| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Given factor, find all roots |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring verification of a given root and polynomial division to find remaining roots. The question explicitly tells students that x=4 is a root, eliminating any problem-solving element. Part (a) is trivial substitution, and part (b) is routine algebraic division followed by solving a quadratic—standard textbook exercise with no novel insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(4) = 2(4)^3 - 6(4)^2 - 7(4) - 4 = 128 - 96 - 28 - 4 = 0\) | B1 | Must show \(128 - 96 - 28 - 4 = 0\); there must be sufficient working to show that \(f(4) = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = (x-4)(2x^2 + 2x + 1)\) | M1A1 | M1: form \((2x^2 + kx + 1)\); uses inspection or long division comparing coefficients and \((x-4)\) (not \((x+4)\)) to obtain quadratic of this form. A1: \((2x^2 + 2x + 1)\) cao |
| \(x = \dfrac{-2 \pm \sqrt{4 - 4(2)(1)}}{2(2)}\) | M1 | Use of correct quadratic formula for their 3TQ, or completes the square: \((2)\!\left(x^2 + x + \frac{1}{2}\right)=0 \Rightarrow (2)\!\left(\left(x+\frac{1}{2}\right)^2 \pm k \pm \frac{1}{2}\right),\ k\neq 0\) |
| \(\Rightarrow x = \dfrac{-2 \pm \sqrt{-4}}{2(2)}\) | Allow attempt at factorisation provided usual conditions satisfied and proceeds as far as \(x = \ldots\) | |
| \(\Rightarrow x = 4,\ \dfrac{-2 \pm 2i}{4}\) | A1 | All three roots stated somewhere in (b). Complex roots must be at least as given; apply isw if necessary |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(4) = 2(4)^3 - 6(4)^2 - 7(4) - 4 = 128 - 96 - 28 - 4 = 0$ | B1 | Must show $128 - 96 - 28 - 4 = 0$; there must be sufficient working to **show** that $f(4) = 0$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = (x-4)(2x^2 + 2x + 1)$ | M1A1 | M1: form $(2x^2 + kx + 1)$; uses inspection or long division comparing coefficients **and** $(x-4)$ (not $(x+4)$) to obtain quadratic of this form. A1: $(2x^2 + 2x + 1)$ cao |
| $x = \dfrac{-2 \pm \sqrt{4 - 4(2)(1)}}{2(2)}$ | M1 | Use of correct quadratic formula for their 3TQ, or completes the square: $(2)\!\left(x^2 + x + \frac{1}{2}\right)=0 \Rightarrow (2)\!\left(\left(x+\frac{1}{2}\right)^2 \pm k \pm \frac{1}{2}\right),\ k\neq 0$ |
| $\Rightarrow x = \dfrac{-2 \pm \sqrt{-4}}{2(2)}$ | | Allow attempt at factorisation provided usual conditions satisfied and proceeds as far as $x = \ldots$ |
| $\Rightarrow x = 4,\ \dfrac{-2 \pm 2i}{4}$ | A1 | All **three** roots stated somewhere in (b). Complex roots must be at least as given; apply isw if necessary |
**Total: 5 marks**1.
$$f ( x ) = 2 x ^ { 3 } - 6 x ^ { 2 } - 7 x - 4$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( 4 ) = 0$
\item Use algebra to solve $\mathrm { f } ( x ) = 0$ completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2012 Q1 [5]}}