C4 Chapter 1: Partial Fractions

Dr J Frost (jfrost@tiffin.kingston.sch.uk)www.drfrostmaths.com

Last modified: 30th August 2015

OverviewAt GCSE you learnt how to combine a sum/difference of fractions into one.We now want to learn how to do the opposite process: split a fraction into a sum of simpler ones, known as partial fractions.

1𝑥+1

+ 𝑥𝑥+2

𝑥2+2𝑥+2(𝑥+1 ) (𝑥+2 )

𝑥+1𝑥 (𝑥−1 )?

Method

Split into partial fractions.Q

METHOD 1: Substitution

If each factor of the denominator is linear, we can split like such (for constants and ):

We don’t like fractions in equations, so we could simplify this to:

METHOD 2: Equating coefficients

We can easily eliminate either or by an appropriate choice of :

If :

If :

Therefore:

Since this is an identity, the coefficients of must match, and the constant terms must match.

Solving simultaneous equations gives same solutions as before.

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Test Your UnderstandingC4 Jan 2011 Q3

Let :

Let :

Therefore Notice we can move the “–” to the front of the fraction.

Note that we don’t technically need this last line from the perspective of the mark scheme, but it’s good to just to be on the safe side

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More than two fractionsThe principle is exactly the same if we have more than two linear factors in the denominator.

Split into partial fractions.Q

When :

When :

When :

So

Bro Tip: While substitution is generally the easier method, I sometimes compare coefficients of just the term to avoid having to deal with fractions. No need to expand; we can see by observation that:

Then is easy to determine given we know and .

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Test Your Understanding

C4 June 2009 Q3

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Exercise 1B/1C

Express the following as partial fractions.

1

a

c

e

g

Exercise 1B

Factorise the denominator first:

Exercise 1C

1

2

a

b

c

a

c

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Repeated linear factorsSuppose we wished to express as . What’s the problem?

Because the denominators are the same, we’d get . There’s no constant values of and we can choose such that because the denominators will still be different.

Split into partial fractions.Q

11𝑥2+14 𝑥+5(𝑥+1 )2 (2𝑥+1 )

≡𝐴𝑥+1

+ 𝐵(𝑥+1 )2

+ 𝐶2𝑥+1

The problem is resolved by having the factor both squared and non-squared.

When : When :

At this point we could substitute something else (e.g. ) but it’s easier to equate terms.

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Test Your Understanding

C4 June 2011 Q1

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Dealing with Improper Fractions

The ‘degree’ of a polynomial is the highest power, e.g. a quadratic has degree 2.

An algebraic fraction is improper if the degree of the numerator is at least the degree of the denominator.

𝑥2−3𝑥+2

𝑥+1𝑥−1

𝑥3−𝑥2+3𝑥2−𝑥

! To split an improper fraction into partial fractions, either:1. Divide algebraically first.2. Or introduce a whole term and deal with identity immediately.

Dealing with Improper FractionsSplit into partial fractions.Q

Dividing algebraically gives:

Turn numerator back:

Let

So

Method 1: Algebraic Division Method 2: Using One Identity(method not in your textbooks but in mark schemes)

Let:

If : If : Comparing coefficients of :

Bropinion: I personally think the second method is easier. And mark schemes present it as “Method 1” – implying more standard!

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Test Your Understanding

C4 Jan 2013 Q3

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Exercise 1EExpress as partial fractions.

1 a

b

c

d

2 a

b

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SummaryIdentify what identity you’d use in each case (no need to identify constants).

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