Question 18
Consider the polynomials $P(z)=z^{3}+(i-a)z^{2}-2biz+3i$ and $Q(z)=z-2i$, where $a,b\in\mathbb{R}$.
Given $\dfrac{P(z)}{Q(z)}$ has a remainder of $a-bi$, evaluate the reasonableness that $\left(z-(a-bi)\right)$ is a factor of $P(z)$.
Two theorems do all the work
The remainder theorem finds $a$ and $b$, and the factor theorem tests the claim. Blue is $P(2i)$ and orange is $a-bi$.
Each line is tagged with the mark it earns.
Find the a and b that give the right remainder
Move the sliders until the remainder $P(2i)$ equals $a-bi$. Then check whether $a-bi$ is a root of $P(z)$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. You can use your calculator for the last substitution, but show the algebra for $P(2i)$.
The remainder theorem says that when you divide $P(z)$ by $z-2i$, the remainder is $P(2i)$. So you do not need long division.
$$P(2i)=a-bi$$
A remainder from long division, $-8i-4(i-a)+4b+3i$, also earns you this mark.
Marker: correctly determines an expression for $P(2i)$ using the remainder theorem
Substitute $z=2i$ into $P(z)$. Work out the powers first: $(2i)^{2}=-4$ and $(2i)^{3}=-8i$.
$$\begin{aligned}P(2i)&=(2i)^{3}+(i-a)(2i)^{2}-2bi(2i)+3i\\&=-8i-4(i-a)+4b+3i\\&=4a+4b-9i\end{aligned}$$
Marker: correctly determines an expression for $P(2i)$ using substitution into $P(z)$
Because $a$ and $b$ are real, you can match the real parts and the imaginary parts separately.
$$4a+4b-9i=a-bi$$
$$\text{Re: }\ a=4a+4b\qquad\text{Im: }\ -b=-9$$
$$b=9,\quad a=-12$$
The imaginary part of $P(2i)$ is always $-9$, so $b$ has to be 9 whatever $a$ is. You keep follow-through marks from here if you slipped earlier.
Marker: forms two simultaneous equations by equating parts
Put the values back into $P(z)$. The factor theorem says $z-(a-bi)$ is a factor only if $P(a-bi)=0$.
$$P(z)=z^{3}+(12+i)z^{2}-18iz+3i$$
$$P(a-bi)=P(-12-9i)\approx1566-285i$$
Use your calculator for this. It is technology-active, and the numbers are large.
Marker: determines $P(a-bi)$ using the values for $a$ and $b$
Now you make the judgement, and you give the reason from your calculation.
Since $P(-12-9i)\neq0$, it is not reasonable that $\left(z-(a-bi)\right)$ is a factor of $P(z)$.
Just writing “not reasonable” is not enough. Your statement has to point to the calculation of $P(a-bi)$.
Marker: evaluates the reasonableness of the statement using mathematical reasoning
Each phrase of the question gave you something
The unknowns $a$ and $b$ appear in the polynomial and in the remainder at the same time. You have to use the remainder theorem to pin them down before you can test the claim with the factor theorem.
The line I look for first is the final statement. The QCAA like “evaluate the reasonableness” questions, and the last mark only goes to a judgement that is backed up by a calculation.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2022 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.