How do i find all zeros of polynomial x^5+4x^3+8x^2+32

so far what what I have is x^3(x^2+4)+8(x^2+4)
= (x^3+8)(x^2+4)
I am stuck otherwise If you can please show me how to approach the rest of find all zeros I greatly appreciateit thanks.

= (x^3+8)(x^2+4)
= (x+2)(x^2- 2x+4)(x-2i)(x+2i)
=(x+2)(x^2- 2x+4) (x-2i)(x+2i)
=(x+2){x-1+i(sqrt 3)}{x-1- i(sqrt 3)} (x-2i)(x+2i)
Therefore all zeros are
x= -2, 2i, -2i, 1+i(sqrt 3),1-i(sqrt 3) ................ans

This factors as: (x + 2)(x^2 + 4)(x^2 - 2x + 4) = 0

You MUST have this set to zero if you're looking for zeros! Otherwise it's just an expression. Best learn the difference.

Now you can look at each set of parentheses equal to zero and determine the zeros of each of those, which you should be able to handle from here.

not sure if this will help u understand
http://www.analyzemath.com/polynomials/p…