Answer
$\dfrac{3x^3 + x^2 - x}{x^2 - 2}$
Restrictions: $x \ne \pm \sqrt 2$
Work Step by Step
Factor all expressions in the original exercise:
$3x + \dfrac{x(x + 5)}{x^2 - 2}$
Before performing the addition, find the least common denominator (LCD) of the two fractions. The LCD is $x^2-2$.
Convert the original fractions to equivalent fractions using the LCD:
$\dfrac{3x(x^2 - 2)}{x^2 - 2} + \dfrac{x(x + 5)}{x^2 - 2}$
Multiply to simplify:
$=\dfrac{3x^3 - 6x}{x^2 - 2} + \dfrac{x^2 + 5x}{x^2 - 2}$
Add the numerators and retain the denominator:
$=\dfrac{3x^3 - 6x + x^2 + 5x}{x^2 - 2}$
Combine like terms:
$=\dfrac{3x^3 + x^2 - x}{x^2 - 2}$
Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$.
Set the denominator equal to $0$ to find restrictions:
$x^2 - 2 = 0$
$x^2 = 2$
Take the square root of each side of the equation:
$x = \pm \sqrt 2$
Restrictions: $x \ne \pm \sqrt 2$
