| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial identity or expansion |
| Difficulty | Moderate -0.5 This is a straightforward application of the Remainder Theorem requiring students to recognize that the remainder when dividing by (x+k) is the constant term (part a), substitute x=-1 to find k (part b), and then factorize a quadratic (part c). All steps are routine with clear signposting, making it slightly easier than average for A-level. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 32 | B1 | If algebraic division used, must clearly state remainder is 32. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(-1) = (-1+k)(3(-1)^2 + 4(-1) - 16) + 32\) | M1 | Attempts \(f(\pm 1)\). Sufficient to see 1 or \(-1\) substituted. May expand \(f(x)\) first (condone algebraic errors before substitution). Alt: algebraically divide by \((x+1)\) to get quotient \(3x^2 + \ldots\) |
| \((-1+k)(3(-1)^2 + 4(-1) - 16) + 32 = 15 \Rightarrow k = \ldots\) | dM1 | Sets \(f(\pm 1) = 15\) and solves for \(k\). Condone rearrangement/arithmetic slips. Alt: set remainder \(= 15\) and find \(k\). |
| \(k = 2^*\) | A1* | No errors in working including brackets. Expect to see \(\pm 17k = \ldots\) as penultimate step. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x+2)(3x^2 + 4x - 16) = \ldots\) | M1 | Attempts to multiply out \((x+2)(3x^2+4x-16)\) to reach \(Ax^3 + Bx^2 + Cx\ (+D)\) with \(A,B,C \neq 0\), no terms in \(k\). Can be awarded if \(k=2\) substituted in part (b) expansion. |
| \(f(x) = 3x^3 + 10x^2 - 8x\) | A1 | |
| \(f(x) = x(3x^2 + 10x - 8) = x(\ldots)(\ldots)\) | dM1 | Attempts to factorise cubic of form \(Ax^3 + Bx^2 + Cx\) to LINEAR × QUADRATIC. Dependent on previous M. |
| \(\{f(x) =\}\ x(3x-2)(x+4)\) | A1 | Must be on one line. ISW after this. |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 32 | B1 | If algebraic division used, must clearly state remainder is 32. |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(-1) = (-1+k)(3(-1)^2 + 4(-1) - 16) + 32$ | M1 | Attempts $f(\pm 1)$. Sufficient to see 1 or $-1$ substituted. May expand $f(x)$ first (condone algebraic errors before substitution). Alt: algebraically divide by $(x+1)$ to get quotient $3x^2 + \ldots$ |
| $(-1+k)(3(-1)^2 + 4(-1) - 16) + 32 = 15 \Rightarrow k = \ldots$ | dM1 | Sets $f(\pm 1) = 15$ and solves for $k$. Condone rearrangement/arithmetic slips. Alt: set remainder $= 15$ and find $k$. |
| $k = 2^*$ | A1* | No errors in working including brackets. Expect to see $\pm 17k = \ldots$ as penultimate step. |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x+2)(3x^2 + 4x - 16) = \ldots$ | M1 | Attempts to multiply out $(x+2)(3x^2+4x-16)$ to reach $Ax^3 + Bx^2 + Cx\ (+D)$ with $A,B,C \neq 0$, no terms in $k$. Can be awarded if $k=2$ substituted in part (b) expansion. |
| $f(x) = 3x^3 + 10x^2 - 8x$ | A1 | |
| $f(x) = x(3x^2 + 10x - 8) = x(\ldots)(\ldots)$ | dM1 | Attempts to factorise cubic of form $Ax^3 + Bx^2 + Cx$ to LINEAR × QUADRATIC. Dependent on previous M. |
| $\{f(x) =\}\ x(3x-2)(x+4)$ | A1 | Must be on one line. ISW after this. |9.
$\mathrm { f } ( x ) = ( x + k ) \left( 3 x ^ { 2 } + 4 x - 16 \right) + 32 , \quad$ where $k$ is a constant (a) Write down the remainder when $\mathrm { f } ( x )$ is divided by $( x + k )$.
When $\mathrm { f } ( x )$ is divided by $( x + 1 )$, the remainder is 15\\
(b) Show that $k = 2$\\
(c) Hence factorise $\mathrm { f } ( x )$ completely.
\section*{9.}
" .\\
$\mathrm { f } ( x ) = ( x + k ) \left( 3 x ^ { 2 } + 4 x - 16 \right) + 32 , \quad$ where $k$ is a constant
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\hfill \mbox{\textit{Edexcel C12 2019 Q9 [8]}}