Chapter 11.5, Problem 52AYU

In Problems 47-54, use the division algorithm to rewrite each improper rational expression as the sum of a polynomial and a proper rational expression. Find the partial fraction decomposition of the proper rational expression. Finally, express the improper rational expression as the sum of a polynomial and the partial fraction decomposition. $\frac{x^4+x^3-x+2}{x^2-2x+1}$

To determine

To find:

i. Rewrite improper rational expression as the sum of polynomial and a proper rational expression using the division algorithm.

ii. The partial fraction decomposition of the proper rational expression

iii. To express the improper rational expression as the sum of polynomial and the partial fraction decomposition.

Answer to Problem 52AYU

Solution:

$(x^2 + 3x + 5) + \frac{6}{(x - 1)} + \frac{3}{{(x - 1)}^2}$

Explanation of Solution

Given:

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

Formula used:

If $Q$ has repeated linear factor of the form ${(x - a)}^n, \text{n} \ge 2$ an integer, then, in the partial fraction decomposition of $P(x)/Q(x)$, takes the form:

$\frac{P(x)}{Q(x)} = \frac{A_1}{(x - a)} + \frac{A_2}{{(x - a)}^2} + \mathrm{...} + \frac{A_n}{{(x - a)}^n}$

where the numbers $A_1, A_2\mathrm{...}{A}_n$ are to be determined.

Calculation:

Divisional Algorithm

From the above division,

$\frac{x^4 + x^3 - x + 2}{x^2 - 2x + 1} = (x^2 + 3x + 5) + \frac{6x - 3}{x^2 - 2x + 1}$   -----Eq (1)

To find the partial fraction expression of $\frac{6x - 3}{x^2 - 2x + 1}$

The denominator $x^2 + 4x + 4$ can written be factored as $(x - 1)(x - 1)$.

The denominator has linear repeated factor and the partial fraction decomposition takes the form:

$\frac{6x - 3}{x^2 - 2x + 1} = \frac{6x - 3}{(x - 1)(x - 1)} = \frac{A}{(x - 1)} + \frac{B}{{(x - 1)}^2}$   -----Eq (2)

$= >6x - 3 = A(x - 1) + B$   -----Eq (3)

Substituting $x = 1$ we get $B = 3$

Equating coefficient of $x$ we get $\text{A} = 6$

$\frac{6x - 3}{x^2 - 2x + 1} = \frac{6}{(x - 1)} + \frac{3}{{(x - 1)}^2}.$.

The partial fraction decomposition of $\frac{6x - 3}{x^2 - 2x + 1}$ is $\frac{6}{(x - 1)} + \frac{3}{{(x - 1)}^2}$

Finally, $\frac{x^4 + x^3 - x + 2}{x^2 - 2x + 1} = (x^2 + 3x + 5) + \frac{6}{(x - 1)} + \frac{3}{{(x - 1)}^2}$