C4 January 2013 Q3

EdexcelOld spec4 marksAlgebraic Fractions

3. Express \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)}\) in partial fractions. (4)

Mark scheme

Scheme	Marks
Method 1: Using one identity \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv A + \dfrac{B}{(x + 2)} + \dfrac{C}{(3x - 1)}\) \(A = 3\) their constant term \(= 3\)	B1
\(9x^2 + 20x - 10 \equiv A(x + 2)(3x - 1) + B(3x - 1) + C(x + 2)\) Forming a correct identity.	B1
Either \(\ x^2:\ 9 = 3A,\quad x:\ 20 = 5A + 3B + C\) \(\qquad\ \ \text{constant}:\ -10 = -2A - B + 2C\) or \(x = -2 \Rightarrow 36 - 40 - 10 = -7B \Rightarrow -14 = -7B \Rightarrow B = 2\) Attempts to find the value of either one of their \(B\) or their \(C\) from their identity.	M1
\(x = \dfrac{1}{3} \Rightarrow 1 + \dfrac{20}{3} - 10 = \dfrac{7}{3}C \Rightarrow -\dfrac{7}{3} = \dfrac{7}{3}C \Rightarrow C = -1\) Correct values for their \(B\) and their \(C\), which are found using a correct identity.	A1
	(4)
	(4 marks)

Notes

1^st B1: Their constant term must be equal to 3 for this mark.
2^nd B1 (M1 on epen): Forming a correct identity. This can be implied by later working.
M1 (A1 on epen): Attempts to find the value of either one of their \(B\) or their \(C\) from their identity. This can be achieved by either substituting values into their identity or comparing coefficients and solving the resulting equations simultaneously.
A1: Correct values for their \(B\) and their \(C\), which are found using a correct identity.

Note : \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv \dfrac{A}{(x + 2)} + \dfrac{B}{(3x - 1)}\), leading to \(9x^2 + 20x - 10 \equiv A(3x - 1) + B(x + 2)\), leading to \(A = 2\) and \(B = -1\) will gain a maximum of B0B0M1A0

Note: You can imply the 2^nd B1 from either \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv \dfrac{A(x + 2)(3x - 1) + B(3x - 1) + C(x + 2)}{(x + 2)(3x - 1)}\)
or \(\dfrac{5x - 4}{(x + 2)(3x - 1)} \equiv \dfrac{B(3x - 1) + C(x + 2)}{(x + 2)(3x - 1)}\)

Method 2: Long Division

Scheme	Marks
\(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv 3 + \dfrac{5x - 4}{(x + 2)(3x - 1)}\) their constant term \(= 3\)	B1
So, \(\dfrac{5x - 4}{(x + 2)(3x - 1)} \equiv \dfrac{B}{(x + 2)} + \dfrac{C}{(3x - 1)}\) \(5x - 4 \equiv B(3x - 1) + C(x + 2)\) Forming a correct identity.	B1
Either \(\ x:\ 5 = 3B + C,\ \text{constant}:\ -4 = -B + 2C\) or \(x = -2 \Rightarrow -10 - 4 = -7B \Rightarrow -14 = -7B \Rightarrow B = 2\) Attempts to find the value of either one of their \(B\) or their \(C\) from their identity.	M1
\(x = \dfrac{1}{3} \Rightarrow \dfrac{5}{3} - 4 = \dfrac{7}{3}C \Rightarrow -\dfrac{7}{3} = \dfrac{7}{3}C \Rightarrow C = -1\) Correct values for their \(B\) and their \(C\), which are found using \(5x - 4 \equiv B(3x - 1) + C(x + 2)\)	A1
So, \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv 3 + \dfrac{2}{(x + 2)} - \dfrac{1}{(3x - 1)}\)	(4)

Alternative Method 1: Initially dividing by \((x + 2)\)

Scheme	Marks
\(\dfrac{9x^2 + 20x - 10}{\text{"}(x + 2)\text{"}(3x - 1)} \equiv \dfrac{9x + 2}{(3x - 1)} - \dfrac{14}{(x + 2)(3x - 1)}\) \(\equiv 3 + \dfrac{5}{(3x - 1)} - \dfrac{14}{(x + 2)(3x - 1)}\) B1: their constant term \(= 3\)	B1
So, \(\dfrac{-14}{(x + 2)(3x - 1)} \equiv \dfrac{B}{(x + 2)} + \dfrac{C}{(3x - 1)}\) \(-14 \equiv B(3x - 1) + C(x + 2)\) B1: Forming a correct identity.	B1
\(\Rightarrow B = 2,\ C = -6\) M1: Attempts to find either one of their \(B\) or their \(C\) from their identity.	M1
So, \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv 3 + \dfrac{5}{(3x - 1)} + \dfrac{2}{(x + 2)} - \dfrac{6}{(3x - 1)}\) and \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv 3 + \dfrac{2}{(x + 2)} - \dfrac{1}{(3x - 1)}\) A1: Correct answer in partial fractions.	A1

Alternative Method 2: Initially dividing by \((3x - 1)\)

Scheme	Marks
\(\dfrac{9x^2 + 20x - 10}{(x + 2)\text{"}(3x - 1)\text{"}} \equiv \dfrac{3x + \frac{23}{3}}{(x + 2)} - \dfrac{\frac{7}{3}}{(x + 2)(3x - 1)}\) \(\equiv 3 + \dfrac{\frac{5}{3}}{(x + 2)} - \dfrac{\frac{7}{3}}{(x + 2)(3x - 1)}\) B1: their constant term \(= 3\)	B1
So, \(\dfrac{-\frac{7}{3}}{(x + 2)(3x - 1)} \equiv \dfrac{B}{(x + 2)} + \dfrac{C}{(3x - 1)}\) \(-\tfrac{7}{3} \equiv B(3x - 1) + C(x + 2)\) B1: Forming a correct identity.	B1
\(\Rightarrow B = \dfrac{1}{3},\ C = -1\) M1: Attempts to find either one of their \(B\) or their \(C\) from their identity.	M1
So, \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv 3 + \dfrac{\frac{5}{3}}{(x + 2)} + \dfrac{\frac{1}{3}}{(x + 2)} - \dfrac{1}{(3x - 1)}\) and \(\dfrac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} \equiv 3 + \dfrac{2}{(x + 2)} - \dfrac{1}{(3x - 1)}\) A1: Correct answer in partial fractions.	A1