resolve 3x-1/(x-2)(x-3) into partial fraction?
Answers
Answered by
2
Step-by-step explanation:
Let
(x
2
+x+1)(x+1)
2
x
3
−3x−2
=
(x+1)
A
+
(x+1)
2
B
+
(x
2
+x+1)
Cx+D
⇒x
3
−3x−2=A(x+1)(x
2
+x+1)+B(x
2
+x+1)+(Cx+D)(x+1)
2
⇒x
3
−3x−2=A(x
3
+2x
2
+2x+1)+B(x
2
+x+1)+C(x
3
+2x
2
+x)+D(x
2
+2x+1)
On comparing coefficients
A+C=1,2A+B+2C+D=0,2A+B+C+2D=−3,A+B+D=−2
⇒A=−3,B=2,C=3,D=−1
Hence
(x
2
+x+1)(x+1)
2
x
3
−3x−2
=
(x+1)
−3
+
(x+1)
2
2
+
(x
2
+x+1)
3x−1
I hope it will help you plz follow
Answered by
0
Answer:
find the partial fraction of 3x+1/(x-1)(x-2)(x-3)
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