X-1/x*2 (x-1) in partial fractions
Answers
x
2
+
x
+
1
=
(
x
+
1
2
)
2
+
3
4
=
(
x
+
1
2
)
2
−
(
√
3
2
i
)
2
=
(
x
+
1
2
−
√
3
2
i
)
(
x
+
1
2
+
√
3
2
i
)
So:
1
x
2
+
x
+
1
=
A
x
+
1
2
−
√
3
2
i
+
B
x
+
1
2
+
√
3
2
i
=
A
(
x
+
1
2
+
√
3
2
i
)
+
B
(
x
+
1
2
−
√
3
2
i
)
x
2
+
x
+
1
=
(
A
+
B
)
x
+
(
A
(
1
2
+
√
3
2
i
)
+
B
(
1
2
−
√
3
2
i
)
)
x
2
+
x
+
1
Equating coefficients, we find:
⎧
⎨
⎩
A
+
B
=
0
A
(
1
2
+
√
3
2
i
)
+
B
(
1
2
−
√
3
2
i
)
=
1
From the first equation we find
B
=
−
A
.
Substitute this in the second equation to get:
A
√
3
i
=
1
Hence:
⎧
⎪
⎨
⎪
⎩
A
=
1
√
3
i
=
−
√
3
3
i
B
=
√
3
3
i
So:
1
x
2
+
x
+
1
=
B
x
+
1
2
+
√
3
2
i
+
A
x
+
1
2
−
√
3
2
i
=
√
3
i
3
(
x
+
1
2
+
√
3
2
i
)
−
√
3
i
3
(
x
+
1
2
−
√
3
2
i
)
Answer:
Step-by-step explanation:
Partial Fractions
When we are writing a rational expression as a sum of partial fractions, we have to first make sure that we use the correct number of parameters. The number of parameters that are chosen is dependent on the number of factors present in the denominator of the expression given.
Answer and Explanation:
Here, we have one linear factor and a repeated linear factor. So,
x
+
1
x
2
(
x
−
1
)
=
a
x
+
b
x
2
+
c
x
−
1
Multiplying the entire equation by
x
2
(
x
−
1
)
gives:
x
+
1
=
a
x
(
x
−
1
)
+
b
(
x
−
1
)
+
c
x
2
Now, if we put
x
=
0
, we get:
0
+
1
=
a
∗
0
(
0
−
1
)
+
b
(
0
−
1
)
+
c
∗
0
2
1
=
−
1
(
b
)
b
=
−
1
Putting
x
=
1
gives:
1
+
1
=
a
∗
1
(
1
−
1
)
+
b
(
1
−
1
)
+
c
∗
1
2
c
=
2
We now put the values of the parameters that we have found back into the equation, expand the equation, regroup it, and then finally compare the coefficients on either side of it.
x
+
1
=
a
x
(
x
−
1
)
+
(
−
1
)
(
x
−
1
)
+
2
x
2
x
+
1
=
a
x
2
−
a
x
−
x
+
1
+
2
x
2
x
+
1
=
x
2
(
a
+
2
)
+
x
(
−
a
−
1
)
+
1
Comparing the two sides of the equation gives:
−
a
−
1
=
1
a
=
−
2
So,
(
x
+
1
)
(
x
2
(
x
−
1
)
)
=
−
2
x
−
1
x
2
+
2
x
−
1