| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Symmetric functions of roots |
| Difficulty | Standard +0.3 This is a standard symmetric functions question requiring direct application of Vieta's formulas. Part (i) uses sum of reciprocals (straightforward algebraic manipulation), part (ii) expands to symmetric functions, and part (iii) uses the identity p³+q³+r³-3pqr=(p+q+r)(p²+q²+r²-pq-pr-qr). While it requires knowing these techniques, all are textbook methods with no novel insight needed, making it slightly easier than average. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(p+q+r=2,\quad pq+pr+qr=4,\quad pqr=5\) | B1 | Identifies correct values for all 3 expressions; allow \(\sum p\), \(\sum pq\) notation |
| \(\frac{2}{p}+\frac{2}{q}+\frac{2}{r} = \frac{2(pq+pr+qr)}{pqr}\) | M1 | Uses correct identity for the sum |
| \(= \frac{8}{5}\) | A1ft | Correct value (follow through on their 2, 4 and 5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x = \frac{2}{y} \Rightarrow \frac{8}{y^3}-\frac{8}{y^2}+\frac{8}{y}-5=0 \Rightarrow 5y^3-8y^2+8y-8=0\) | B1 | Obtains correct cubic in \(y\) |
| \(\frac{2}{p}+\frac{2}{q}+\frac{2}{r} = -\frac{-8}{5}\) | M1 | Uses correct method |
| \(= \frac{8}{5}\) | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((p-4)(q-4)(r-4) = (pq-4p-4q+16)(r-4)\) \(= pqr - 4pq-4pr-4qr+16p+16q+16r-64\) | M1, A1 | Attempt to expand — must involve sum, pair sum and product; correct expansion |
| \(= pqr - 4(pq+pr+qr)+16(p+q+r)-64\) \(= 5-4(4)+16(2)-64 = -43\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((x+4)^3 - 2(x+4)^2 + 4(x+4) - 5 = 0\) | M1 | Substitutes \(x+4\) for \(x\) in the given cubic |
| \(= \ldots 64 + \ldots - 32 + \ldots 16 + \ldots - 5 = 43\) | A1 | Calculates correct constant term |
| \(\therefore (p-4)(q-4)(r-4) = -43\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(p^3+q^3+r^3 = (p+q+r)^3 - 3(p+q+r)(pq+pr+qr)+3pqr\) or equivalent identities | M1 | Establishes correct identity in terms of sum, pair sum and product, and substitutes to reach numerical expression for \(p^3+q^3+r^3\) |
| \(= 2^3 - 3(2)(4)+3(5) = -1\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(p^3-2p^2+4p-5=0\), etc. \(\Rightarrow p^3+q^3+r^3 = 2(p^2+q^2+r^2)-4(p+q+r)+15\) \(= 2\!\left((p+q+r)^2-2(pq+pr+qr)\right)-4(p+q+r)+15\) | M1 | Establishes correct identity in terms of sum, pair sum and product |
| \(= 2(2^2-2(4))-4(2)+15 = -1\) | A1 | Correct value |
# Question 2:
## Part (i)
| Working/Answer | Marks | Guidance |
|---|---|---|
| $p+q+r=2,\quad pq+pr+qr=4,\quad pqr=5$ | B1 | Identifies correct values for all 3 expressions; allow $\sum p$, $\sum pq$ notation |
| $\frac{2}{p}+\frac{2}{q}+\frac{2}{r} = \frac{2(pq+pr+qr)}{pqr}$ | M1 | Uses correct identity for the sum |
| $= \frac{8}{5}$ | A1ft | Correct value (follow through on their 2, 4 and 5) |
**Alternative:**
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = \frac{2}{y} \Rightarrow \frac{8}{y^3}-\frac{8}{y^2}+\frac{8}{y}-5=0 \Rightarrow 5y^3-8y^2+8y-8=0$ | B1 | Obtains correct cubic in $y$ |
| $\frac{2}{p}+\frac{2}{q}+\frac{2}{r} = -\frac{-8}{5}$ | M1 | Uses correct method |
| $= \frac{8}{5}$ | A1ft | |
**(3 marks)**
## Part (ii)
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(p-4)(q-4)(r-4) = (pq-4p-4q+16)(r-4)$ $= pqr - 4pq-4pr-4qr+16p+16q+16r-64$ | M1, A1 | Attempt to expand — must involve sum, pair sum and product; correct expansion |
| $= pqr - 4(pq+pr+qr)+16(p+q+r)-64$ $= 5-4(4)+16(2)-64 = -43$ | A1 | Correct value |
**Alternative:**
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(x+4)^3 - 2(x+4)^2 + 4(x+4) - 5 = 0$ | M1 | Substitutes $x+4$ for $x$ in the given cubic |
| $= \ldots 64 + \ldots - 32 + \ldots 16 + \ldots - 5 = 43$ | A1 | Calculates correct constant term |
| $\therefore (p-4)(q-4)(r-4) = -43$ | A1 | Correct value |
**(3 marks)**
## Part (iii)
| Working/Answer | Marks | Guidance |
|---|---|---|
| $p^3+q^3+r^3 = (p+q+r)^3 - 3(p+q+r)(pq+pr+qr)+3pqr$ or equivalent identities | M1 | Establishes correct identity in terms of sum, pair sum and product, and substitutes to reach numerical expression for $p^3+q^3+r^3$ |
| $= 2^3 - 3(2)(4)+3(5) = -1$ | A1 | Correct value |
**Alternative:**
| Working/Answer | Marks | Guidance |
|---|---|---|
| $p^3-2p^2+4p-5=0$, etc. $\Rightarrow p^3+q^3+r^3 = 2(p^2+q^2+r^2)-4(p+q+r)+15$ $= 2\!\left((p+q+r)^2-2(pq+pr+qr)\right)-4(p+q+r)+15$ | M1 | Establishes correct identity in terms of sum, pair sum and product |
| $= 2(2^2-2(4))-4(2)+15 = -1$ | A1 | Correct value |
**(2 marks)**
**(8 marks total)**\begin{enumerate}
\item The roots of the equation
\end{enumerate}
$$x ^ { 3 } - 2 x ^ { 2 } + 4 x - 5 = 0$$
are $p , q$ and $r$.\\
Without solving the equation, find the value of\\
(i) $\frac { 2 } { p } + \frac { 2 } { q } + \frac { 2 } { r }$\\
(ii) $( p - 4 ) ( q - 4 ) ( r - 4 )$\\
(iii) $p ^ { 3 } + q ^ { 3 } + r ^ { 3 }$
\hfill \mbox{\textit{Edexcel CP2 2019 Q2 [8]}}