| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Two unknowns with show-that step |
| Difficulty | Moderate -0.3 This is a standard multi-part Factor/Remainder Theorem question requiring systematic application of well-known techniques: substituting x=2 and x=-1, solving simultaneous equations, polynomial division, and checking discriminant. While it has multiple parts (7 marks total), each step follows directly from the previous with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Sets \(f(\pm2) = 0\) | M1 | The \(=0\) may be implied by later working. |
| \(f(2)=0 \Rightarrow 24+4a+2b-10=0 \Rightarrow 4a+2b=-14 \Rightarrow 2a+b=-7\) | A1* | Must have at least one intermediate line. This is a given result. |
| Answer | Marks | Guidance |
|---|---|---|
| Sets \(f(\pm1) = -36\) | M1 | If division attempted, look for minimum \(\frac{3x^2+(a-3)x\ldots}{x+1\, |
| \(f(-1)=-36 \Rightarrow -3+a-b-10=-36 \Rightarrow a-b=-23\) | A1 | Equation does not need to be simplified but indices must be correct. |
| Solves simultaneously to get both \(a\) and \(b\) | dM1 | Dependent on previous M. |
| \(a=-10,\quad b=13\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Divides \(f(x)\) by \((x-2)\) to get a quadratic | M1 | If by division, look for first two terms \(\frac{3x^2+(a+6)x\ldots}{x-2\, |
| \(f(x)=(x-2)(3x^2-4x+5)\) or \(Q(x)=3x^2-4x+5\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Calculates \(b^2-4ac\) on their \(Q(x)\) or solves \(Q(x)=0\) | M1 | Scored for attempt at finding number of roots of their \(Q(x)\). |
| \((3x^2-4x+5)\) has no roots as \(b^2-4ac = 16-60 < 0\). Hence \(f(x)\) has 1 root. | A1* | cso. All aspects correct including \(f(x)=(x-2)(3x^2-4x+5)\) with proof that \(3x^2-4x+5\) has no roots and hence \(f(x)\) has only 1 root at \(x=2\). |
## Question 7(a):
Sets $f(\pm2) = 0$ | M1 | The $=0$ may be implied by later working.
$f(2)=0 \Rightarrow 24+4a+2b-10=0 \Rightarrow 4a+2b=-14 \Rightarrow 2a+b=-7$ | A1* | Must have at least one intermediate line. This is a given result.
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## Question 7(b):
Sets $f(\pm1) = -36$ | M1 | If division attempted, look for minimum $\frac{3x^2+(a-3)x\ldots}{x+1\,|\,3x^3+ax^2+bx-10}$ before setting remainder equal to $-36$.
$f(-1)=-36 \Rightarrow -3+a-b-10=-36 \Rightarrow a-b=-23$ | A1 | Equation does not need to be simplified but indices must be correct.
Solves simultaneously to get both $a$ and $b$ | dM1 | Dependent on previous M.
$a=-10,\quad b=13$ | A1 | —
---
## Question 7(c)(i):
Divides $f(x)$ by $(x-2)$ to get a quadratic | M1 | If by division, look for first two terms $\frac{3x^2+(a+6)x\ldots}{x-2\,|\,3x^3+ax^2+bx-10}$ following through on their $a$.
$f(x)=(x-2)(3x^2-4x+5)$ or $Q(x)=3x^2-4x+5$ | A1 | —
---
## Question 7(c)(ii):
Calculates $b^2-4ac$ on their $Q(x)$ or solves $Q(x)=0$ | M1 | Scored for attempt at finding number of roots of their $Q(x)$.
$(3x^2-4x+5)$ has no roots as $b^2-4ac = 16-60 < 0$. Hence $f(x)$ has 1 root. | A1* | cso. All aspects correct including $f(x)=(x-2)(3x^2-4x+5)$ **with proof** that $3x^2-4x+5$ has no roots and hence $f(x)$ has only 1 root at $x=2$.7.
$$f ( x ) = 3 x ^ { 3 } + a x ^ { 2 } + b x - 10 \text {, where } a \text { and } b \text { are constants. }$$
Given that $( x - 2 )$ is a factor of $\mathrm { f } ( x )$,
\begin{enumerate}[label=(\alph*)]
\item use the factor theorem to show that $2 a + b = - 7$
Given also that when $\mathrm { f } ( x )$ is divided by $( x + 1 )$ the remainder is - 36
\item find the value of $a$ and the value of $b$.\\
$\mathrm { f } ( x )$ can be written in the form
$$\mathrm { f } ( x ) = ( x - 2 ) \mathrm { Q } ( x ) \text {, where } \mathrm { Q } ( x ) \text { is a quadratic function. }$$
\item \begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { Q } ( x )$.
\item Prove that the equation $\mathrm { f } ( x ) = 0$ has only one real root.
You must justify your answer and show all your working.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q7 [10]}}