Summary Cheat Sheet for Edexcel IAL P4

Pure Mathematics 4
Partial Fractions, Integrating Partial Fractions, and Binomial Expansion (Ch. 2, 4, 6.5)
Partial Fractions 5 𝐴 𝐵 𝟓𝒙$𝟑
[C4 June 2005 Q3(a)] Express (𝟐𝒙(𝟑)(𝒙$𝟐) as partial fractions.
≡ +
(𝑥 + 1)(𝑥 − 4) 𝑥 + 1 𝑥 − 4
5𝑥 + 3 𝐴 𝐵
= +
(2𝑥 − 3)(𝑥 + 2) (2𝑥 − 3) (𝑥 + 2)
Methodology ①: Factors = 0
e.g. for A(x-4), when x=4, A=0; thus can find B. 5𝑥 + 3 = 𝐴(𝑥 + 2) + 𝐵(2𝑥 − 3)

when x=-2 when x=1.5
Methodology ②: Compare Coefficients 5(−2) + 3 = 𝐴(−2 + 2) + 𝐵(2(−2) − 3) 3 3 3
5( ) + 3 = 𝐴 5( ) + 26 + 𝐵 52( ) − 36
e.g. for 𝐴(𝑥 − 𝑛)! , you can compare the coefficient of 𝑥 ! to find A. (be −7 = 0 − 7𝐵 2 2 2
𝐵=1 10.5 = 3.5𝐴 + 0
aware of all terms might contain powers) 𝐴=3


Remarks: 5𝑥 + 3 3 1
Remember to rewrite your impartial fractions! ∴ = +
(2𝑥 − 3)(𝑥 + 2) (2𝑥 − 3) (𝑥 + 2)

Improper When the denominator’s highest order is equal or lower than that of [C4 June 2010 Q5(a)]
Fractions as numerator, or vice versa, it is an improper fraction. 2𝑥 ! + 5𝑥 − 10 𝐵 𝐶
≡𝐴+ +
Impartial (𝑥 − 1)(𝑥 + 2) 𝑥−1 𝑥+2
89$! (:9$;)! 9"
Fractions Examples of Improper Fractions: :9(! , (<(;9)(9(=) , =(9
Find Values of A, B, C.
!9 " $:9(=B
9 " (<9$; 9 ;9$8 expanding denominator gives
Examples of Proper Fractions: (9(:)!
, (>(9)" , 9 " $9(!
(9$!)(9(8) 89(C
using long division gives 2 + 9 " (<9(!
In order to set improper fractions as impartial fractions, you must use 3𝑥 − 6 𝐵 𝐶
long division! = +
(𝑥 − 1)(𝑥 + 2) 𝑥 − 1 𝑥 + 2
3𝑥 − 6 = 𝐵(𝑥 + 2) + 𝐶(𝑥 − 1)
Tip: When splitting as impartial fraction, the denominators’ power will
determine how many times the factor will appear on the denominator when x=-2 when x=1
with the power ‘n’ and ‘n-1’ until 1, 3(−2) − 6 = 𝐵(−2 + 2) + 𝐶(−2 − 1) 3(1) − 6 = 𝐵(1 + 2) + 𝐶(1 − 1)
9 ? @ A −12 = −3𝐶 −3 = 3𝐵
i.e. (>(9)! = (>(9)! + (>(9)" + (>(9) 𝐶=4 𝐵 = −1
∴ 𝐴 = 2; 𝐵 = −1; 𝐶 = 4

2𝑥 ! + 5𝑥 − 10 1 4
∴ =2− +
(𝑥 − 1)(𝑥 + 2) 𝑥−1 𝑥+2

, Binomial In P2, the rule for binomial expansion is only valid for natural numbers. [C4 June 2010 Q5(b)]
Expansion For numbers such as fractions or negative numbers, you have to use this Find the expansion in ascending order x, up and including 𝒙𝟐 .
(1+x)^n formula: (from (a), question above)
2𝑥 ! + 5𝑥 − 10 1 4
𝒏(𝒏#𝟏) 𝒏(𝒏#𝟏)(𝒏#𝟐) 𝒏(𝒏#𝟏)…(𝒏#𝒓+𝟏) =2− +
(𝟏 + 𝒙)𝒏 = 𝟏 + 𝒏𝒙 + 𝒙𝟐 + 𝒙𝟑 + ⋯ + 𝒙𝒓 + ⋯. (𝑥 − 1)(𝑥 + 2) 𝑥−1 𝑥+2
𝟐! 𝟑! 𝒓!


The expansion is only valid for −1 < 𝑥 < 1 𝑜𝑟 |𝑥| < 1, 𝑛 𝜖 ℝ (because it 1 4
2− + = 2 − (𝑥 − 1)(= + 4(𝑥 + 2)(=
is an estimation, the series above will create an infinite number of terms) 𝑥−1 𝑥+2

The expansion of (𝑥 − 1)(= The expansion of 4(𝑥 + 2)(=
Tip: Most Exam Questions only require you to expand up to 𝑥 8
−(1 − 𝑥)(= ≈ 1 #, 1 #,
Binomial What if the factors is not (1+x)? It can be summed up in below: ("#)("%) 4 *2 ,1 + 𝑥/ 0 = 2 ,1 + 𝑥/
−(1 + (−1)(−𝑥) + %! (−𝑥)% . 2 2
Expansion , (#,)(#-) , -
(a+bx)^n 𝑏 D 𝑏 D = −(1 + 𝑥 + 𝑥 ! ) ≈ 2(1 + (−1) 3- 𝑥4 + -!
3- 𝑥4 .
(𝑎 + 𝑏𝑥) = C𝑎 ?1 + 𝑥@ D = 𝑎D ?1 + 𝑥@
D
1 1
𝑎 𝑎 = 2(1 − 𝑥 + 𝑥 ! )
2 4
E F 1 !
This expansion is valid for EF 𝑥E < 1, |𝑥| < EE E , 𝑎 𝜖 ℝ, 𝑏 𝜖 ℝ =2−𝑥+ 𝑥
2
Partial You can use Partial Fractions to simplify more difficult expansions. 1 4 1
Fractions and 2− + = 2 − (−(1 + 𝑥 + 𝑥 ! )) + ?2 − 𝑥 + 𝑥 ! @
𝑥−1 𝑥+2 2
Binomial 1
= 2 + 1 + 𝑥 + 𝑥! + 2 − 𝑥 + 𝑥!
Expansion 2
3 !
=5+ 𝑥
2


Integrating You can use Partial Fractions to integrate more difficult expressions. [C4 June 2005 Q3(b)]
Partial Hence find the exact value of
𝟔
Fractions (𝟓𝒙 + 𝟐)
F 𝒅𝒙
𝟐 (𝟐𝒙 − 𝟑)(𝒙 + 𝟐)

From above (Q3(a)),
(𝟓𝒙 + 𝟐) 3 1
= +
(𝟐𝒙 − 𝟑)(𝒙 + 𝟐) (2𝑥 − 3) (𝑥 + 2)
𝟔
3 1 3 6
∴F + = K ln(2𝑥 − 3) + ln(𝑥 + 2)N
𝟐 (2𝑥 − 3) (𝑥 + 2) 2 2
= ln 216 − ln 4 = ln 54