Page 41 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

When x = 1, 2 + 3 = A(3) + B(0)
5 = 3A
1 A = 5
3
When x = –2, –4 + 3 = A(0) + B(–3)
–1 = –3B
B = 1
3
2x + 3 5 1
Hence, ≡ +
(x – 1)(x + 2) 3(x – 1) 3(x + 2)

Denominator with prime quadratic factors
We have learnt that for a partial fraction with a linear denominator, its numerator is a constant and the partial
fraction is a proper rational function. In the same way, if the denominator of a partial fraction is a quadratic
function, its numerator must be a constant or a linear function such that the partial fraction is a proper rational
function.
Hence, if the denominator of a proper rational function has a prime quadratic factor in the form
ax + bx + c, where a, b and c are constants, the given partial fraction must be of the form Ax + B , where
2
2
ax + bx + c
A and B are suitable constants.
Example 35

Express 3x + 4 in partial fractions.
2
(x + 2)(x – x + 1)
Solution: Let 3x + 4 ≡ A + Bx + C
2
(x + 2)(x – x + 1) x + 2 x – x + 1
2
2
A(x – x + 1) + (Bx + C)(x + 2)
≡
2
(x + 2)(x – x + 1)
2
3x + 4 ≡ A (x – x + 1) + (Bx + C)(x + 2)

When x = –2, –6 + 4 = A(4 + 2 + 1) + 0
–2 = 7A
2
A = –
7
2
Equating coefficients of x : 0 = A + B
2
B =
7
Equating the constants: 4 = A + 2C
2
= – + 2C
7
30
2C =
7
15
C =
7
Hence, 3x + 4 ≡ – 2 + 2x + 15
2
(x + 2)(x – x + 1) 7(x + 2) 7(x – x + 1)
2
Notice that the two partial fractions are proper rational functions with the numerator
of one of them as a linear function, i.e. 2x + 15.
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01a STPM Math T T1.indd 38 3/28/18 4:20 PM