3+2i; 1-√3 and -1

For a polynomial to have integral coefficients, if it has a root
of the form a+bi, it also has a-bi as a root, and vice-versa.
Alos if a polynomial has a root of the form a+√b, it also 
has a-√b as a root, and vice-versa.

So the complete set of roots are

3+2i, 3-2i, 1-√3, 1-√3 and -1.

We make 5 equations with x= before each one:

x=3+2i, x=3-2i, x=1-√3, x=1+√3, x=-1

Get zero on the right of each one:

x-3-2i=0, x-3+2i=0, x-1+√3=0, x=1-√3, x+1=0

Then we multiply equals by equals:

(x-3-2i)(x-3+2i)(x-1+√3)(x-1-√3)(x+1)=0

Then you multiply that out.  You can make
it a little easier by grouping like this:

[(x-3)-2i][(x-3)+2i][(x-1)+√3][(x-1)-√3](x+1)=0

But it's a big job.  I won't go through the
details here, to let you get some practice.  
The result is:



---------------------------------------

The other one has a fraction and a cube root.

,, 

In this one, before you mutiply them you must get them so that
there are no fractions or cube roots, only integers.
So you must clear  of fractions, getting  
and cube both sides of , getting 

So you have these 

,, 

Get 0 on the right of each:

,, 

Multiply equals by equals:



Multiply that out and the result is



Edwin