## Recommended

###### Partial Fractions

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

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
= 3ww+w.DuloMix.co+m 19

(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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