Specialist Mathematics · Mock exam 2026 All mock exam questions
Mock exam · Paper 1 · Simple familiar

Question 11

[4 marks] Technology-free
The question

Consider the polynomial $P(z)=z^{3}-3z^{2}+7z-5$, where $z\in C$.

a)

Use the factor theorem to show that $(z-1)$ is a factor of $P(z)$.

[1 mark] b)

Determine the quadratic factor $Q(z)$ such that $P(z)=(z-1)\,Q(z)$.

[1 mark] c)

Use your result from Question 11b) to determine the solutions of $P(z)=0$.

[2 marks]
Watch the situation first

One root from the factor theorem, two from the square

Watch the running total for $P(1)$ land on 0. Then match coefficients column by column to get $Q(z)$. Last, watch the two non-real solutions sit 2 above and 2 below $z=1$.

Part a): put z = 1 into P(z) and keep a running total The total lands on 0, so P(1) = 0 and (z − 1) is a factor Part b): match coefficients with (z − 1)(z2 + bz + c) The spare z column checks out, so Q(z) = z2 − 2z + 5 Part c): complete the square, so (z − 1)2 = −4 z − 1 = ±2i, so step 2 up and 2 down from 1 Keep z = 1 from part a): three solutions in all −1 1 2 3 −3 −2 −1 1 2 3 O Re(z) Im(z) Solutions of P(z) = 0 z = 1 +2i z = 1 + 2i −2i z = 1 − 2i P(1) = 13 − 3(1)2 + 7(1) − 5 = 1 − 3 + 7 − 5 Running total +1 −3 +7 −5 −3 −2 −1 0 1 2 3 4 5 6 1 − 3 = −2 −2 + 7 = 5 5 − 5 = 0 P(1) = 0 z3 − 3z2 + 7z − 5 = (z − 1)(z2 + bz + c) z3 z2 z 1 P(z) right side 1 −3 7 −5 1 b − 1 c − b −c z2: b − 1 = −3, so b = −2 constant: −c = −5, so c = 5 check the z term: c − b = 5 − (−2) = 7 ✓ Q(z) = z2 − 2z + 5 z2 − 2z + 5 = (z2 − 2z + 1) + 4 = (z − 1)2 + 4 (z − 1)2 + 4 = 0 (z − 1)2 = −4 z − 1 = ±2i z = 1 + 2i or z = 1 − 2i Solutions of P(z) = 0: z = 1, z = 1 + 2i, z = 1 − 2i

The Argand plane is drawn to scale, with equal scales on the axes. The two non-real solutions are a conjugate pair, one each side of the real axis.

Now for the mathematics

Choose the quadratic factor and watch the solutions

Set $b$ and $c$ in $Q(z)=z^2+bz+c$. The readout expands $(z-1)\,Q(z)$ and checks each coefficient against $P(z)$. Match all three coefficients, then switch routes to watch each method solve $Q(z)=0$. Try the two common wrong factors from b) too, $z^2+2z+5$ and $z^2-2z-5$, to see where they lead in c).

Re(z) Im(z) O
$z=1$, from the factor $(z-1)$ Solutions of $Q(z)=0$ When every coefficient matches
Choose your route for part c)
Solutions of (z − 1) Q(z) = 0

$(z-1)\,Q(z)$ doesn’t match $P(z)$.

All three coefficients match, so $Q(z)=z^2-2z+5$. Its solutions are $1+2i$ and $1-2i$, and with $z=1$ from part a) that makes three solutions of $P(z)=0$.

The constant matches, but $(z-1)\left(z^{2}+2z+5\right)=$ $z^{3}+z^{2}+3z-5\neq P(z)$. Check the sign of $b$: the $z^2$ column needs $b-1=-3$. Under follow-through, this factor still earns both marks in c), but not the mark in b).

The $z^2$ coefficient matches, but $(z-1)\left(z^{2}-2z-5\right)=$ $z^{3}-3z^{2}-3z+5$. The constant on the right side is $-c$, so $-c=-5$ gives $c=5$.

Not yet. Start with the $z^2$ column: $b-1$ has to equal $-3$.

The $z^2$ column matches. Now the constant column: $-c$ has to equal $-5$.

Marking guide · 4 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Choose a method

Step 1 · part a)1 mark

By the factor theorem, $(z-1)$ is a factor exactly when $P(1)=0$. Substitute $z=1$.

$$P(1)=1-3+7-5=0$$

So $(z-1)$ is a factor of $P(z)$.

Marker: correctly evaluates $P(1)$ and concludes that $(z-1)$ is a factor

Step 2 · part b)1 mark

Write $P(z)$ as $(z-1)\left(z^{2}+bz+c\right)$ and match coefficients.

$$P(z)=(z-1)\left(z^{2}-2z+5\right)$$

$Q(z)=z^{2}-2z+5$
Why?

Two polynomials are equal for every $z$ only if matching coefficients are equal. Expanding gives $z^{3}+(b-1)z^{2}+(c-b)z-c$, so $b-1=-3$ gives $b=-2$ and $-c=-5$ gives $c=5$.

Marker: correctly determines the quadratic factor

Step 3 · part c)1 mark

Set $Q(z)=0$, complete the square and take square roots over $C$.

$$\begin{aligned}z^{2}-2z+5&=0\\(z-1)^{2}&=-4\\z&=1\pm2i\end{aligned}$$

Marking

The quadratic formula or completing the square is accepted. Follow-through (FT) applies in Question 11c) from an incorrect quadratic factor in Question 11b).

Marker: correctly determines the non-real solutions

Step 4 · part c)1 mark

Add the real solution from $z-1=0$ to the two non-real solutions.

$z=1$,$z=1+2i$,$z=1-2i$

Marker: correctly states all three solutions

Step 1 · part a)1 mark

By the factor theorem, $(z-1)$ is a factor exactly when $P(1)=0$. Substitute $z=1$.

$$P(1)=1-3+7-5=0$$

So $(z-1)$ is a factor of $P(z)$.

Marker: correctly evaluates $P(1)$ and concludes that $(z-1)$ is a factor

Step 2 · part b)1 mark

Write $P(z)$ as $(z-1)\left(z^{2}+bz+c\right)$ and match coefficients.

$$P(z)=(z-1)\left(z^{2}-2z+5\right)$$

$Q(z)=z^{2}-2z+5$
Why?

Two polynomials are equal for every $z$ only if matching coefficients are equal. Expanding gives $z^{3}+(b-1)z^{2}+(c-b)z-c$, so $b-1=-3$ gives $b=-2$ and $-c=-5$ gives $c=5$.

Marker: correctly determines the quadratic factor

Step 3 · part c)1 mark

Solve $Q(z)=0$ with the quadratic formula, using $a=1$, $b=-2$ and $c=5$.

$$\begin{aligned}z&=\frac{2\pm\sqrt{4-20}}{2}\\&=\frac{2\pm4i}{2}\\&=1\pm2i\end{aligned}$$

Marking

The quadratic formula or completing the square is accepted. Follow-through (FT) applies in Question 11c) from an incorrect quadratic factor in Question 11b).

Marker: correctly determines the non-real solutions

Step 4 · part c)1 mark

Add the real solution from $z-1=0$ to the two non-real solutions.

$z=1$,$z=1+2i$,$z=1-2i$

Marker: correctly states all three solutions

Putting it all together

Each phrase of the question gave you something

“where $z\in C$”
The solutions can be non-real. When a square equals $-4$, you take both complex square roots, $\pm 2i$.
“Use the factor theorem to show that $(z-1)$ is a factor”
Substitute $z=1$, not $z=-1$. Show $P(1)=0$ step by step, then write the conclusion.
“Determine the quadratic factor $Q(z)$ such that $P(z)=(z-1)\,Q(z)$”
Write $Q(z)=z^2+bz+c$, expand and match coefficients: $b=-2$ and $c=5$, so $Q(z)=z^2-2z+5$.
“Use your result from Question 11b)”
Solve $Q(z)=0$ by completing the square or the quadratic formula: $z=1\pm 2i$.
“determine the solutions of $P(z)=0$”
All three: $z=1$ from $z-1=0$, plus $z=1+2i$ and $z=1-2i$.
Common errors

If I had to guess where you might lose marks…

  • Substituting $z=-1$ instead of $z=1$ in a). You get $P(-1)=-1-3-7-5=-16\neq 0$, and wrongly conclude that $(z-1)$ is not a factor. The root that matches the factor $(z-a)$ is $z=a$.
  • A sign error in the quadratic factor. You get $Q(z)=z^{2}+2z+5$. Expanding, $(z-1)\left(z^{2}+2z+5\right)=z^{3}+z^{2}+3z-5\neq P(z)$, and c) then gives $z=-1\pm 2i$. Under follow-through this earns both marks in c), but not the mark in b).
  • The wrong constant in the quadratic factor. You match the constant terms as $c=-5$ instead of $-c=-5$ and get $z^{2}-2z-5$. Its discriminant is 24, so c) gives the real roots $z=1\pm\sqrt{6}$. Expanding shows the error: $(z-1)\left(z^{2}-2z-5\right)=z^{3}-3z^{2}-3z+5$.
  • Halving only the real part. In the quadratic formula you write $\dfrac{2\pm 4i}{2}$ as $1\pm 4i$. Check: $Q(1+4i)=-12\neq 0$.
  • Dropping the $i$. You drop the $i$ from $\sqrt{-16}$ and get $\dfrac{2\pm 4}{2}=3$ or $-1$. Neither is a root: $P(3)=16$ and $P(-1)=-16$. Leaving out $z=1$ from your final list in c) is also common, and it loses the last mark.

Only mark this done when you could do it without help. Reading the solution does not count.

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A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.

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