Page 40 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions
Denominator with linear factors

Consider the addition of two proper rational functions as follows.
f(x) = 2 + 3 1
x + 1 x + 2
The function f(x) can be expressed as a single rational function with a common denominator, i.e.
f(x) = 2 + 3
x + 1 x + 2
= 2(x + 2) + 3(x + 1)
(x + 1)(x + 2)

= 5x + 7
(x + 1)(x + 2)
Notice that the addition of two proper rational functions results in a single proper rational function.

From the operation of addition as illustrated above, we see that the inverse process can also be carried out,
5x + 7
i.e. expressing the proper rational function (x + 1)(x + 2) as the sum of two rational functions with denominators
(x + 1) and (x + 2) respectively.

5x + 7 A B
Let ≡ + , where A and B are constants.
(x + 1)(x + 2) x + 1 x + 2
Hence 5x + 7 ≡ A(x + 2) + B(x + 1)
(x + 1)(x + 2) (x + 1)(x + 2)
5x + 7 ≡ A(x + 2) + B(x + 1)
= (A + B)x + (2A + B)
Equating coefficients of x: 5 = A + B ………… 
Equating constants: 7 = 2A + B ………… 
 – : 2 = A
Substituting A = 2 into ,
5 = 2 + B
B = 3

This process is known as expressing the function 5x + 7 in partial fractions with terms 2 and
3 . (x + 1)(x + 2) x + 1
x + 2
Notice that these two partial fractions are proper rational functions with the numerator as a constant and the
denominator as a linear function.

Example 34

Express 2x + 3 in partial fractions.
(x – 1)(x + 2)
Solution: Let 2x + 3 ≡ A + B
(x – 1)(x + 2) x – 1 x + 2
A(x + 2) + B(x – 1)
≡
(x – 1)(x + 2)
2x + 3 ≡ A(x + 2) + B(x – 1) This is an identity and
is valid for all x ∈ R

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