Question 978698
First try check for root of 1 gives  {{{(z-1)(z^3+z^2-2)}}}.  Be sure that your dividend of synthetic division shows ALL degrees of z.


continuing, again if root of 1 is checked, you find factorization {{{(z-1)(z-1)(z^2+2z+2)}}}.  You might have made a visual omission on paper or through keyboard, or maybe forgot about accounting for degrees of z in the division process.


NOW to deal with the quadratic factor,
roots are {{{(-2+- sqrt(4-4*2))/2}}}
{{{-1+- i}}}.



To conitnue,
{{{(z-1)^2(z-(-1-i))(x-(-1+i))}}}, and you simply simplify this obviously complex factorization or keep as linear and quadratic factorization...
{{{(z-1)^2(z+1+i)(z+1-i)}}}
{{{highlight((z-1)^2((z+1)+i)((z+1)-i))}}}
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Either you want it that way, or you want ...
{{{highlight((z-1)^2(z^2+2z+2))}}}.