Answer to Question #339510 in Calculus for Momo

Question #339510

5. Decompose


(i)


x2 +x +1


(x +3)(x2 −x +1)



(ii)


x4 −x3 −2x2 +4x +1


x (x −1)2

1
Expert's answer
2022-05-16T16:39:57-0400

(i)


x2+x+1(x+3)(x2x+1)=Ax+3+Bx+Cx2x+1\dfrac{x^2+x+1}{(x +3)(x^2 −x +1)}=\dfrac{A}{x+3}+\dfrac{Bx+C}{x^2 −x +1}

=A(x2x+1)+(Bx+C)(x+3)(x+3)(x2x+1)=\dfrac{A(x^2-x+1)+(Bx+C)(x+3)}{(x +3)(x^2 −x +1)}

=Ax2Ax+A+Bx2+3Bx+Cx+3C(x+3)(x2x+1)=\dfrac{Ax^2-Ax+A+Bx^2+3Bx+Cx+3C}{(x +3)(x^2 −x +1)}

x2:A+B=1x^2:A+B=1

x1:A+3B+C=1x^1:-A+3B+C=1

x0:A+3C=1x^0:A+3C=1

A=13CA=1-3C

B=3CB=3C

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


A=713A=\dfrac{7}{13}

B=613B=\dfrac{6}{13}

C=213C=\dfrac{2}{13}

x2+x+1(x+3)(x2x+1)=713x+3+613x+213x2x+1\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)


x4x32x2+4x+1x(x1)2=x2(x22x+1)x(x1)2\dfrac{x^4 −x^3 −2x^2 +4x +1}{x(x -1)^2}=\dfrac{x^2(x^2-2x+1)}{x(x -1)^2}

+x(x22x+1)x(x1)2+x2+3x+1x(x1)2+\dfrac{x(x^2-2x+1)}{x(x -1)^2}+\dfrac{-x^2+3x+1}{x(x -1)^2}

=x+1+x2+3x+1x(x1)2=x+1+\dfrac{-x^2+3x+1}{x(x -1)^2}

x2+3x+1x(x1)2=Ax+Bx1+C(x1)2\dfrac{-x^2+3x+1}{x(x -1)^2}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^2}

=A(x1)2+Bx(x1)+Cxx(x1)2=\dfrac{A(x-1)^2+Bx(x-1)+Cx}{x(x -1)^2}

=Ax22Ax+A+Bx2Bx+Cxx(x1)2=\dfrac{Ax^2-2Ax+A+Bx^2-Bx+Cx}{x(x -1)^2}

x2:A+B=1x^2:A+B=-1

x1:2AB+C=3x^1:-2A-B+C=3

x0:A=1x^0:A=1

A=1A=1

B=2B=-2

C=3C=3

x2+3x+1x(x1)2=1x+2x1+3(x1)2\dfrac{-x^2+3x+1}{x(x -1)^2}=\dfrac{1}{x}+\dfrac{-2}{x-1}+\dfrac{3}{(x-1)^2}

Therefore

x4x32x2+4x+1x(x1)2\dfrac{x^4 −x^3 −2x^2 +4x +1}{x(x -1)^2}

=x+1+1x+2x1+3(x1)2=x+1+\dfrac{1}{x}+\dfrac{-2}{x-1}+\dfrac{3}{(x-1)^2}


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!

Leave a comment