+128475 I used x instead of z.....same difference
If 2 + 3i is a root, then so is 2 - 3i
So
( x - (2 + 3i) ) ( x - (2- 3i) will divide the polyomial.....simplify.....
x^2 - (2 + 3i)x - (2 - 3i)x + (2 + 3i) (2 - 3i)
x^2 -2x -3ix - 2x + 3ix + (4 + 9)
x^2 -4x + 13
Using synthetic division
x^2 - x - 2
x^2 - 4x + 13 [ x^4 - 5x^3 +15x^2 -5x - 26 ]
x^4 -4x^3 +13x^2
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-x^3 + 2x^2 - 5x - 26
-x^3 +4x^2 -13x
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-2x^2 + 8x - 26
-2x^2 + 8x - 26
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The remaining polynomial is x^2 - x - 2 which factors as ( x + 1) (x - 2)
Seting each linear factor to 0 and solving for x produces the other two roots x = - 1 and x = 2
