Question

resolve into partial fraction. 2 X + 3 / x
$2x + 3 \div x { - 2x - 3}$
​

Answer 1

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Resolve into partial fractions:

x(x−1)(2x+3)

2x

3

+x

2

−x−3

.

Medium

Solution

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x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x

A

+

(x−1)

B

+

(2x+3)

C

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x(x−1)(2x+3)

A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x(x−1)(2x+3)

A(2x

2

+x−3)+B(2x

2

+3x)+C(x

2

−x)

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x(x−1)(2x+3)

2Ax

2

+Ax−3A+2Bx

2

+3Bx+Cx

2

−Cx

2x

3

+x

2

−x−3=(2A+2B+C)x

2

+(A+3B−C)x−3A

⇒ 2A+2B+C=1 --- ( 1 )

⇒ A+3B−C=−1 ---- ( 2 )

⇒ 3A=3 --- ( 3 )

By solving ( 1 ), ( 2 ) and ( 3 ) we get,

A=1,B=

5

−3

,C=

5

1

∴

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x

1

−

5(x−1)

3

+

5(2x+3)

1

Answer 2

x(x−1)(2x+3)

2x

3

+x

2

−x−3

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x

A

+

(x−1)

B

+

(2x+3)

C

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x(x−1)(2x+3)

A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x(x−1)(2x+3)

A(2x

2

+x−3)+B(2x

2

+3x)+C(x

2

−x)

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x(x−1)(2x+3)

2Ax

2

+Ax−3A+2Bx

2

+3Bx+Cx

2

−Cx

2x

3

+x

2

−x−3=(2A+2B+C)x

2

+(A+3B−C)x−3A

⇒ 2A+2B+C=1 --- ( 1 )

⇒ A+3B−C=−1 ---- ( 2 )

⇒ 3A=3 --- ( 3 )

By solving ( 1 ), ( 2 ) and ( 3 ) we get,

A=1,B=

5

−3

,C=

5

1

∴

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=

x

1

−

5(x−1)

3

+

5(2x+3)

1