Write the partial fraction decomposition of the rational expression.
$\frac { 1 } { 100 x ^ { 2 } - 81 }$
A) $\frac { 1 } { 18 } \left( \frac { x - 10 } { 10 x - 9 } - \frac { x - 10 } { 10 x + 9 } \right)$
B) $\frac { 1 } { 100 x ^ { 2 } } - \frac { 1 } { 81 }$
C) $\frac { 1 } { 18 } \left( \frac { 1 } { 10 x - 9 } - \frac { 1 } { 10 x + 9 } \right)$
D) $\frac { 1 } { 9 } \left( \frac { 1 } { 10 x + 9 } - \frac { 1 } { 10 x - 9 } \right)$
E) $\frac { 1 } { 9 } \left( \frac { 1 } { 10 x - 9 } - \frac { 1 } { 10 x + 9 } \right)$

Question

Quizplus · Accepted Answer

We can factor the denominator as a difference of squares: $100x^2-81=(10x+9)(10x-9)$. Then we can write:

$$\frac{1}{100x^2-81}=\frac{A}{10x-9}+\frac{B}{10x+9}$$

for some constants $A$ and $B$. Multiplying both sides by the denominator, we get:

$$1=A(10x+9)+B(10x-9)$$

We can solve for $A$ and $B$ by plugging in convenient values of $x$. For example, if we plug in $x=0$, we get $1=9A-9B$, so $A-B=\frac{1}{9}$. If we plug in $x=1$, we get $1=19A+1B$, so $19A+B=1$. Solving this system of equations (for example, by multiplying the first equation by 19 and adding to the second equation) gives $A=\frac{1}{18}$ and $B=-\frac{1}{18}$. Therefore,

$$\frac{1}{100x^2-81}=\frac{1}{18}\left(\frac{1}{10x-9}-\frac{1}{10x+9}\right)$$

which matches choice (C).

Write the Partial Fraction Decomposition of the Rational Expression 1100x2−81\frac { 1 } { 100 x ^ { 2 } - 81 }100x2−811​

Write the Partial Fraction Decomposition of the Rational Expression $\frac { 1 } { 100 x ^ { 2 } - 81 }$