Resolve into partial friction (3x-2)/((X+1)(X+1)
Answers
Answered by
20
Answer:
The table give general form of the partial fractions :
So ,
(1−2x)(1−x
2
)
1+3x+2x
2
=
(1−2x)(1−x)(1+x)
1+3x+2x
2
=>
(1−2x)(1−x)(1+x)
1+3x+2x
2
=
1−2x
A
+
1−x
B
+
1+x
C
=>1+3x+2x
2
=A(1−x)(1+x)+B(1−2x)(1+x)+C(1−2x)(1−x)
For x=1,=>1+3(1)+2(1)
2
=A(0)+B(1−2)(1+1)+C(0)
=>6=−2B
=>B=−3 −(1)
For x=−1,=>1+3(−1)+2(−1)
2
=A(0)+B(0)+C(1−2(−1))(1−(−1))
=>0=6C
=>C=0 −(2)
For x=
2
1
,=>1+3(
2
1
)+2(
2
1
)
2
=A(
2
1
)(1+
2
1
)+B(0)+C(0)
=>3=
4
3
A
=>A=4 −(3)
Therefore , from (1),(2) and (3) =>
(1−2x)(1−x)(1+x)
1+3x+2x
2
=
1−2x
4
Answered by
8
3x−2(x+1)2(x−3)2=A(x+1)2+Bx+1+C(x−3)2+Dx−3
Multiply buy the least common denominator ( (x+1)2(x−3)2 ) to eliminate the fractions.
3x−2=A(x−3)2+B(x+1)(x−3)2+C(x+1)2+D(x+1)2(x−3)
Multiply the binomials (sorry about the long equations)
3x−2=A(x2−6x+9)+B(x3−5x2+3x+9)+...
...C(x2+2x+1)+D(x3−x2−5x−3)
Distribute A, B, C, and D:
3x−2=Ax2−6Ax+9A+Bx3−5Bx2+3Bx+9B+...
...Cx2+2Cx+C+Dx3−Dx2−5Dx−3D
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