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

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

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