Answer
$\dfrac{y(y + 3)}{12(y + 4)}$
Restrictions: $x \ne 0; y \ne -4, 3$
Work Step by Step
Let's first take a look at the problem to see what we can cancel out from the numerator and denominator. We can factor out the polynomials to make this easier:
Let's look at the numerator first:
$xy^3 - 9xy$
Factor out $xy$ from the expression:
$xy(y^2 - 9)$
We can factor the binomial as the difference of two squares:
$xy(y + 3)(y - 3)$
Let's look at the polynomial in the denominator:
$12xy^2 + 12xy - 144x$
We can factor out $12x$:
$12x(y^2 + y - 12)$
We can factor the trinomial inside the parentheses further:
$12x(y + 4)(y - 3)$
Let's put the factors back into the rational expression:
$\frac{xy(y + 3)(y - 3)}{12x(y + 4)(y - 3)}$
Factor out common factor $y - 3$:
$\frac{xy(y + 3)}{12x(y + 4)}$
We can still factor out an $x$ from both numerator and denominator to cancel them out:
$\frac{y(y + 3)}{12(y + 4)}$
To check what restrictions we have for the variable, we need to find which values of the variable will make the denominator of the original expression equal $0$, which would make the fraction undefined. Let's set the denominator equal to $0$, and then solve:
$12x(y + 4)(y - 3) = 0$
Set each factor equal to zero, according to the zero product property:
First factor:
$12x = 0$
Divide each side by $12$:
$x = 0$
Second factor:
$y + 4 = 0$
Subtract $4$ from each side:
$y = -4$
Third factor:
$y - 3 = 0$
Add $3$ to each side:
$y = 3$
Restrictions: $x \ne 0; y \ne -4, 3$
