z=2+j is one root of the equation z4 - 2z³- z²+ 2z + 10 = 0

We use synthetic division:

   2+j|1    -2      -1         2         10
      |      2+j    -1+2j     -6+2j     -10                
       1       j    -2+2j     -4+2j       0

So we have factored the polynomial as

 [z - (2+j)][z³+ jz²+ (-2+2j)z + (-4+2j)z] = 0

Next we factor the cubic polynomial, since we know that if
2+j is a root, its conjugate 2-j is also a root


   2-j|1       j    -2+2j     -4+2j
      |      2-j     4-2j      4-2j
       1     2       2         0

So we have now factored the polynomial as

 [z - (2+j)][z - (2-j)](z²+ 2z + 2) = 0

We set each equal to 0

Setting the first two factors = 0 just gives z = 2+j and z = 2-j.

Setting the third factor = 0 gives

     z²+ 2z + 2 = 0















z = -1±j

So the roots other than the given one, 2+j, are 2-j, -1+j and -1-j

Edwin