Answer
$\dfrac{5x^2 + 6x + 12}{(x - 3)(x + 2)(x + 2)}$
Restriction: $x \ne -2, 3$
Work Step by Step
Factor all expressions in the original exercise:
$\dfrac{5x}{(x - 3)(x + 2)} + \dfrac{4}{(x + 2)(x + 2)}$
Before performing the subtraction, find the least common denominator (LCD) of the two fractions. The LCD is $(x-3)(x+2)(x+2)$. Convert the original fractions to equivalent fractions using the LCD:
$=\dfrac{5x(x + 2)}{(x - 3)(x + 2)(x + 2)} - \dfrac{4(x - 3)}{(x + 2)(x + 2)(x - 3)}$
Multiply to simplify:
$=\dfrac{5x^2 + 10x}{(x - 3)(x + 2)(x + 2)} - \dfrac{4x - 12}{(x - 3)(x + 2)(x + 2)}$
Subtract the fractions:
$=\dfrac{5x^2 + 10x - (4x - 12)}{(x - 3)(x + 2)(x + 2)}$
$=\dfrac{5x^2 + 10x - 4x + 12}{(x - 3)(x + 2)(x + 2)}$
$=\dfrac{5x^2 + 6x + 12}{(x - 3)(x + 2)(x + 2)}$
Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$.
Set the factors in the denominators equal to $0$ to find restrictions:
First factor:
$x - 3 = 0$
$x = 3$
Second factor:
$x + 2 = 0$
$x = -2$
Thus, the restrictions are: $x \ne -2, 3$
