The key to this problem is to recognize that any complex root to a polynomial must be accompanied by its complex conjugate root as well. Thus, if #a+bi# is a root to a function, so too must #a-bi#.
We're given 4 roots: #2i#, #2i# (again), #4-i#, and #isqrt(3)#. Because of the need for complex conjugates, we must also have #-2i#, #-2i# (again), #4+i#, and #-isqrt(3)# as roots as well.
Thus, we know that the function #f(x)#,in factored form, must look like this:
#f(x)=(x-2i)(x+2i)(x-2i)(x+2i)(x-(4-i))(x-(4+i))(x-isqrt(3))(x+isqrt(3))#
By multiplying this out and simplifying we should remove all imaginary units and arrive at a polynomial of real number coefficients:
#f(x) = (x^2+4)(x^2+4)(x^2-8x+17)(x^2+3)#
# = (x^4+8x^2+16)(x^2-8x+17)(x^2+3)#
# = (x^6-8x^5+17x^4+8x^4-64x^3+136x^2+16x^2-128x+272)(x^2+3)#
# = (x^6-8x^5+25x^4-64x^3+152x^2-128x+272)(x^2+3) #
# = (x^8-8x^7+25x^6-64x^5+152x^4-128x^3+272x^2+3x^6-24x^5+75x^4-192x^3+456x^2-384x+816) #
# = x^8-8x^7 + 28x^6-88x^5+227x^4-320x^3+728x^2-384x+816#
