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In Problems 13–46, write the partial fraction decomposition of each rational expression.

1(2x+3)(4x-1)

Short Answer

Expert verified

The partial fraction decomposition of a rational expression is

1(2x+3)(4x-1)=-17(2x+3)+27(4x-1)

Step by step solution

01

Step 1. Given information

Given rational expression is

1(2x+3)(4x-1)

02

Step 2. Partial fraction decomposition  

partial fraction decomposition of a rational expression

P(x)Q(x)=A1x-a1+A2x-a2+⋯+Anx-an1(2x+3)(4x-1)=A(2x+3)+B(4x-1)⋯(i)1(2x+3)(4x-1)=A(4x-1)(2x+3)(4x-1)+B(2x+3)(2x+3)(4x-1)1=A(4x-1)+B(2x+3)0x+1=(4A+2B)x+(3B-A)⋯(ii)

03

Step 3. Values of coefficients and constants of the numerator  

Compare the constants in equation ii

1=3B-AA=3B-1

Compare the coefficient of xin equation ii and Substitute the expression for A

0=4A+2B0=4(3B-1)+2BB=27

SoA=327-1=-17

04

Step 4. partial fraction decomposition of a rational expression   

Substitute the value of A, B,and C in the equation i

1(2x+3)(4x-1)=A(2x+3)+B(4x-1)1(2x+3)(4x-1)=-17(2x+3)+27(4x-1)

So the partial fraction decomposition of a rational expression is1(2x+3)(4x-1)=-17(2x+3)+27(4x-1)

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