Q. 29

Question

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

x2+2x+3(x+1)(x2+2x+4)

Step-by-Step Solution

Verified
Answer

The partial fraction decomposition of a rational expression is 

x2+2x+3(x+1)(x2+2x+4)=23x+1+13(x+1)(x2+2x+4)

1Step 1. Given information

Given rational expression is

x2+2x+3(x+1)(x2+2x+4)

2Step 2. Partial fraction decomposition

partial fraction decomposition of a rational expression 

P(x)Q(x)=A1x-a1+A2x-a2++Anx-anx2+2x+3(x+1)(x2+2x+4)=Ax+1+Bx+C(x2+2x+4)   (i)x2+2x+3(x+1)(x2+2x+4)=A(x2+2x+4)(x+1)(x2+2x+4)+Bx+Cx+1(x+1)(x2+2x+4)x2+2x+3=A(x2+2x+4)+Bx+Cx+1x2+2x+3=Ax2+2Ax+4A+Bx2+Bx+Cx+Cx2+2x+3=(A+B)x2+(2A+B+C)x+(4A+C)   (ii)

3Step 3. Values of numerator coefficients and constants

Compare the coefficient of x2in equation ii 

A+B=1B=1-A

Compare the constants in equation ii

3=4A+CC=3-4A

Compare the coefficient of in equation ii and Substitute the expression for B and C

2=2A+B+C2=2A+(1-A)+(3-4A)A=23

so

B=1-23=13C=3-423=13

4Step 4. partial fraction decomposition of a rational expression

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

x2+2x+3(x+1)(x2+2x+4)=Ax+1+Bx+C(x2+2x+4)x2+2x+3(x+1)(x2+2x+4)=23x+1+13x+13(x2+2x+4)x2+2x+3(x+1)(x2+2x+4)=23x+1+13(x+1)(x2+2x+4)

So the partial fraction decomposition of a rational expression is x2+2x+3(x+1)(x2+2x+4)=23x+1+13(x+1)(x2+2x+4)