Edexcel C3 2013 June — Question 1 4 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2013
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolynomial Division & Manipulation
TypePartial Fraction Form via Division
DifficultyModerate -0.5 This is a straightforward polynomial long division question requiring a standard algorithmic procedure to divide the quartic by the quadratic, then identify coefficients. It's slightly easier than average because it's purely mechanical with no problem-solving or conceptual insight needed—just careful arithmetic following a well-practiced technique.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02y Partial fractions: decompose rational functions

  1. Given that
$$\frac { 3 x ^ { 4 } - 2 x ^ { 3 } - 5 x ^ { 2 } - 4 } { x ^ { 2 } - 4 } \equiv a x ^ { 2 } + b x + c + \frac { d x + e } { x ^ { 2 } - 4 } , \quad x \neq \pm 2$$ find the values of the constants \(a , b , c , d\) and \(e\).
(4)

Question 1:
By Division method:
\(\frac{3x^2 - 2x + 7}{x^2(+0x)-4\overline{)3x^4 - 2x^3 - 5x^2 + (0x) - 4}}\)
Working shown:
- \(3x^4 + 0x^3 - 12x^2\)
- \(-2x^3 + 7x^2 + 0x\)
- \(-2x^3 + 0x^2 + 8x\)
- \(7x^2 - 8x - 4\)
- \(7x^2 + 0x - 28\)
- \(-8x + 24\)
AnswerMarks Guidance
\(a = 3\)B1 Stating \(a=3\); can also be scored by the coefficient of \(x^2\) in \(3x^2 - 2x + 7\)
Long division reaching as far as \(3x^4 + 0x^3 - 12x^2\) then \(-2x^3 + \ldots\)M1 Using long division by \(x^2 - 4\) getting as far as the \(x\) term; coefficients need not be correct. Award if whole number part seen as \(\ldots x^2 + \ldots x\) following working. Long division by \((x+2)\) scores nothing until \((x-2)\) has been divided into new quotient.
Two of \(b=-2\), \(c=7\), \(d=-8\), \(e=24\)A1 Answers may be embedded within division sum and can be implied
All four of \(b=-2\), \(c=7\), \(d=-8\), \(e=24\)A1 Need to see \(a=\ldots\), \(b=\ldots\) or values embedded in rhs for all 4 marks
*Note: Accept a correct long division for 3 out of 4 marks scoring B1M1A1A0*
Alt 1 — By Multiplication:
\(3x^4 - 2x^3 - 5x^2 - 4 \equiv (ax^2 + bx + c)(x^2 - 4) + dx + e\)
AnswerMarks Guidance
Compares \(x^4\) terms: \(a = 3\)B1 Can also be scored for writing \(3x^4 = ax^4\)
Compares coefficients to obtain numerical value of one further constant: \(-2 = b\), \(-5 = -4a + c \Rightarrow c = \ldots\)M1 Multiply out expression to get \(*\); condone sign slips only. Compare coefficient of any term (other than \(x^4\)) to obtain numerical value of one further constant — in practice a valid attempt at \(b\) or \(c\). Method may be implied by a correct additional constant to \(a\).
Two of \(b=-2\), \(c=7\), \(d=-8\), \(e=24\)A1
All four of \(b=-2\), \(c=7\), \(d=-8\), \(e=24\)A1
## Question 1:

**By Division method:**

$\frac{3x^2 - 2x + 7}{x^2(+0x)-4\overline{)3x^4 - 2x^3 - 5x^2 + (0x) - 4}}$

Working shown:
- $3x^4 + 0x^3 - 12x^2$
- $-2x^3 + 7x^2 + 0x$
- $-2x^3 + 0x^2 + 8x$
- $7x^2 - 8x - 4$
- $7x^2 + 0x - 28$
- $-8x + 24$

| $a = 3$ | B1 | Stating $a=3$; can also be scored by the coefficient of $x^2$ in $3x^2 - 2x + 7$ |

| Long division reaching as far as $3x^4 + 0x^3 - 12x^2$ then $-2x^3 + \ldots$ | M1 | Using long division by $x^2 - 4$ getting as far as the $x$ term; coefficients need not be correct. Award if whole number part seen as $\ldots x^2 + \ldots x$ following working. Long division by $(x+2)$ scores nothing until $(x-2)$ has been divided into new quotient. |

| Two of $b=-2$, $c=7$, $d=-8$, $e=24$ | A1 | Answers may be embedded within division sum and can be implied |

| All four of $b=-2$, $c=7$, $d=-8$, $e=24$ | A1 | Need to see $a=\ldots$, $b=\ldots$ or values embedded in rhs for all 4 marks |

*Note: Accept a correct long division for 3 out of 4 marks scoring B1M1A1A0*

---

**Alt 1 — By Multiplication:**

$3x^4 - 2x^3 - 5x^2 - 4 \equiv (ax^2 + bx + c)(x^2 - 4) + dx + e$

| Compares $x^4$ terms: $a = 3$ | B1 | Can also be scored for writing $3x^4 = ax^4$ |

| Compares coefficients to obtain numerical value of one further constant: $-2 = b$, $-5 = -4a + c \Rightarrow c = \ldots$ | M1 | Multiply out expression to get $*$; condone sign slips only. Compare coefficient of any term (other than $x^4$) to obtain numerical value of one further constant — in practice a valid attempt at $b$ or $c$. Method may be implied by a correct additional constant to $a$. |

| Two of $b=-2$, $c=7$, $d=-8$, $e=24$ | A1 | |

| All four of $b=-2$, $c=7$, $d=-8$, $e=24$ | A1 | |

---
\begin{enumerate}
  \item Given that
\end{enumerate}

$$\frac { 3 x ^ { 4 } - 2 x ^ { 3 } - 5 x ^ { 2 } - 4 } { x ^ { 2 } - 4 } \equiv a x ^ { 2 } + b x + c + \frac { d x + e } { x ^ { 2 } - 4 } , \quad x \neq \pm 2$$

find the values of the constants $a , b , c , d$ and $e$.\\
(4)\\

\hfill \mbox{\textit{Edexcel C3 2013 Q1 [4]}}