2.7 Partial fractions 53
To derive this result, write
It is required therefore that
x 1 + 121 = 1 A(x 1 + 1 4) 1 + 1 B(x 1 − 1 3)
for allvalues of x. The values of Aand Bcan be obtained from this equation of the
numerators by the ‘method of equating coefficients’ described in Example 2.23.
Alternatively, they are obtained directly by making suitable choices of the variable x.
Thus
when x 1 = 1 3: 5 1 = 17 A and A 1 = 1527
when x 1 = 1 −4: 2 1 = 17 B and B 1 = 1227
0 Exercises 61–63
EXAMPLE 2.29Three linear factors in the denominator
Then,
3 x
21 + 14 x 1 − 121 = 1 A(x 1 − 1 2)(x 1 + 1 3) 1 + 1 B(x 1 − 1 1)(x 1 + 1 3) 1 + 1 C(x 1 − 1 1)(x 1 − 1 2)
so that
when x 1 = 1 1: 5 1 = 11 − 4 A and A 1 = 1 − 524
when x 1 = 1 2: 18 1 = 15 B and B 1 = 11825
when x 1 = 1 −3: 13 1 = 120 C and C 1 = 113220
Therefore
0 Exercise 64
342
123
1
20
25
1
72
2
13
2xx
xx x x x x
+−
−− +
=−
−
−
()( )() ++
3=
−++−++−−
−
Ax x Bx x Cx x
xx
()()()()()()
()(
23 13 12
1 −−+23)( )x
342
123 1 2 3
2xx
xx x
A
x
B
x
C
x
+−
−− +
=
−
−
()( )() +
x
xx
A
x
B
x
Ax Bx
x
−+
=
−
=
++ −
−
2
34 3 4
43
()() 3
()()
()(xx+4)