How do you find all the zeros of #9x^4 + 3x^3  7x^2 + 7x  9 = 0#?
1 Answer
See explanation...
Explanation:
By the rational root theorem, any rational zeros of
That means that the only possible rational zeros are:
#+1/9# ,#+1/3# ,#+1# ,#+3# ,#+9#
In addition, note that
So the only possible rational zeros are:
#1/9, 1/3, 1, 3, 9#
Evaluating
At this stage we have two main choices. We can seek an algebraic solution, which is doable, but lengthy and complicated. Alternatively, we can seek a numerical solution.
One simple method to seek numerical values for the zeros is Newton's:

Start with an approximation
#a_0# for the zero of#f(x)# . 
Refine the approximation using the formula:
#a_(i+1) = a_i  (f(x))/(f'(x))# 
Repeat until you get the desired numerical accuracy.
In our example:
#f'(x) = 36x^3+9x^214x+7#
So our iterative step is:
#a_(i+1) = a_i  (9x^4+3x^37x^2+7x9)/(36x^3+9x^214x+7)#
#=a_i  (9x^43x^3+7x^27x+9)/(36x9x^2+14x7)#
The slight complication is that the zeros we are looking for are all nonReal Complex ones, so to use this formula in a spreadsheet (which typically do not understand Complex numbers), you need to split the Real and imaginary parts into separate columns and express the Complex arithmetic in terms of those parts.
This takes a little patience to set up, but once you have it, you can try different initial approximations,
#x~~0.48389+0.98869i#
#x~~0.65056+0.63411i#