Partial fractions 15
The denominator is of the same degree as the numerator.
Thus dividing out gives:
1
x 2 − 3x + 2
x 2
+ 1
x 2 − 3x + 2
—————
3x − 1
———
For more on polynomial division, see Section 1.4,
page 6.
Hence
x 2 + 1
x 2 − 3x + 2
≡ 1 +
3x − 1
x 2 − 3x + 2
≡ 1 +
3x − 1
(x − 1)(x − 2)
Let
3x − 1
(x − 1)(x − 2)
≡
A
(x − 1)
+
B
(x − 2)
≡
A(x − 2) + B(x − 1)
(x − 1)(x − 2)
Equating numerators gives:
3x − 1 ≡ A(x − 2) + B(x − 1)
Let x = 1. Then 2 = −A
i.e.
A = −2
Let x = 2. Then 5 = B
Hence
3x − 1
(x − 1)(x − 2)
≡
−2
(x − 1)
+
5
(x − 2)
Thus
x 2 + 1
x 2 − 3x + 2
≡ 1−
2
(x−1)
+
5
(x−2)
Problem 4. Express
x 3 − 2x 2 − 4x − 4
x 2 + x − 2
in partial
fractions.
The numerator is of higher degree than the denominator.
Thus dividing out gives:
x − 3
x 2 + x − 2
x 3 − 2x 2 − 4x − 4
x 3 + x 2 − 2x
——————
− 3x 2 − 2x − 4
− 3x 2 − 3x + 6
———————
x − 10
Thus
x 3 − 2x 2 − 4x − 4
x 2 + x − 2
≡ x − 3 +
x − 10
x 2 + x − 2
≡ x − 3 +
x − 10
(x + 2)(x − 1)
Let
x − 10
(x + 2)(x − 1)
≡
A
(x + 2)
+
B
(x − 1)
≡
A(x − 1) + B(x + 2)
(x + 2)(x − 1)
Equating the numerators gives:
x − 10 ≡ A(x − 1) + B(x + 2)
Let x =−2. Then −12 = −3 A
i.e.
A = 4
Let x = 1. Then
−9 = 3B
i.e.
B = −3
Hence
x − 10
(x + 2)(x − 1)
≡
4
(x + 2)
−
3
(x − 1)
Thus
x 3 − 2 x 2 − 4x − 4
x 2 + x − 2
≡ x − 3 +
4
(x + 2)
−
3
(x − 1)
Now try the following exercise
Exercise 8 Further problems on partial
fractions with linear factors
Resolve the following into partial fractions.
1.
12
x 2 − 9
2
(x − 3)
−
2
(x + 3)
2.
4(x − 4)
x 2 − 2x − 3
5
(x + 1)
−
1
(x − 3)
3.
x 2 − 3x + 6
x(x − 2)(x − 1)
3
x
+
2
(x − 2)
−
4
(x − 1)
4.
3(2x 2 − 8x − 1)
(x + 4)(x + 1)(2x − 1)
7
(x + 4)
−
3
(x + 1)
−
2
(2x − 1)