)(2))^26(color(blue)(2))+12#
#color(white)(f(2)) = 96968+812+12 = 0#
So
#3x^56x^4x^3+2x^26x+12 = (x2)(3x^4x^26)#
Then:
#12(3x^4x^26) = 36x^412x^272#
#color(white)(12(3x^4x^26)) = (6x^2)^22(6x^2)+173#
#color(white)(12(3x^4x^26)) = (6x^21)^2(sqrt(73))^2#
#color(white)(12(3x^4x^26)) = ((6x^21)sqrt(73))((6x^21)+sqrt(73))#
#color(white)(12(3x^4x^26)) = (6x^21sqrt(73))(6x^21+sqrt(73))#
Hence:
#x = +sqrt(1/6(1+sqrt(73)))" "# or#" "x = +sqrt(1/6(sqrt(73)1))i#