| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| 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/Remainder Theorem question requiring students to set up two simultaneous equations from the given conditions (f(-1)=0 and f(1/2)=-15), solve for a and b, then factorize. It's slightly easier than average because it's a routine textbook exercise with clear steps and no conceptual surprises, though it does require careful algebraic manipulation. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(f(\pm1)\) or \(f(\pm\frac{1}{2})\) or uses long division | M1 | One attempt sufficient; do not need to equate to 0 and \(-15\) |
| \(6(-1)^3+a(-1)^2+b(-1)-5=0\) or \(-6+a-b-5=0\) or \(a-b=11\) | A1 | Any equivalent form; mark earned even if "=0" not explicitly seen |
| \(6(\frac{1}{2})^3+a(\frac{1}{2})^2+b(\frac{1}{2})-5=-15\) or \(\frac{6}{8}+\frac{a}{4}+\frac{b}{2}-5=-15\) or \(a+2b=-43\) | A1 | Must be accurate but may be unsimplified; NB using 15 instead of \(-15\) is A0 |
| Solve simultaneous equations to obtain \(a=-7\) and \(b=-18\) | M1 A1 | Both \(a\) and \(b\) correct; correct answers without working can earn M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6x^3+ax^2+bx-5=(x+1)(6x^2+\ldots x+\ldots)\) | M1 | Recognises \((x+1)\) is factor; use of \((x-1)\) is M0 |
| \(6x^3-7x^2-18x-5=(x+1)(6x^2-13x-5)\) | A1 | Correct quadratic \((6x^2-13x-5)\) |
| \((6x^2-13x-5)=(ax+b)(cx+d)\) where \(ac=\text{"6"}\) and \(bd=\text{"}\pm5\text{"}\) | M1 | Attempt to factorise quadratic |
| \(=(x+1)(2x-5)(3x+1)\) | A1 | Any correct equivalent form e.g. \(2(x+1)(x-\frac{5}{2})(3x+1)\) |
## Question 10:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $f(\pm1)$ or $f(\pm\frac{1}{2})$ or uses long division | M1 | One attempt sufficient; do not need to equate to 0 and $-15$ |
| $6(-1)^3+a(-1)^2+b(-1)-5=0$ or $-6+a-b-5=0$ or $a-b=11$ | A1 | Any equivalent form; mark earned even if "=0" not explicitly seen |
| $6(\frac{1}{2})^3+a(\frac{1}{2})^2+b(\frac{1}{2})-5=-15$ or $\frac{6}{8}+\frac{a}{4}+\frac{b}{2}-5=-15$ or $a+2b=-43$ | A1 | Must be accurate but may be unsimplified; NB using 15 instead of $-15$ is A0 |
| Solve simultaneous equations to obtain $a=-7$ and $b=-18$ | M1 A1 | Both $a$ and $b$ correct; correct answers without working can earn M1A1 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6x^3+ax^2+bx-5=(x+1)(6x^2+\ldots x+\ldots)$ | M1 | Recognises $(x+1)$ is factor; use of $(x-1)$ is M0 |
| $6x^3-7x^2-18x-5=(x+1)(6x^2-13x-5)$ | A1 | Correct quadratic $(6x^2-13x-5)$ |
| $(6x^2-13x-5)=(ax+b)(cx+d)$ where $ac=\text{"6"}$ and $bd=\text{"}\pm5\text{"}$ | M1 | Attempt to factorise quadratic |
| $=(x+1)(2x-5)(3x+1)$ | A1 | Any correct equivalent form e.g. $2(x+1)(x-\frac{5}{2})(3x+1)$ |
---10.
$$f ( x ) = 6 x ^ { 3 } + a x ^ { 2 } + b x - 5$$
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 ( $2 x - 1$ ) the remainder is - 15
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and the value of $b$.
\item Factorise $\mathrm { f } ( x )$ completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2015 Q10 [9]}}