Question #223096

Decompose the following into partial fractions

3)5x2+20x+6 / x3+2x2+x


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


1
Expert's answer
2021-09-06T07:23:00-0400

Solution 3:

5x2+20x+6x(x+1)2=Ax+Bx+1+C(x+1)2\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}  

5x2+20x+6=A(x+1)2+Bx(x+1)+Cx\\ \Rightarrow 5 x^{2}+20 x+6=A(x+1)^{2}+B x(x+1)+C x  

Equating constants

6=A6=\mathrm{A}  

Equating coefficients of  x2\mathrm{x}^{2}

5=A+BB=15=A+B\\ B=-1

Equating coefficients of  xx

20=2A+B+C20=121+CC=920=2 A+B+C\\ \Rightarrow 20=12-1+C \\ \Rightarrow C=9  

5x2+20x+6x(x+1)2=6x1x+1+9(x+1)2\therefore \frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1)^{2}}



Solution 4:

x2+2x+3(x+1)(x2+2x+4)=x2+2x+41(x+1)(x2+2x+4)=1x+11(x+1)(x2+2x+4)\frac{x^2+2x+3}{(x+1)(x^2+2x+4)}\\ =\frac{x^2+2x+4-1}{(x+1)(x^2+2x+4)}\\ =\frac{1}{x+1}-\frac{1}{(x+1)(x^2+2x+4)}\\


Now decomposing 1(x+1)(x2+2x+4)\frac{1}{(x+1)(x^2+2x+4)}\\ as: Ax+1+Bx+Cx2+2x+4\frac{A}{x+1}+\frac{Bx+C}{x^2+2x+4}

1(x+1)(x2+2x+4)=Ax+1+Bx+Cx2+2x+41=A(x2+2x+4)+(Bx+C)(x+1)1=(A+B)x2+(2A+B+C)x+4A+C\therefore \frac{1}{(x+1)(x^2+2x+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+2x+4}\\ \Rightarrow 1=A(x^2+2x+4)+(Bx+C)(x+1)\\ \Rightarrow 1=(A+B)x^2+(2A+B+C)x+4A+C

A+B=0;2A+B+C=0;4A+C=1A=B2B+B+C=0B=C\therefore A+B=0;2A+B+C=0;4A+C=1\\ \Rightarrow A=-B \\ \therefore -2B+B+C=0\\ \Rightarrow B=C\\

So, 4(B)+B=14(-B)+B=1\\

B=13\Rightarrow B=-\frac{1}{3}\\

A=13,C=13\therefore A=\frac{1}{3},C=-\frac{1}{3}

So, 1(x+1)(x2+2x+4)=13(x+1)x+13(x2+2x+4)\frac{1}{(x+1)(x^2+2x+4)}=\frac{1}{3(x+1)}-\frac{x+1}{3(x^2+2x+4)}

Hence, x2+2x+3(x+1)(x2+2x+4)=43(x+1)x+13(x2+2x+4)\frac{x^2+2x+3}{(x+1)(x^2+2x+4)}=\frac{4}{3(x+1)}-\frac{x+1}{3(x^2+2x+4)}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS