Part II: The Answers
560
Answers
901–1,001
Then perform the decomposition:
x
x
x
Ax B
x
Cx D
x
2
2
2
2
2
2
1
3
1
3
+
+
(
) +
(
)
=
+
+
(
)
+
+
+
(
)
Multiply both sides of the equation by (x
2
+ 1)(x
2
+ 3) to get
x
Ax B x
Cx D x
2
2
2
2
3
1
+ =
+
(
) +
(
) + +
(
) +
(
)
Expand the right side of the equation and collect like terms:
x
A x
Ax Bx
B Cx Cx Dx
D
x
x
x
A C x
B D
2
3
2
3
2
3
2
3
2
3
3
0
1
0
2
+ =
+
+
+
+
+
+
+
+
+
+ =
+
+
+
(
)
(
)x x
A C x
B D
2
3
3
+
+
+
+
(
)
(
)
Equating coefficients gives you the equations A + C = 0, B + D = 1, 3A + C = 0, and
3B + D = 2. Because A = –C, you get 3(–C) + C = 0, so –2C = 0, or C = 0; that means
A = 0 as well. Likewise, because B = 1 – D, you get 3(1 – D) + D = 2, so 3 – 2D = 2, or
D = 1
2
; you also get B = − =
1 1
2
1
2
. With these values, you get the following solution:
x
x
x
x
x
x
x
x
2
4
2
2
2
2
2
2
2
4
3
2
1
3
1 2
1
1 2
3
+
+
+
=
+
+
(
) +
(
)
=
+
(
)
+
+
(
)
949.
1
5
4
5
2
2
2
x
x
+
−
+
(
)
Begin by performing the decomposition:
x
x
Ax B
x
Cx D
x
2
2
2
2
2
2
1
5
5
5
+
+
(
)
=
+
+
+
+
+
(
)
Multiplying both sides of the equation by (x
2
+ 5)
2
gives you
x
x
x
x
Ax B
x
Cx D
x
2
2
2
2
2
2
2
2
2
2
5
1
5
5
5
5
+
(
)
+
+
(
)



 



 
=
+
(
)
+
+
+
+
+
(
)


 
 



 
+ =
+
+
(
) + +
x
A x B x
C x D
2
2
1
5
(
)
(
)
Expand the right side of the equation and regroup:
x
A x
Ax Bx
B Cx D
x
x
x
Ax
Bx
A C x
B D
2
3
2
3
2
3
2
1
5
5
0
1
0 1
5
5
+ =
+
+
+
+
+
+
+
+ =
+
+
+
+
+
(
)
(
) )
By equating coefficients, you find A = 0, B = 1, 5A + C = 0, and 5B + D = 0. Using A = 0 and
5A + C = 0, you get 5(0) + C = 0, or C = 0. Likewise, using B = 1 and 5B + D = 1, you find
that 5(1) + D = 1 so that D = –4. Therefore, the fraction decomposition becomes
x
x
x
x
x
x
2
2
2
2
2
2
2
2
2
1
5
1
5
4
5
1
5
4
5
+
+
(
)
=
+
+
−
+
(
)
=
+
−
+
(
)