Answer to Question #223096 in Analytic Geometry for yarris

Question

Decompose the following into partial fractions

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

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

Answer 1

Solution 3:

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

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

Equating constants

$6=\mathrm{A}$  

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

$5=A+B\\ B=-1$

Equating coefficients of  $x$

$20=2 A+B+C\\ \Rightarrow 20=12-1+C \\ \Rightarrow C=9$  

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

Solution 4:

$\frac{x^2+2x+3}{(x+1)(x^2+2x+4)}\\ =\frac{x^2+2x+4-1}{(x+1)(x^2+2x+4)}\\ =\frac{1}{x+1}-\frac{1}{(x+1)(x^2+2x+4)}\\$

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

$\therefore \frac{1}{(x+1)(x^2+2x+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+2x+4}\\ \Rightarrow 1=A(x^2+2x+4)+(Bx+C)(x+1)\\ \Rightarrow 1=(A+B)x^2+(2A+B+C)x+4A+C$

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

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

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

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

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

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