Question 10
The polynomial $P(z)=z^3-2iz^2+z-2i$ can be expressed in factorised form as $P(z)=(z-i)\left(z^2+biz+2\right)$, where $b\in Z$.
Determine the value of $b$.
Not quite. With $b=2$ the $z$ term becomes $4z$, not $z$. Expand the brackets and match the coefficient of $z$. In 2024, 7.3% of Queensland students chose (A).
Not quite. With $b=1$ the $z^2$ terms cancel, but $P(z)$ has $-2iz^2$. Watch the sign when you multiply $-i$ by $biz$. In 2024, 11.6% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2024, 68.4% of Queensland students chose (C).
Not quite. With $b=-2$ the $z$ term disappears, but $P(z)$ has $+z$. Check the product $(-i)(biz)=bz$. In 2024, 12.3% of Queensland students chose (D).
Choose a value and compare the expanded product with $P(z)$ term by term.
Expand the factorised form. Multiply each term in the first bracket by the whole second bracket, and use $i^2=-1$.
$$(z-i)(z^2+biz+2)=z^3+biz^2+2z-iz^2-bi^2z-2i$$
Collect like terms. The $-bi^2z$ term becomes $+bz$.
$$=z^3+(b-1)iz^2+(2+b)z-2i$$
Match the coefficients with $P(z)=z^3-2iz^2+z-2i$. The $z$ terms give the quickest equation. The $z^2$ terms should then agree as a check.
$$2+b=1\qquad (b-1)i=-2i$$
$$(z-i)(z^2+biz+2)=z^3+(b-1)iz^2+(2+b)z-2i$$
Matching the coefficient of $z$:
$$2+b=1\ \Rightarrow\ b=-1$$
Check with the $z^2$ term: $(b-1)i=-2i$.
How Queensland students answered in 2024
From the QCAA Specialist Mathematics subject report for 2024. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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