Resolve Into partial fraction of 2x, +3
x2 (x-1)
Answers
Step-by-step explanation:
Answer
Resolve into partial fractions
x(x−1)(2x+3)
2x
3
+x
2
−x−3
=1+
x(x−1)(2x+3)
2x−3
Now assume,
x(x−1)(2x+3)
2x−3
=
x
A
+
x−1
B
+
2x+3
C
After simplifying this, we get
x(x−1)(2x+3)
2x−3
=
x(x−1)(2x+3)
A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)
A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)=2x−3
Equating the coefficients, we get
2A+2B+C=0,A+3B−C=0 and −3A=−3
On Solving for A,B and C , we get
A=1,B=
5
−1
,C=
5
−8
∴
x(x−1)(2x+3)
2x
3
+x
2
−x−3
=1+
x
1
−
5(x−1)
1
−
5(2x+3)
8
Answer:
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Step-by-step explanation:
Answer
Resolve into partial fractions
x(x−1)(2x+3)
2x
3
+x
2
−x−3
=1+
x(x−1)(2x+3)
2x−3
Now assume,
x(x−1)(2x+3)
2x−3
=
x
A
+
x−1
B
+
2x+3
C
After simplifying this, we get
x(x−1)(2x+3)
2x−3
=
x(x−1)(2x+3)
A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)
A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)=2x−3
Equating the coefficients, we get
2A+2B+C=0,A+3B−C=0 and −3A=−3
On Solving for A,B and C , we get
A=1,B=
5
−1
,C=
5
−8
∴
x(x−1)(2x+3)
2x
3
+x
2
−x−3
=1+
x
1
−
5(x−1)
1
−
5(2x+3)
8