| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Factor theorem and finding roots |
| Difficulty | Standard +0.8 This is a multi-part question requiring factor theorem application, verification of a complex root, and then a non-trivial substitution step (z² for x) requiring careful algebraic manipulation. Part (c) demands understanding that if α is a root of p(x), then z²=α gives z=±√α, involving complex square roots. This goes beyond routine polynomial factorization into problem-solving territory typical of Further Maths Pure. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(x=-3\) to obtain value of \(p(-3)\) | M1 | |
| Obtain \(p(-3)=0\) and hence given result | A1 | |
| Alternative method: Divide \(p(x)\) by \((x+3)\) to obtain quotient \(x^2 \pm 2x + \ldots\) | M1 | |
| Obtain quotient \(x^2 + 2x + 25\), with zero remainder and hence given result | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(z = -1 + 2\sqrt{6}\)i and attempt expansions of \(z^2\) and \(z^3\) | M1 | \(z^2 = -23 - 4\sqrt{6}\)i, \(z^3 = -1 + 6\sqrt{6}\)i \(+ 72 - 48\sqrt{6}\)i |
| Use \(i^2 = -1\) | M1 | Seen at least once |
| Obtain \(p(z) = 0\) and hence given result | A1 | SC B1 if there is no evidence of working for the square or the cube. Total 1/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use roots \(z = -1 + 2\sqrt{6}\)i to form quadratic factor | M1 | \(z^2 + 2z + 25\) |
| Divide \(p(z)\) by *their* quadratic factor | M1 | |
| Obtain zero remainder and hence given result | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Set *their* quadratic factor from (a) equal to zero | M1 | |
| Solve for \(z\) | M1 | Need to see method here as answer is given |
| Obtain \(z = -1 + 2\sqrt{6}\)i (and \(z = -1 - 2\sqrt{6}\)i) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(z = -1 + 2\sqrt{6}\)i into *their* quadratic factor and attempt expansion of \(z^2\) | M1 | |
| Use \(i^2 = -1\) | M1 | |
| Obtain 0 and hence given result | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(z_1 = \sqrt{3}\)i and \(z_2 = -\sqrt{3}\)i | B1 | |
| Expand \((x + \text{i}y)^2 = -1 + 2\sqrt{6}\)i and compare real and imaginary parts | M1 | Allow for use of \(z^2 = -1 - 2\sqrt{6}\)i |
| Obtain \(x^2 - y^2 = -1\) and \(xy = \sqrt{6}\) | A1 | |
| Solve to obtain \(x\) and \(y\) | M1 | |
| Obtain \(z_3 = \sqrt{2} + \sqrt{3}\)i and \(z_4 = -\sqrt{2} - \sqrt{3}\)i | A1 | |
| Use \(z^2 = -1 - 2\sqrt{6}\)i to obtain \(z_5\) and \(z_6\) | M1 | Allow for use of \(z^2 = -1 + 2\sqrt{6}\)i |
| Obtain \(z_5 = \sqrt{2} - \sqrt{3}\)i and \(z_6 = -\sqrt{2} + \sqrt{3}\)i | A1 |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x=-3$ to obtain value of $p(-3)$ | M1 | |
| Obtain $p(-3)=0$ and hence given result | A1 | |
| **Alternative method:** Divide $p(x)$ by $(x+3)$ to obtain quotient $x^2 \pm 2x + \ldots$ | M1 | |
| Obtain quotient $x^2 + 2x + 25$, with zero remainder and hence given result | A1 | |
| **Total** | **2** | |
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $z = -1 + 2\sqrt{6}$i and attempt expansions of $z^2$ and $z^3$ | M1 | $z^2 = -23 - 4\sqrt{6}$i, $z^3 = -1 + 6\sqrt{6}$i $+ 72 - 48\sqrt{6}$i |
| Use $i^2 = -1$ | M1 | Seen at least once |
| Obtain $p(z) = 0$ and hence given result | A1 | SC B1 if there is no evidence of working for the square or the cube. Total 1/3 |
**Alternative Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Use roots $z = -1 + 2\sqrt{6}$i to form quadratic factor | M1 | $z^2 + 2z + 25$ |
| Divide $p(z)$ by *their* quadratic factor | M1 | |
| Obtain zero remainder and hence given result | A1 | |
**Alternative Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Set *their* quadratic factor from (a) equal to zero | M1 | |
| Solve for $z$ | M1 | Need to see method here as answer is given |
| Obtain $z = -1 + 2\sqrt{6}$i (and $z = -1 - 2\sqrt{6}$i) | A1 | |
**Alternative Method 3:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $z = -1 + 2\sqrt{6}$i into *their* quadratic factor and attempt expansion of $z^2$ | M1 | |
| Use $i^2 = -1$ | M1 | |
| Obtain 0 and hence given result | A1 | |
**Total: 3 marks**
---
## Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $z_1 = \sqrt{3}$i and $z_2 = -\sqrt{3}$i | B1 | |
| Expand $(x + \text{i}y)^2 = -1 + 2\sqrt{6}$i and compare real and imaginary parts | M1 | Allow for use of $z^2 = -1 - 2\sqrt{6}$i |
| Obtain $x^2 - y^2 = -1$ and $xy = \sqrt{6}$ | A1 | |
| Solve to obtain $x$ and $y$ | M1 | |
| Obtain $z_3 = \sqrt{2} + \sqrt{3}$i and $z_4 = -\sqrt{2} - \sqrt{3}$i | A1 | |
| Use $z^2 = -1 - 2\sqrt{6}$i to obtain $z_5$ and $z_6$ | M1 | Allow for use of $z^2 = -1 + 2\sqrt{6}$i |
| Obtain $z_5 = \sqrt{2} - \sqrt{3}$i and $z_6 = -\sqrt{2} + \sqrt{3}$i | A1 | |
**Total: 7 marks**10 The polynomial $x ^ { 3 } + 5 x ^ { 2 } + 31 x + 75$ is denoted by $\mathrm { p } ( x )$.
\begin{enumerate}[label=(\alph*)]
\item Show that $( x + 3 )$ is a factor of $\mathrm { p } ( x )$.
\item Show that $z = - 1 + 2 \sqrt { 6 } \mathrm { i }$ is a root of $\mathrm { p } ( z ) = 0$.
\item Hence find the complex numbers $z$ which are roots of $\mathrm { p } \left( z ^ { 2 } \right) = 0$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q10 [12]}}