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A3/PF/ 1
Partial Fractions
In this chapter you will learn how to express a given algebraic fraction into its components which
are known as partial fractions
Recall we combine two fractions by the “common denominator” method, to get a single fraction
7
1
2
13 2 2
23 23 2 3
Partial fractions is simply the reverse of this process! As shown below:
7
1
2
23 2 3 Step 1: Condition; check if the degree of the polynomial in the numerator is less than the degree of
the polynomial in the denominator. For example
7
2x 5
The degree of the numerator comes from the highest power of x, i.e. x3, and the degree of
the denominator is two(x2). This is known as an improper faction. Hence to perform partial
fractions, we need to convert it into a proper fraction. This can be done via long division.
Step 2: Fully factorise the denominator if it is not already factorised, for example:
3x 2
3x 2
4x 5 5 1
Step 3: express in partial fraction formats. There are three different formats used depending on the
factors of the denominator.
TYPE 1
The denominator is factorised into linear factors only, and no repeated factors.
Linear factor means the highest degree of the polynomial is one. The format for this case is:
where f(x) is a polynomial, and A, B and C are real numbers(unknowns) to be found.
Examples:
2x 1
23x 5 2 3x 5
7
3x 25 3x 2 5 7
3x 25 1 3x 2 5 1
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TYPE 2 The denominator is factorised into linear factors with repeated factors and the
format for this case is:
where f(x) is a polynomial, and A,B and C are real numbers to be found.
Examples:
2x 1
3x 5 3x 5 3x 5
7
3x 25 3x 2 5 5 2x 1
3x 5
3x 5 3x 5
3x 5
TYPE 3 The denominator has a quadratic factor that cannot be further factorised and the
corresponding format is:
where f(x) is a polynomial, and A,B and C are real numbers to be found.
Examples:
2x 1
3x 5 1 3x 5 1
2x 1
3x 5 13 3x 5 13
Step 4: Solve for unknowns A, B and C by combining the partial fractions together and using
polynomial identity
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Example 1
Express the following in partial fractions
2x 1
23x 5
Step 1: degree of numerator's polynomial is lower than the denominator, this is a proper fraction
hence proceed
Step 2: denominator is already factorised, hence proceed
Step 3: express in suitable format
2x 1
23x 5 2 3x 5
Step 4: Solve for unknowns
Multiplying both sides by (x+2)(3x-5), we get:
2 1 3 5 2 -----(1)
from polynomial identity, we can substitute values of into the (1) to eliminate one
unknown to find the other:
Substitute 2,
5 11
5
11
with A now known, we can substitute any value of to find B:
Substitute 1,
1 2 3
7
5 11
Therefore
5
7
2x 1
23x 5 11 2 113x 5
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Example 2
Express the following in partial fractions
3
3x 21 Step 1 – 3, and hence we have:
2
3x 21 3x 2 1 1 Step 4: Multiply both sides by 3 21 , we get
2 1 3 21 3 2
Substitute 1, we eliminate A & B;
3 = 5C
C=
Substitute 2
= A(1 , we eliminate B & C;
2
)
A = Substitute 0, we have,
2 = + 2B + !
B = Therefore
3
12
4
3
3x 21 253x 2 251 51 Exercise 1:
Express the following in partial fractions:
a)
" # "$
"#
!
!%
%& "
b) "$"# $ c) "# $" d) "# %#