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

Question

asked Mar 3, 2021 15.3k views

1 Answer

Leonid Dashko · Answer 1 · 2021-03-09T02:16:09+0000

Answer:

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:

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