Answer to Question #331664 in Algebra for Busi

Question #331664

Decompose



(i)x^4-x^3-2x^2+4x+1/x(x-1)^2




(5)



into partial fractions (show all the steps)

1
Expert's answer
2022-04-27T14:35:26-0400

First, let's detach the whole part of the fraction:

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

Second, decompose the fraction "\\frac{-2x^2+4x+1}{x(x-1)^2}" to partial fractions:

"\\frac{-2x^2+4x+1}{x(x-1)^2}=\\frac{A}{x}+\\frac{B}{x-1}+\\frac{C}{(x-1)^2}"

Let's multiply both parts of this equation by "x(x-1)^2" and equate the coefficients at equal powers of x:

"-2x^2+4x+1=A(x-1)^2+Bx(x-1)+\\\\\n+Cx=Ax^2-2Ax+A+Bx^2-Bx+\\\\\n+Cx=(A+B)x^2+\\\\\n+(-2A-B+C)x+A\\\\\n\\begin{cases}\nA+B=-2\\\\\n-2A-B+C=4\\\\\nA=1\n\\end{cases}\\Rarr\\\\\n\\Rarr\\begin{cases}\nA=1\\\\\nB=-2-A=-2-1=-3\\\\\n-B+C=4+2A=4+2\\cdot1=6\n\\end{cases}\\Rarr\\\\\n\\Rarr\\begin{cases}\nA=1\\\\\nB=-3\\\\\nC=6+B=6+(-3)=3\n\\end{cases}"


"\\frac{x^4-x^3-2x^2+4x+1}{x(x-1)^2}=x+1+\\frac{1}{x-1}+\\\\\n+\\frac{1}{x}-\\frac{3}{x-1}+\\frac{3}{(x-1)^2}=\\\\\nx+1+\\frac{1}{x}-\\frac{2}{x-1}+\\frac{3}{(x-1)^2}"


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