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Write the partial fraction decomposition fro the ration expression below

-2x + 2y +3z = 7

x - y = -5

y + 4z = -1

  1. (5x - 2)/(x-1)^2
  2. (x^(3) + x^(2) + x + 2)/(x^(4) + x^(2))

Verified Answer :

LammettHash

\(\dfrac{5x-2}{(x-1)^2}=\dfrac a{x-1}+\dfrac b{(x-1)^2}\)
\(5x-2=a(x-1)+b\)

Since \(5x-2=5x-5+3=5(x-1)+3\), it follows that \(a=5\) and \(b=3\), so

\(\dfrac{5x-2}{(x-1)^2}=\dfrac5{x-1}+\dfrac3{(x-1)^2}\)

\(\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac{x^3+x+x+2}{x^2(x^2+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac{cx+d}{x^2+1}\)
\(x^3+x^2+x+2=ax(x^2+1)+b(x^2+1)+(cx+d)x^2\)
\(x^3+x^2+x+2=(a+c)x^3+(b+d)x^2+ax+b\)

When \(x=0\), you find that \(b=2\). It's also clear that \(a=1\). So the remaining constants must be \(1+c=1\implies c=0\) and \(2+d=1\implies d=-1\), and so you get

\(\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac1x+\dfrac2{x^2}-\dfrac1{x^2+1}\)

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