Answer to Question #325139 in Calculus for Busi

Question

Decompose

(i)

(x^2+X+1)/(X+3)(x^2-x+1)

into partial fractions (show all the steps).

Answer 1

$\frac{x^2+x+1}{(x+3)(x^2-x+1)}=$ $\frac{A}{x+3}+\frac{Bx+C}{x^2-x+1}$

$x^2+x+1=A(x^2-x+1)+(Bx+C)(x+3)$

Equating coefficients, we get:

$1=A+B$ (1)

$1=-A+3B+C$ (2)

$1=A+3C$ (3)

Substituting (1) into (2) gives

$1=-A+3(1-A)+C=-4A+C+3$ (4)

Now let’s substitute (3) into (4)

$1=-4(1-3C)+C+3=13C-1$

$C=\frac{2}{13}$

$A=1-3C=1-\frac{6}{13}=\frac{7}{13}$

$B=1-A=1-\frac{7}{13}=\frac{6}{13}$

$\frac{x^2+x+1}{(x+3)(x^2-x+1)}=$$\frac{1}{13}\frac{7}{x+3}+\frac{1}{13}\frac{6x+2}{x^2-x+1}$