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Since w=1+2i is a root,




Comparing the real and imaginary coefficients,

-2a+54 = 0 ~\Rightarrow ~ a=27,

-11(27)+2+b= 0 ~\Rightarrow ~ b=295,

\therefore a=27, b=295

Since all coefficient of the polynomial is real, by conjugate root theorem, if 1+2i is root, then 1-2i is also a root.

27z^{3}+5z^{2}+17z+295 \equiv (z^{2}-2z+5)(cz+d),

by comparing coefficients, c=27, d=59

\mathrm{Roots}=1+2i, 1-2i, -\frac{59}{27}

KS Comments:

Since they want to see all workings, let us not skip any steps here and do dilligently. Do check your answers with the GC though. It is good if students emphasise that the coefficients of polynomial is real here, to indicate your understanding. It is shocking that some students confused a root with a factor though.

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