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

Answer:

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

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

Step-by-step explanation:

A.

Given that:

$\dfrac{x^4 -2x^3 +x^2+2x -1}{x^3 -2x +1}$

$\dfrac{x^4 -2x^3 +x^2+2x -1}{x^3 -2x +1} =\dfrac{(x^4 -2x^3+x^2) +(2x-1)}{x^3 -2x +1}$

$=\dfrac{(x^4 -2x^3+x^2)}{x^3 -2x +1} + \dfrac{(2x-1)}{x^3 -2x +1}$

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

$\dfrac{x^4-2x^3+x^2+2x-1}{x^2-2x+1 }= x^2 + \dfrac{2x-1}{(x-1)^2}---(1)$

Further, the partial fraction decomposition of $\dfrac{2x-1}{(x-1)^2}$ can be written as:

$\dfrac{2x-1}{(x-1)^2} = \dfrac{A}{x-1}+ \dfrac{B}{(x-1)^2}$

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

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

B.

Given that:

$\dfrac{x^2-1}{x^3+x^2+x}$

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

Therefore, the partial fraction of decomposition is:

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