Page 42 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions
Denominator with repeated linear factors

If one of the terms in the denominator of a proper rational function is a repeated linear factor such as
2
(x + 1) , this factor can be considered to be a quadratic factor. So, the partial fraction concerned is of the 1
form Ax + B .
(x + 1) 2
2
Suppose we want to express 3x + 6x – 1 in partial fractions.
(x – 1)(x + 1) 2
2
Then, 3x + 6x – 1 ≡ A + Bx + C
(x – 1)(x + 1) 2 x – 1 (x + 1) 2
2
≡ A(x + 1) + (Bx + C)(x – 1)
(x – 1)(x + 1) 2
2
3x + 6x – 1 ≡ A(x + 1) + (Bx + C)(x – 1)
2
When x = 1, 3 + 6 – 1 = 4A
8 = 4A
A = 2
2
Equating coefficients of x : 3 = A + B
3 = 2 + B
B = 1
Equating constants: –1 = A – C
–1 = 2 – C
C = 3
2
Therefore, 3x + 6x – 1 ≡ 2 + x + 3
(x – 1)(x + 1) 2 x – 1 (x + 1) 2
Notice that the partial fractions obtained is correct, except that it is not in its simplest form, because
x + 3 ≡ (x + 1) + 2
(x + 1) 2 (x + 1) 2
≡ 1 + 2
x + 1 (x + 1) 2

Hence, the partial fractions in its simplest form is
2
3x + 6x – 1 ≡ 2 + 1 + 2
(x – 1)(x + 1) 2 x – 1 x + 1 (x + 1) 2
2
This means that the function 3x + 6x – 1 can be expressed as partial fractions in the following way.
(x – 1)(x + 1) 2
2
Let 3x + 6x – 1 ≡ A + B + C
(x – 1)(x + 1) 2 x – 1 x + 1 (x + 1) 2
2
≡ A(x + 1) + B(x – 1)(x + 1) + C(x – 1)
(x – 1)(x + 1) 2
2
Hence, 3x + 6x – 1 ≡ A(x +1) + B(x – 1)(x + 1) + C(x – 1)
2
When x = 1, 3 + 6 – 1 = 4A
8 = 4A
A = 2
When x = –1, 3 – 6 – 1 = –2C

–4 = –2C
C = 2

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01a STPM Math T T1.indd 39 3/28/18 4:20 PM