| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single polynomial, two remainder/factor conditions |
| Difficulty | Moderate -0.3 This is a standard Factor and Remainder Theorem question requiring straightforward application of f(-1)=0 and f(2)=-12 to form two simultaneous equations in a and b, then factorising the resulting cubic. It's slightly easier than average because the method is routine and well-practiced, though it does require multiple steps including solving simultaneous equations and factorising a cubic. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(\pm1) = \ldots\) or \(f(\pm2) = \ldots\) | M1 | Attempts \(f(\pm1)\) or \(f(\pm2)\) |
| \(a(-1)^3 - 8(-1)^2 + b(-1) + 6 = 0\) | A1 | Allow un-simplified but do not condone missing brackets unless later work implies a correct expression |
| \(a(2)^3 - 8(2)^2 + b(2) + 6 = -12\) | A1 | Allow un-simplified |
| \(a + b = -2,\ 4a + b = 7 \Rightarrow a = 3,\ b = -5\) | M1A1 | M1: Solves two linear equations in \(a\) and \(b\) simultaneously; A1: Correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((ax^3 - 8x^2 + bx + 6) \div (x+1) \rightarrow\) remainder \(f(a,b)\) or \((ax^3 - 8x^2 + bx + 6) \div (x-2) \rightarrow\) remainder \(g(a,b)\) | M1 | Attempts long division by either expression to obtain a remainder in terms of \(a\) and \(b\) |
| \(-a - b - 2 = 0\) | A1 | Allow un-simplified but do not condone missing brackets unless later work implies a correct expression |
| \(8a + 2b - 26 = -12\) | A1 | Allow un-simplified |
| \(a + b = -2,\ 4a + b = 7 \Rightarrow a = 3,\ b = -5\) | M1A1 | M1: Solves simultaneously; A1: Correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x+1)(ax^2 + kx + \ldots)\) | M1 | Uses \((x+1)\) as a factor and obtains at least the first 2 terms of a quadratic with an \(ax^2\) term and an \(x\) term. May be by inspection or long division |
| \((x+1)(3x^2 - 11x + 6)\) | A1 | Correct quadratic factor |
| \(3x^2 - 11x + 6 = (3x-2)(x-3)\) | M1 | Attempt to factorise their 3 term quadratic; \((x+1)\) must have been used as a factor. Note: \(3x^2-11x+6=(x-\frac{2}{3})(x-3)\) scores M0; \(3x^2-11x+6=3(x-\frac{2}{3})(x-3)\) is fine for M1 |
| \((f(x) =)(x+1)(3x-2)(x-3)\) or \((f(x) =)3(x+1)(x-\frac{2}{3})(x-3)\) | A1 | Fully correct factorisation. Factors need to appear together all on one line and no commas in between |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(\pm1) = \ldots$ or $f(\pm2) = \ldots$ | M1 | Attempts $f(\pm1)$ or $f(\pm2)$ |
| $a(-1)^3 - 8(-1)^2 + b(-1) + 6 = 0$ | A1 | Allow un-simplified but do not condone missing brackets unless later work implies a correct expression |
| $a(2)^3 - 8(2)^2 + b(2) + 6 = -12$ | A1 | Allow un-simplified |
| $a + b = -2,\ 4a + b = 7 \Rightarrow a = 3,\ b = -5$ | M1A1 | M1: Solves two linear equations in $a$ and $b$ simultaneously; A1: Correct values |
**Alternative by long division:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(ax^3 - 8x^2 + bx + 6) \div (x+1) \rightarrow$ remainder $f(a,b)$ or $(ax^3 - 8x^2 + bx + 6) \div (x-2) \rightarrow$ remainder $g(a,b)$ | M1 | Attempts long division by either expression to obtain a remainder in terms of $a$ and $b$ |
| $-a - b - 2 = 0$ | A1 | Allow un-simplified but do not condone missing brackets unless later work implies a correct expression |
| $8a + 2b - 26 = -12$ | A1 | Allow un-simplified |
| $a + b = -2,\ 4a + b = 7 \Rightarrow a = 3,\ b = -5$ | M1A1 | M1: Solves simultaneously; A1: Correct values |
**(5 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x+1)(ax^2 + kx + \ldots)$ | M1 | Uses $(x+1)$ as a factor and obtains at least the first 2 terms of a quadratic with an $ax^2$ term and an $x$ term. May be by inspection or long division |
| $(x+1)(3x^2 - 11x + 6)$ | A1 | Correct quadratic factor |
| $3x^2 - 11x + 6 = (3x-2)(x-3)$ | M1 | Attempt to factorise their 3 term quadratic; $(x+1)$ must have been used as a factor. Note: $3x^2-11x+6=(x-\frac{2}{3})(x-3)$ scores M0; $3x^2-11x+6=3(x-\frac{2}{3})(x-3)$ is fine for M1 |
| $(f(x) =)(x+1)(3x-2)(x-3)$ or $(f(x) =)3(x+1)(x-\frac{2}{3})(x-3)$ | A1 | Fully correct factorisation. Factors need to appear together all on one line and no commas in between |
**(4 marks) — Total 9**
---6.
$$f ( x ) = a x ^ { 3 } - 8 x ^ { 2 } + b x + 6$$
where $a$ and $b$ are constants.
When $\mathrm { f } ( x )$ is divided by ( $x + 1$ ) there is no remainder.
When $\mathrm { f } ( x )$ is divided by $( x - 2 )$ the remainder is - 12
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and the value of $b$.
\item Factorise $\mathrm { f } ( x )$ completely.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q6 [9]}}