| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Complex roots with real coefficients |
| Difficulty | Moderate -0.3 This is a straightforward application of the complex conjugate root theorem. Part (a) is routine substitution to verify a root, and part (b) requires knowing that the conjugate must also be a root, then finding the real root by division or sum of roots. Standard technique with no novel insight required, making it slightly easier than average. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(-1+\sqrt{5}\)i in the equation and attempt expansions of \(x^2\) and \(x^3\) | M1 | All working must be seen. Allow M1 if small errors in \(1-2\sqrt{5}\)i\(-5\) or \(1-\sqrt{5}\)i\(-\sqrt{5}\)i\(-5\) and \(4-2\sqrt{5}\)i\(+10\) or \(4-4\sqrt{5}\)i\(+2\sqrt{5}\)i\(+10\) |
| Use \(i^2=-1\) correctly at least once | M1 | \(1-5\) or \(4+10\) seen |
| Complete the verification correctly | A1 | \(2(14-2\sqrt{5}\)i\()+(-4-2\sqrt{5}\)i\()+6(-1+\sqrt{5}\)i\()-18=0\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State second root \(-1-\sqrt{5}\)i | B1 | |
| Carry out a complete method for finding a quadratic factor with zeros \(-1+\sqrt{5}\)i and \(-1-\sqrt{5}\)i | M1 | |
| Obtain \(x^2+2x+6\) | A1 | |
| Obtain root \(x=\dfrac{3}{2}\) | A1 | OE |
| Alternative: State second root \(-1-\sqrt{5}\)i | B1 | |
| \((x+1-\sqrt{5}\)i\()(x+1+\sqrt{5}\)i\()(2x+a)=2x^3+x^2+6x-18\) | M1 | |
| \((1-\sqrt{5}\)i\()(1+\sqrt{5}\)i\()\,a=-18\) | A1 | |
| \(6a=-18,\ a=-3\) leading to \(x=\dfrac{3}{2}\) | A1 | OE |
| Alternative: State second root \(-1-\sqrt{5}\)i | B1 | |
| POR \(= 6\), SOR \(= -2\) | M1 | |
| Obtain \(x^2+2x+6\) | A1 | |
| Obtain root \(x=\dfrac{3}{2}\) | A1 | OE |
| Alternative: State second root \(-1-\sqrt{5}\)i | B1 | |
| POR \((-1-\sqrt{5}\)i\()(-1+\sqrt{5}\)i\()a=9\) | M1 A1 | |
| Obtain root \(x=\dfrac{3}{2}\) | A1 | OE |
| Alternative: State second root \(-1-\sqrt{5}\)i | B1 | |
| SOR \((-1-\sqrt{5}\)i\()+(-1+\sqrt{5}\)i\()+a=-\dfrac{1}{2}\) | M1 A1 | |
| Obtain root \(x=\dfrac{3}{2}\) | A1 | OE |
| 4 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $-1+\sqrt{5}$i in the equation and attempt expansions of $x^2$ and $x^3$ | M1 | All working must be seen. Allow M1 if small errors in $1-2\sqrt{5}$i$-5$ or $1-\sqrt{5}$i$-\sqrt{5}$i$-5$ and $4-2\sqrt{5}$i$+10$ or $4-4\sqrt{5}$i$+2\sqrt{5}$i$+10$ |
| Use $i^2=-1$ correctly at least once | M1 | $1-5$ or $4+10$ seen |
| Complete the verification correctly | A1 | $2(14-2\sqrt{5}$i$)+(-4-2\sqrt{5}$i$)+6(-1+\sqrt{5}$i$)-18=0$ |
| | **3** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State second root $-1-\sqrt{5}$i | B1 | |
| Carry out a complete method for finding a quadratic factor with zeros $-1+\sqrt{5}$i and $-1-\sqrt{5}$i | M1 | |
| Obtain $x^2+2x+6$ | A1 | |
| Obtain root $x=\dfrac{3}{2}$ | A1 | OE |
| **Alternative:** State second root $-1-\sqrt{5}$i | B1 | |
| $(x+1-\sqrt{5}$i$)(x+1+\sqrt{5}$i$)(2x+a)=2x^3+x^2+6x-18$ | M1 | |
| $(1-\sqrt{5}$i$)(1+\sqrt{5}$i$)\,a=-18$ | A1 | |
| $6a=-18,\ a=-3$ leading to $x=\dfrac{3}{2}$ | A1 | OE |
| **Alternative:** State second root $-1-\sqrt{5}$i | B1 | |
| POR $= 6$, SOR $= -2$ | M1 | |
| Obtain $x^2+2x+6$ | A1 | |
| Obtain root $x=\dfrac{3}{2}$ | A1 | OE |
| **Alternative:** State second root $-1-\sqrt{5}$i | B1 | |
| POR $(-1-\sqrt{5}$i$)(-1+\sqrt{5}$i$)a=9$ | M1 A1 | |
| Obtain root $x=\dfrac{3}{2}$ | A1 | OE |
| **Alternative:** State second root $-1-\sqrt{5}$i | B1 | |
| SOR $(-1-\sqrt{5}$i$)+(-1+\sqrt{5}$i$)+a=-\dfrac{1}{2}$ | M1 A1 | |
| Obtain root $x=\dfrac{3}{2}$ | A1 | OE |
| | **4** | |7
\begin{enumerate}[label=(\alph*)]
\item Verify that $- 1 + \sqrt { 5 } \mathrm { i }$ is a root of the equation $2 x ^ { 3 } + x ^ { 2 } + 6 x - 18 = 0$.
\item Find the other roots of this equation.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q7 [7]}}