All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$(1+2i)^{3}$

$=(1+2i)^{2}(1+2i)$

$=[1+4i+(2i)^{2}](1+2i)$

$=(1+4i-4)(1+2i)$

$=(4i-3)(1+2i)$

$=4i-3+4i(2i)-3(2i)$

$=4i-3-8-6i$

$=-11-2i$

(ii)
Since $w=1+2i$ is a root,
$a(1+2i)^{3}+5(1+2i)^{2}+17(1+2i)+b=0$

$a(-11-2i)+5(-3+4i)+17(1+2i)+b=0$

$-11a-2ai-15+20i+17+34i+b=0$

$-11a+2+b-2ai+54i=0$

Comparing the real and imaginary coefficients,

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

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

$\therefore a=27, b=295$

(iii)
Since all coefficient of the polynomial is real, by conjugate root theorem, if $1+2i$ is root, then $1-2i$ is also a root.
$(z-1-2i)(z-1+2i)=z^{2}-2z+5$

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

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.

### One Comment

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