Question

Write out the form of the partial fraction decomposition of the function (as in this example). A) x4 − 2x3 + x2 + 2x − 1/x2 − 2x + 1B) x2 - 1/x3 + x2 + x

Answer 1

A.
`\mathbf{(x^4-2x^3+x^2+2x-1)/(x^2-2x+1 )= x^2 + (A)/(x-1)+ (B)/((x-1)^2)}`

B.
`\mathbf{(x^2-1)/(x(x^2+x+1)) = (A)/(x)+ (Bx+C)/(x^2 +x+1)}`

Explanation:

A.

Given that:

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

`(x^4 -2x^3 +x^2+2x -1)/(x^3 -2x +1) =((x^4 -2x^3+x^2) +(2x-1))/(x^3 -2x +1)`

`=((x^4 -2x^3+x^2))/(x^3 -2x +1) + ((2x-1))/(x^3 -2x +1)`

`= (x^2(x^2-2x+1))/(x^2-2x+1)+ (2x-1)/((x-1)^2)`

`(x^4-2x^3+x^2+2x-1)/(x^2-2x+1 )= x^2 + (2x-1)/((x-1)^2)---(1)`

Further, the partial fraction decomposition of
`(2x-1)/((x-1)^2)` can be written as:

`(2x-1)/((x-1)^2) = (A)/(x-1)+ (B)/((x-1)^2)`

From equation one, replacing the values; we have the partial fraction decomposition of the function to be:

`\mathbf{(x^4-2x^3+x^2+2x-1)/(x^2-2x+1 )= x^2 + (A)/(x-1)+ (B)/((x-1)^2)}`

B.

Given that:

`(x^2-1)/(x^3+x^2+x)`

`(x^2-1)/(x^3+x^2+x) = (x^2-1)/(x(x^2+x+1))`

Therefore, the partial fraction of decomposition is:

`\mathbf{(x^2-1)/(x(x^2+x+1)) = (A)/(x)+ (Bx+C)/(x^2 +x+1)}`