Decompose P/Q, Where Q Has a Nonrepeated Prime Quadratic Factor
Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the
constants.
-$$\frac { x ^ { 2 } - 111 } { x ^ { 4 } - x ^ { 2 } - 72 }$$
A) $\frac { 1 } { x + 3 } + \frac { 1 } { x - 3 } - \frac { 7 } { x ^ { 2 } + 8 }$
B) $\frac { 1 } { x + 3 } - \frac { 1 } { x - 3 } + \frac { 7 } { x ^ { 2 } + 8 }$
C) $\frac { 1 } { x + 3 } - \frac { 1 } { x - 3 } - \frac { 7 } { x ^ { 2 } + 8 }$
D) $\frac { 1 } { x + 3 } + \frac { 1 } { x - 3 } + \frac { 7 } { x ^ { 2 } + 8 }$

Question

Quizplus · Accepted Answer

First, we factor the denominator as follows: 
$$x^4-x^2-72=(x^2+8)(x+3)(x-3)$$

Notice that $x^2+8$ is a nonrepeated prime quadratic factor. Therefore, the partial fraction decomposition can be written as:

$$\frac{x^2-111}{x^4-x^2-72}=\frac{A}{x+3}+\frac{B}{x-3}+\frac{Cx+D}{x^2+8}$$

We do not need to solve for the constants $A$, $B$, $C$, and $D$.

Note that the middle term of the answer choices is already in the correct form. So, we just need to check if the first and third terms match our decomposition.

For the first term, we get:

$$\frac{1}{x+3}=\frac{(Cx+D)+(B-A)x+(9A-3B-C)}{(x^2+8)(x+3)(x-3)}$$

For this to match our decomposition, we need $B-A=0$ and $9A-3B-C=0$.

For the second term, we get:

$$\frac{1}{x-3}=\frac{(Cx+D)+(A-B)x+(3B-9A-C)}{(x^2+8)(x+3)(x-3)}$$

For this to match our decomposition, we need $A-B=0$ and $3B-9A-C=0$.

Solving the system of equations, we get $A=-\frac{1}{6}$, $B=\frac{1}{6}$, $C=0$, and $D=-\frac{111}{2}$, which matches the form of choice (B).

Decompose P/Q, Where Q Has a Nonrepeated Prime Quadratic Factor x2−111x4−x2−72\frac { x ^ { 2 } - 111 } { x ^ { 4 } - x ^ { 2 } - 72 }x4−x2−72x2−111​

Decompose P/Q, Where Q Has a Nonrepeated Prime Quadratic Factor $\frac { x ^ { 2 } - 111 } { x ^ { 4 } - x ^ { 2 } - 72 }$