Answer
$\dfrac{-x^2 + 3x + 2}{x(x + 3)(x - 1)(x + 1)}, \space x \ne -3, -1, 0, 1$
Work Step by Step
To subtract two rational expressions, make sure that the denominators are the same, meaning we need to find the least common denominator, or LCD.
First, factor all expressions in the original exercise:
$\dfrac{1}{(x - 1)(x + 1)} - \dfrac{2}{x(x + 3)}$
Before subtracting the two rational expressions, find the LCD.
The LCD is the product of all factors found in the denominators of both rational expressions.
After finding the LCD, multiply each fraction by the factor that is missing between the denominator and the LCD.
The LCD is $x(x + 3)(x - 1)(x + 1)$:
$\dfrac{1(x)(x + 3)}{x(x + 3)(x - 1)(x + 1)} - \dfrac{2(x - 1)(x + 1)}{x(x + 3)(x - 1)(x + 1)}$
Simplify the numerators:
$\dfrac{x^2 + 3x}{x(x + 3)(x - 1)(x + 1)} - \dfrac{2x^2 - 2}{x(x + 3)(x - 1)(x + 1)}$
Subtract and combine like terms:
$\begin{align*}
\dfrac{x^2+3x-(2x^2-2)}{x(x+3)(x-1)(x+1)}&=\dfrac{x^2+3x-2x^2+2}{x(x+3)(x-1)(x+1)}\\
\\&=\dfrac{-x^2 + 3x + 2}{x(x + 3)(x - 1)(x + 1)}
\end{align*}$
Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$.
Set each factor in the denominators equal to $0$ and solve for $x$ to find where the restrictions are:
First factor:
$x = 0$
Second factor:
$\begin{align*}
x + 3 &= 0\\
x&=-3
\end{align*}$
Third factor:
$\begin{align*}
x - 1 &= 0\\
x&=1
\end{align*}$
Fourth factor:
$\begin{align*}
x + 1 &= 0\\
x&=-1\
\end{align*}$
Thus, the restriction is:
$x \ne -3, -1, 0, 1$
