| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.3 This is a standard P2 factor/remainder theorem question requiring systematic application of well-known results. Part (a) uses the factor theorem at x=-1/2, part (b) uses the remainder theorem at x=2, part (c) solves simultaneous equations and factorises. All steps are routine with no novel insight required, making it slightly easier than average. |
| 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 |
| Sets \(f\!\left(-\frac{1}{2}\right)=0\): \(4\times\left(-\frac{1}{2}\right)^3+a\left(-\frac{1}{2}\right)^2-29\left(-\frac{1}{2}\right)+b=0\) | M1 | Condone sign slips; \(f\!\left(\frac{1}{2}\right)=0\) is M0 |
| \(\Rightarrow \frac{1}{4}a+b+14=0 \Rightarrow a+4b=-56\) | A1* | At least one intermediate simplified line required e.g. \(-\frac{1}{2}+\frac{1}{4}a+\frac{29}{2}+b=0\); incorrect lines after start of proof score A0* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets \(f(2)=-25\): \(4\times2^3+a\times2^2-29\times2+b=-25\) | M1 | Condone sign slips; \(f(-2)=-25\) is M0 |
| \(4a+b=1\) | A1 | Or equivalent e.g. \(a+\frac{1}{4}b-\frac{1}{4}=0\); constant terms must be collected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Solves \(a+4b=-56\) simultaneously with \(4a+b=1\) | M1 | Allow slips; award when values appear for both \(a\) and \(b\) |
| \(a=4,\ b=-15\) | A1 | Both correct; may be written down from calculator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4x^3+4x^2-29x-15=(2x+1)(2x^2+x-15)\) | M1, A1 | |
| \(=(2x+1)(2x-5)(x+3)\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to divide or factorise out \((2x+1)\) from \(4x^3+4x^2-29x-15\) | M1 | For factorisation: correct first and last terms e.g. \(4x^3 + "a"x^2 - 29x + "b" = (2x+1)(2x^2+kx\pm"b")\); For division: quadratic quotient of form \(2x^2+px+q\) with correct work to find term in \(x\) |
| Correct quadratic factor \(2x^2+x-15\) | A1 | |
| \((2x+1)(2x-5)(x+3)\) | A1 | cso - M1,A1 must have been awarded and full product must be seen |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets $f\!\left(-\frac{1}{2}\right)=0$: $4\times\left(-\frac{1}{2}\right)^3+a\left(-\frac{1}{2}\right)^2-29\left(-\frac{1}{2}\right)+b=0$ | M1 | Condone sign slips; $f\!\left(\frac{1}{2}\right)=0$ is M0 |
| $\Rightarrow \frac{1}{4}a+b+14=0 \Rightarrow a+4b=-56$ | A1* | At least one intermediate simplified line required e.g. $-\frac{1}{2}+\frac{1}{4}a+\frac{29}{2}+b=0$; incorrect lines after start of proof score A0* |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets $f(2)=-25$: $4\times2^3+a\times2^2-29\times2+b=-25$ | M1 | Condone sign slips; $f(-2)=-25$ is M0 |
| $4a+b=1$ | A1 | Or equivalent e.g. $a+\frac{1}{4}b-\frac{1}{4}=0$; constant terms must be collected |
## Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solves $a+4b=-56$ simultaneously with $4a+b=1$ | M1 | Allow slips; award when values appear for both $a$ and $b$ |
| $a=4,\ b=-15$ | A1 | Both correct; may be written down from calculator |
## Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4x^3+4x^2-29x-15=(2x+1)(2x^2+x-15)$ | M1, A1 | |
| $=(2x+1)(2x-5)(x+3)$ | A1 | cso |
# Question 4(c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to divide or factorise out $(2x+1)$ from $4x^3+4x^2-29x-15$ | M1 | For factorisation: correct first and last terms e.g. $4x^3 + "a"x^2 - 29x + "b" = (2x+1)(2x^2+kx\pm"b")$; For division: quadratic quotient of form $2x^2+px+q$ with correct work to find term in $x$ |
| Correct quadratic factor $2x^2+x-15$ | A1 | |
| $(2x+1)(2x-5)(x+3)$ | A1 | cso - M1,A1 must have been awarded and full product must be seen |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying on calculator technology are not acceptable.
$$f ( x ) = 4 x ^ { 3 } + a x ^ { 2 } - 29 x + b$$
where $a$ and $b$ are constants.\\
Given that $( 2 x + 1 )$ is a factor of $\mathrm { f } ( x )$,\\
(a) show that
$$a + 4 b = - 56$$
Given also that when $\mathrm { f } ( x )$ is divided by $( x - 2 )$ the remainder is - 25\\
(b) find a second simplified equation linking $a$ and $b$.\\
(c) Hence, using algebra and showing your working,\\
(i) find the value of $a$ and the value of $b$,\\
(ii) fully factorise $\mathrm { f } ( x )$.
\hfill \mbox{\textit{Edexcel P2 2023 Q4 [9]}}