| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| 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 and Vieta's formulas. Students need to recognize that -3+2i must also be a root, then use sum/product of roots to find a and b. While it requires multiple steps, it's a standard textbook exercise with no novel insight required, making it slightly easier than average. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| \(-2i - 3\) | B1 | \(-2i - 3\) seen anywhere in solution |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-(2i-3))(x-(-2i-3)) = x^2 + 6x + 13\) or \(x = -3 \pm 2i \Rightarrow (x+3)^2 = -4\); \(= x^2 + 6x + 13\) | M1; A1 | Must follow from part (a). Any incorrect signs for part (a) in initial statement award M0; accept any equivalent expanded expression |
| \((x-4)(x^2 + 6x + 13) \{= x^3 + ax^2 + bx - 52\}\) | M1 | \((x - 3^{\text{rd}}\text{ root})(\text{their quadratic})\). Could be found by comparing coefficients from long division |
| \(a = 2,\ b = -11\) or \(x^3 + 2x^2 - 11x - 52\) | A1 | At least one of \(a=2\) or \(b=-11\) |
| Both \(a = 2\) and \(b = -11\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Sum \(= (2i-3) + (-2i-3) = -6\); Product \(= (2i-3) \times (-2i-3) = 13\); quadratic is \(x^2 + 6x + 13\) | M1; A1 | \(x^2 - (\text{sum roots})x + (\text{product roots}) = 0\) or \(x^2 - 2\text{Re}(\alpha)x + |
| \((x-4)(x^2+6x+13)\{= x^3+ax^2+bx-52\}\) | M1 | \((x - 3^{\text{rd}}\text{ root})(\text{their quadratic})\) |
| \(a=2,\ b=-11\) | A1, A1 | At least one correct; both correct |
| Answer | Marks | Guidance |
|---|---|---|
| \((2i-3)^3 + a(2i-3)^2 + b(2i-3) - 52 = 0\); \(5a - 3b = 43\) and \(6a - b = 23\) | M1; A1 | Substitutes \(2i-3\), equates both real and imaginary parts. Also accept \(16a + 4b = -12\) |
| Solve simultaneously | M1 | Solves to find at least one of \(a=\ldots\) or \(b=\ldots\) |
| \(a=2,\ b=-11\) | A1, A1 | At least one correct; both correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(b = 4(2i-3) + 4(-2i-3) + (2i-3)(-2i-3)\); \(a = -(\text{sum of 3 roots}) = -(4 + 2i - 3 - 2i - 3)\) | M1; A1; M1 | Attempts sum of product pairs; all pairs correct; adds up all 3 roots |
| \(a=2,\ b=-11\) | A1, A1 | At least one correct; both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \(f(4) = 0\); \(16a + 4b = -12\); \(a = -(\text{sum of 3 roots}) = -(4 + 2i - 3 - 2i - 3)\) | M1; A1; M1 | Adds up all 3 roots |
| \(a=2,\ b=-11\) | A1, A1 | At least one correct; both correct |
# Question 6:
## Part (a):
| $-2i - 3$ | B1 | $-2i - 3$ seen anywhere in solution |
## Part (b) - Way 1:
| $(x-(2i-3))(x-(-2i-3)) = x^2 + 6x + 13$ or $x = -3 \pm 2i \Rightarrow (x+3)^2 = -4$; $= x^2 + 6x + 13$ | M1; A1 | Must follow from part (a). Any incorrect signs for part (a) in initial statement award M0; accept any equivalent expanded expression |
| $(x-4)(x^2 + 6x + 13) \{= x^3 + ax^2 + bx - 52\}$ | M1 | $(x - 3^{\text{rd}}\text{ root})(\text{their quadratic})$. Could be found by comparing coefficients from long division |
| $a = 2,\ b = -11$ or $x^3 + 2x^2 - 11x - 52$ | A1 | At least one of $a=2$ or $b=-11$ |
| Both $a = 2$ and $b = -11$ | A1 | |
## Part (b) - Way 2:
| Sum $= (2i-3) + (-2i-3) = -6$; Product $= (2i-3) \times (-2i-3) = 13$; quadratic is $x^2 + 6x + 13$ | M1; A1 | $x^2 - (\text{sum roots})x + (\text{product roots}) = 0$ or $x^2 - 2\text{Re}(\alpha)x + |\alpha^2| = 0$ |
| $(x-4)(x^2+6x+13)\{= x^3+ax^2+bx-52\}$ | M1 | $(x - 3^{\text{rd}}\text{ root})(\text{their quadratic})$ |
| $a=2,\ b=-11$ | A1, A1 | At least one correct; both correct |
## Part (b) - Way 3:
| $(2i-3)^3 + a(2i-3)^2 + b(2i-3) - 52 = 0$; $5a - 3b = 43$ and $6a - b = 23$ | M1; A1 | Substitutes $2i-3$, equates both real and imaginary parts. Also accept $16a + 4b = -12$ |
| Solve simultaneously | M1 | Solves to find at least one of $a=\ldots$ or $b=\ldots$ |
| $a=2,\ b=-11$ | A1, A1 | At least one correct; both correct |
## Part (b) - Way 4:
| $b = 4(2i-3) + 4(-2i-3) + (2i-3)(-2i-3)$; $a = -(\text{sum of 3 roots}) = -(4 + 2i - 3 - 2i - 3)$ | M1; A1; M1 | Attempts sum of product pairs; all pairs correct; adds up all 3 roots |
| $a=2,\ b=-11$ | A1, A1 | At least one correct; both correct |
## Part (b) - Way 5:
| Uses $f(4) = 0$; $16a + 4b = -12$; $a = -(\text{sum of 3 roots}) = -(4 + 2i - 3 - 2i - 3)$ | M1; A1; M1 | Adds up all 3 roots |
| $a=2,\ b=-11$ | A1, A1 | At least one correct; both correct |
---6. Given that 4 and $2 \mathrm { i } - 3$ are roots of the equation
$$x ^ { 3 } + a x ^ { 2 } + b x - 52 = 0$$
where $a$ and $b$ are real constants,
\begin{enumerate}[label=(\alph*)]
\item write down the third root of the equation,
\item find the value of $a$ and the value of $b$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2017 Q6 [6]}}