Answer to Question #339510 in Calculus for Momo

Question

5. Decompose

(i)

x2 +x +1

(x +3)(x2 −x +1)

(ii)

x4 −x3 −2x2 +4x +1

x (x −1)2

Answer 1

(i)

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

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

$=\dfrac{Ax^2-Ax+A+Bx^2+3Bx+Cx+3C}{(x +3)(x^2 −x +1)}$

$x^2:A+B=1$

$x^1:-A+3B+C=1$

$x^0:A+3C=1$

$A=1-3C$

$B=3C$

$-1+3C+9C+C=1$

$A=\dfrac{7}{13}$

$B=\dfrac{6}{13}$

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

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

(ii)

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

$+\dfrac{x(x^2-2x+1)}{x(x -1)^2}+\dfrac{-x^2+3x+1}{x(x -1)^2}$

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

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

$=\dfrac{A(x-1)^2+Bx(x-1)+Cx}{x(x -1)^2}$

$=\dfrac{Ax^2-2Ax+A+Bx^2-Bx+Cx}{x(x -1)^2}$

$x^2:A+B=-1$

$x^1:-2A-B+C=3$

$x^0:A=1$

$A=1$

$B=-2$

$C=3$

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

Therefore

$\dfrac{x^4 −x^3 −2x^2 +4x +1}{x(x -1)^2}$

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