Answer
$\dfrac{x^2 - 24}{3x(x + 3)}$
Restrictions: $x \ne -3, 0$
Work Step by Step
Before subtracting the two expressions, we need to find the least common denominator. Let's factor the denominators completely first:
$\dfrac{x}{3(x + 3)} - \dfrac{8}{x(x + 3)}$
The least common multiple or denominator is the product of the prime factors of those polynomials in the denominator expressed to the greatest power present, which is $3x(x + 3)$.
Now, we have to multiply the numerator of each polynomial with whatever factor is lacking between its denominator and the least common multiple:
$\dfrac{(x)(x)}{3x(x + 3)} - \dfrac{(8)(3)}{3x(x + 3)}$
Multiply out the terms to simplify:
$\dfrac{x^2}{3x(x + 3)} - \dfrac{24}{3x(x + 3)}$
Now, we subtract the numerators and keep the denominator as-is:
$\dfrac{x^2 - 24}{3x(x + 3)}$
Now, we need to find the restrictions by seeing what values of the variable will make the denominator equal to zero because if the denominator of any rational expression is zero, the expression becomes undefined.
To find the restrictions on the variable, set each factor in the denominator equal to zero:
$3x(x + 3) = 0$
First factor:
$3x = 0$
Divide each side by $3$:
$x = 0$
Second factor:
$x + 3 = 0$
Subtract $3$ from each side:
$x = -3$
Third factor:
$x + 4 = 0$
Subtract $4$ from each side:
$x = -4$
Restrictions: $x \ne -3, 0$
