Introduction
• In this chapter you will learn to add
fractions with different denominators
(a recap)
• You will learn to work backwards and
split an algebraic fraction into
components called ‘Partial Fractions’
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Partial Fractions
1 3
You can add and subtract several Calculate: +
3 8
fractions as long as they share a 8 3
common denominator × ×
8 3
8 9
+
You will have seen this plenty of times 24 24
already! If you want to combine
fractions you must make the
denominators equivalent… 17
=
24
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1A
Partial Fractions
Calculate: 2 1
You can add and subtract several −
𝑥 + 3 𝑥 + 1
fractions as long as they share a 𝑥 + 1 𝑥 + 3
common denominator × ×
𝑥 + 1 𝑥 + 3
2(𝑥 + 1) 1(𝑥 + 3)
−
You will have seen this plenty of times (𝑥 + 3)(𝑥 + 1) (𝑥 + 3)(𝑥 + 1)
already! If you want to combine Multiply
fractions you must make the brackets
denominators equivalent… 2𝑥 + 2 𝑥 + 3
−
(𝑥 + 3)(𝑥 + 1) (𝑥 + 3)(𝑥 + 1)
Group
terms
𝑥 − 1
=
(𝑥 + 3)(𝑥 + 1)
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1A
Partial Fractions
You can split a fraction with two
linear factors into Partial
Fractions
𝑥 − 1 2 1
For example: = − when split up into Partial Fractions
(𝑥 + 3)(𝑥 + 1) 𝑥 + 3 𝑥 + 1
11 𝐴 𝐵
= + when split up into Partial Fractions
(𝑥 − 3)(𝑥 + 2) 𝑥 − 3 𝑥 + 2
You need to be able to calculate the values of A and B…
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1B
Partial Fractions
6𝑥 − 2
You can split a fraction with two
linear factors into Partial (𝑥 − 3)(𝑥 + 1)
Split the Fraction into its 2 linear
Fractions parts, with numerators A and B
𝐴 𝐵
Split (𝑥 − 3) + (𝑥 + 1)
Cross-multiply to make the
6𝑥 − 2 𝐴(𝑥 + 1) 𝐵(𝑥 − 3) denominators the same
(𝑥 − 3)(𝑥 + 1) (𝑥 − 3)(𝑥 + 1) + (𝑥 − 3)(𝑥 + 1)
Group together as one fraction
into Partial Fractions 𝐴 𝑥 + 1 + 𝐵(𝑥 − 3)
=
(𝑥 − 3)(𝑥 + 1)
This has the same denominator as
the initial fraction, so the
numerators must be the same
6𝑥 − 2 = A 𝑥 + 1 + 𝐵(𝑥 − 3)
If x = -1: −8 = −4𝐵
2 = 𝐵
If x = 3: 16 = 4𝐴
4 = 𝐴
You now have the values of A and
B and can write the answer as
4 2 Partial Fractions
=
(𝑥 − 3) +
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1B
Partial Fractions
You can also split fractions
with more than 2 linear factors
in the denominator
4 𝐴 𝐵 𝐶
For example: = + +
𝑥 + 1 𝑥 − 3 (𝑥 + 4) 𝑥 + 1 𝑥 − 3 𝑥 + 4
when split up into Partial Fractions
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1C
Partial Fractions
6𝑥2 + 5𝑥 − 2
You can also split fractions 𝑥(𝑥 − 1)(2𝑥 + 1) Split the Fraction into
with more than 2 linear factors its 3 linear parts
in the denominator 𝐴 𝐵 𝐶
+ +
𝑥 𝑥 − 1 2𝑥 + 1
Cross Multiply to make
Split the denominators equal
𝐴(𝑥 − 1)(2𝑥 + 1) 𝐵(𝑥)(2𝑥 + 1) 𝐶(𝑥)(𝑥 − 1)
6𝑥2 + 5𝑥 − 2 + +
𝑥(𝑥 − 1)(2𝑥 + 1) 𝑥(𝑥 − 1)(2𝑥 + 1) 𝑥(𝑥 − 1)(2𝑥 + 1)
𝑥(𝑥 − 1)(2𝑥 + 1) Put the fractions
together
into Partial fractions 𝐴 𝑥 − 1 2𝑥 + 1 + 𝐵 𝑥 2𝑥 + 1 + 𝐶(𝑥)(𝑥 − 1)
𝑥(𝑥 − 1)(2𝑥 + 1)
The numerators
must be equal
6𝑥2 + 5𝑥 − 2 = 𝐴 𝑥 − 1 2𝑥 + 1 + 𝐵 𝑥 2𝑥 + 1 + 𝐶(𝑥)(𝑥 − 1)
If x = 1 9 = 3𝐵
3 = 𝐵
If x = 0 −2 = −𝐴
2 = 𝐴
If x = -0.5 −3 = 0.75𝐶
−4 = 𝐶
You can now fill in
the numerators
2 3 4
= + −
𝑥 𝑥 − 1 2𝑥 + 1
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1C
Partial Fractions
You can also split fractions 𝑥3 − 4𝑥2 + 𝑥 + 6
Try substituting factors to
with more than 2 linear factors make the expression 0
(1)3−4 1 2 + (1) + 6 = 4
in the denominator
Split (−1)3−4 −1 2 + (−1) + 6 = 0
4𝑥2 − 21𝑥 + 11 Therefore (x + 1) is a factor…
𝑥3 − 4𝑥2 + 𝑥 + 6 Divide the expression by (x + 1)
𝑥2 − 5𝑥 + 6
into Partial fractions
𝑥 + 1 𝑥3 − 4𝑥2 + 𝑥 + 6
You will need to 𝑥3 + 𝑥2
factorise the
denominator first… −5𝑥2 + 𝑥 + 6
−5𝑥2 − 5𝑥
6𝑥 + 6
6𝑥 + 6
0
𝑥3 − 4𝑥2 + 𝑥 + 6 = (𝑥 + 1)(𝑥2 − 5𝑥 + 6)
You can now factorise
the quadratic part
𝑥3 − 4𝑥2 + 𝑥 + 6 = (𝑥 + 1)(𝑥 − 2)(𝑥 − 3)
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1C
Partial Fractions
4𝑥2 − 21𝑥 + 11 4𝑥2 − 21𝑥 + 11
You can also split fractions =
𝑥3 − 4𝑥2 + 𝑥 + 6 (𝑥 + 1)(𝑥 − 2)(𝑥 − 3) Split the fraction into
with more than 2 linear factors its 3 linear parts
in the denominator 𝐴 𝐵 𝐶
+ +
𝑥 + 1 𝑥 − 2 𝑥 − 3
Split Cross
multiply
𝐴(𝑥 − 2)(𝑥 − 3) 𝐵(𝑥 + 1)(𝑥 − 3) 𝐶(𝑥 + 1)(𝑥 − 2)
4𝑥2 − 21𝑥 + 11 + +
(𝑥 + 1)(𝑥 − 2)(𝑥 − 3) (𝑥 + 1)(𝑥 − 2)(𝑥 − 3) (𝑥 + 1)(𝑥 − 2)(𝑥 − 3)
𝑥3 − 4𝑥2 + 𝑥 + 6 Group the
fractions
into Partial fractions 𝐴 𝑥 − 2 𝑥 − 3 + 𝐵 𝑥 + 1 𝑥 − 3 + 𝐶(𝑥 + 1)(𝑥 − 2)
(𝑥 + 1)(𝑥 − 2)(𝑥 − 3) The
numerators
must be
equal
4𝑥2 − 21𝑥 + 11 = 𝐴 𝑥 − 2 𝑥 − 3 + 𝐵 𝑥 + 1 𝑥 − 3 + 𝐶(𝑥 + 1)(𝑥 − 2)
If x = 2 −15 = −3𝐵
5 = 𝐵
If x = 3 16 = −4C
−4 = 𝐶
If x = -1 36 = 12A
3 = 𝐴
Replace A,
B and C
3 5 4
= + −
𝑥 + 1 𝑥 − 2 𝑥 − 3
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1C
Partial Fractions
You need to be able to split a
fraction that has repeated linear
roots into a Partial Fraction
3𝑥2 − 4𝑥 + 2 𝐴 𝐵 𝐶
For example: = + + when split up into
𝑥 + 1 (𝑥 − 5)2 (𝑥 + 1) (𝑥 − 5) (𝑥 − 5)2 Partial Fractions
The repeated root is
included once ‘fully’ and
once ‘broken down’
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1D
Partial Fractions
11𝑥2 + 14𝑥 + 5
You need to be able to split a (𝑥 + 1)2(2𝑥 + 1) Split the fraction into
fraction that has repeated linear its 3 parts
roots into a Partial Fraction 𝐴 𝐵 𝐶
+ +
(𝑥 + 1) (𝑥 + 1)2 (2𝑥 + 1)
Split Make the denominators
equivalent
11𝑥2 + 14𝑥 + 5 𝐴(𝑥 + 1)(2𝑥 + 1) 𝐵(2𝑥 + 1) 𝐶(𝑥 + 1)2
+ +
(𝑥 + 1)2(2𝑥 + 1) (𝑥 + 1)2(2𝑥 + 1) (𝑥 + 1)2(2𝑥 + 1) (𝑥 + 1)2(2𝑥 + 1)
Group up
into Partial fractions 𝐴 𝑥 + 1 2𝑥 + 1 + 𝐵 2𝑥 + 1 + 𝐶(𝑥 + 1)2
=
(𝑥 + 1)2(2𝑥 + 1) The numerators
will be the same
11𝑥2 + 14𝑥 + 5 = 𝐴 𝑥 + 1 2𝑥 + 1 + 𝐵 2𝑥 + 1 + 𝐶(𝑥 + 1)2
If x = -1 2 = −𝐵
−2 = 𝐵
If x = -0.5 0.75 = 0.25𝐶
At this point there is no way to 3 = 𝐶
cancel B and C to leave A by
substituting a value in If x = 0 5 = 1𝐴 + 1𝐵 + 1C
Choose any value for x (that 5 = 𝐴 − 2 + 3
hasn’t been used yet), and use 4 = 𝐴
Sub in the values
the values you know for B and C of A, B and C
to leave A 4 2 3
= − +
(𝑥 + 1) www(𝑥.D+ulo1M)2ix.com(2𝑥 + 1) 16
1D
Partial Fractions
You can split an improper fraction into
Partial Fractions. You will need to divide
the numerator by the denominator first
to find the ‘whole’ part
A regular fraction being
22 1 3 split into 2 ‘components’
= +
35 5 7
57 1 3
= 2 + + A top heavy (improper) fraction
20 4 5 will have a ‘whole number part
before the fractions
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1E
Partial Fractions
3𝑥2 − 3𝑥 − 2 = 3𝑥2 − 3𝑥 − 2
You can split an improper fraction into (𝑥 − 1)(𝑥 − 2) 𝑥2 − 3𝑥 + 2
Partial Fractions. You will need to divide Divide the numerator by
the numerator by the denominator first the denominator to find
3 the ‘whole’ part
to find the ‘whole’ part
𝑥2 − 3𝑥 + 2 3𝑥2 − 3𝑥 − 2
Split 3𝑥2 − 9𝑥 + 6
3𝑥2 − 3𝑥 − 2 6𝑥 − 8 Now rewrite the original
fraction with the whole
(𝑥 − 1)(𝑥 − 2) part taken out
3𝑥2 − 3𝑥 − 2 6𝑥 − 8
= 3 +
into Partial fractions (𝑥 − 1)(𝑥 − 2) (𝑥 − 1)(𝑥 − 2)
Split the fraction into 2
parts (ignore the whole
𝐴 𝐵 part for now)
Remember, Algebraically an +
‘improper’ fraction is one where (𝑥 − 1) (𝑥 − 2)
Make denominators
the degree (power) of the equivalent and group up
numerator is equal to or exceeds 𝐴 𝑥 − 2 + 𝐵(𝑥 − 1)
=
that of the denominator (𝑥 − 1)(𝑥 − 2)
The numerators will be
the same
6𝑥 − 8 = 𝐴 𝑥 − 2 + 𝐵(𝑥 − 1)
If x = 2 4 = 𝐵
If x = 1 −2 = −𝐴
2 = 𝐴
2 4
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(𝑥 − 1) (𝑥 − 2) 1E
Summary
• We have learnt how to split Algebraic
Fractions into ‘Partial fractions’
• We have also seen how to do this when
there are more than 2 components,
when one is repeated and when the
fraction is ‘improper’
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