| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Verify factor then factorise/solve |
| Difficulty | Moderate -0.8 This is a straightforward application of remainder and factor theorems with standard polynomial division. Part (a) requires substituting x=-3/2, part (b) verifying f(-2)=0, and part (c) performing routine polynomial division followed by factoring a quadratic. All steps are textbook procedures with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure and arithmetic involved. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = 6x^3 + 13x^2 - 4\) | ||
| (a) \(f\left(-\frac{3}{2}\right) = 6\left(-\frac{3}{2}\right)^3 + 13\left(-\frac{3}{2}\right)^2 - 4 = 5\) | M1 | Attempting \(f\left(-\frac{3}{2}\right)\) or \(f\left(\frac{3}{2}\right)\) |
| A1 cao | 5 | |
| [2] | ||
| (b) \(f(-2) = 6(-2)^3 + 13(-2)^2 - 4 = 0\), and so \((x+2)\) is a factor | M1 | Attempts \(f(-2)\) |
| A1 | Must correctly show \(f(-2) = 0\) and give conclusion in part (b) only. No simplification of terms required here | |
| [2] | ||
| (c) \(f(x) = \{(x+2)\}(6x^2 + x - 2)\) ... \((x + 2)(2x - 1)(3x + 2)\) | M1 A1 | |
| [2] | ||
| [8] |
| Answer | Marks |
|---|---|
| Note | Details |
| Note | Long division scores no marks in part (a). The remainder theorem is required. |
| M1 | Attempting \(f\left(-\frac{3}{2}\right)\) or \(f\left(\frac{3}{2}\right)\). \(6\left(-\frac{3}{2}\right)^3 + 13\left(-\frac{3}{2}\right)^2 - 4\) or \(6\left(\frac{3}{2}\right)^3 + 13\left(\frac{3}{2}\right)^2 - 4\) is sufficient |
| A1 | 5 cao |
| M1 | Attempting \(f(-2)\). (This is not given for f(2)) |
| A1 | Must correctly show \(f(-2) = 0\) and give a conclusion in part (b) only. No simplification of terms is required here. Stating "hence factor" or "it is a factor" or a "tick" "QED" are possible conclusions. Also a conclusion can be implied from a preamble, e.g. "If \(f(-2)=0\), \((x+2)\) is a factor...." |
| Long division scores no marks in part (b). The factor theorem is required. | |
| 1st M1 | Attempting to divide by \((x+2)\) leading to a quotient which is quadratic with at least two terms beginning with first term of \(\pm 6x^2\) + linear or constant term. Or \(f(x) = (x+2)(6x^2 + \text{linear and/or constant term})\) (This may be seen in part (b) where candidates did not use factor theorem and might be referred to here) |
| 1st A1 | \((6x^2 + x - 2)\) seen as quotient or as factor. If there is an error in the division resulting in a remainder give A0, but allow recovery to gain next two marks if \((6x^2 + x - 2)\) is used |
| 2nd M1 | For a valid attempt to factorise their three term quadratic. \((x+2)(2x-1)(3x+2)\) and needs all three factors on the same line. Ignore subsequent work (such as a solution to a quadratic equation). |
| A1 | |
| Special cases | Calculator methods: Award M1A1M1A1 for correct answer \((x+2)(2x-1)(3x+2)\) with no working. Award M1A0M1A0 for either \((x+2)(2x+1)(3x+2)\) or \((x+2)(2x-1)(3x-2)\) or \((x+2)(2x-1)(3x+2)\) with no working. (At least one bracket incorrect) Award M1A1M1A1 for \(x = -2, -\frac{1}{2}, -\frac{2}{3}\) followed by \((x+2)(2x-1)(3x+2)\). Award SC: M1A0M1A0 for \(x = -2, \frac{1}{2}, -\frac{2}{3}\) followed by \((x+2)(2x-\frac{1}{2})(x + \frac{2}{3})\). |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = 6x^3 + 13x^2 - 4$ | | |
| **(a)** $f\left(-\frac{3}{2}\right) = 6\left(-\frac{3}{2}\right)^3 + 13\left(-\frac{3}{2}\right)^2 - 4 = 5$ | M1 | Attempting $f\left(-\frac{3}{2}\right)$ or $f\left(\frac{3}{2}\right)$ |
| | A1 cao | 5 |
| | [2] | |
| **(b)** $f(-2) = 6(-2)^3 + 13(-2)^2 - 4 = 0$, and so $(x+2)$ is a factor | M1 | Attempts $f(-2)$ |
| | A1 | Must correctly show $f(-2) = 0$ and give conclusion in part (b) only. No simplification of terms required here |
| | [2] | |
| **(c)** $f(x) = \{(x+2)\}(6x^2 + x - 2)$ ... $(x + 2)(2x - 1)(3x + 2)$ | M1 A1 | |
| | [2] | |
| | [8] | |
**Question 4 Notes:**
| Note | Details |
|---|---|
| Note | Long division scores no marks in part (a). The remainder theorem is required. |
| M1 | Attempting $f\left(-\frac{3}{2}\right)$ or $f\left(\frac{3}{2}\right)$. $6\left(-\frac{3}{2}\right)^3 + 13\left(-\frac{3}{2}\right)^2 - 4$ or $6\left(\frac{3}{2}\right)^3 + 13\left(\frac{3}{2}\right)^2 - 4$ is sufficient |
| A1 | 5 cao |
| M1 | Attempting $f(-2)$. (This is not given for f(2)) |
| A1 | Must correctly show $f(-2) = 0$ and give a conclusion in part (b) only. No simplification of terms is required here. Stating "hence factor" or "it is a factor" or a "tick" "QED" are possible conclusions. Also a conclusion can be implied from a preamble, e.g. "If $f(-2)=0$, $(x+2)$ is a factor...." |
| Long division scores no marks in part (b). The factor theorem is required. | |
| 1st M1 | Attempting to divide by $(x+2)$ leading to a quotient which is quadratic with at least two terms beginning with first term of $\pm 6x^2$ + linear or constant term. Or $f(x) = (x+2)(6x^2 + \text{linear and/or constant term})$ (This may be seen in part (b) where candidates did not use factor theorem and might be referred to here) |
| 1st A1 | $(6x^2 + x - 2)$ seen as quotient or as factor. If there is an error in the division resulting in a remainder give A0, but allow recovery to gain next two marks if $(6x^2 + x - 2)$ is used |
| 2nd M1 | For a valid attempt to factorise their three term quadratic. $(x+2)(2x-1)(3x+2)$ and needs all three factors on the same line. Ignore subsequent work (such as a solution to a quadratic equation). |
| A1 | |
| Special cases | Calculator methods: Award M1A1M1A1 for correct answer $(x+2)(2x-1)(3x+2)$ with no working. Award M1A0M1A0 for either $(x+2)(2x+1)(3x+2)$ or $(x+2)(2x-1)(3x-2)$ or $(x+2)(2x-1)(3x+2)$ with no working. (At least one bracket incorrect) Award M1A1M1A1 for $x = -2, -\frac{1}{2}, -\frac{2}{3}$ followed by $(x+2)(2x-1)(3x+2)$. Award SC: M1A0M1A0 for $x = -2, \frac{1}{2}, -\frac{2}{3}$ followed by $(x+2)(2x-\frac{1}{2})(x + \frac{2}{3})$. |
---4.
$$f ( x ) = 6 x ^ { 3 } + 13 x ^ { 2 } - 4$$
\begin{enumerate}[label=(\alph*)]
\item Use the remainder theorem to find the remainder when $\mathrm { f } ( x )$ is divided by ( $2 x + 3$ ).
\item Use the factor theorem to show that $( x + 2 )$ is a factor of $\mathrm { f } ( x )$.
\item Factorise $\mathrm { f } ( x )$ completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2016 Q4 [8]}}