Answer
$\dfrac{x + 1}{x - 4}$
Restrictions: $x \ne -3, \frac{1}{2}, 2, 4$
Work Step by Step
Dividing by a fraction means multiplying by its reciprocal. Let's rewrite this problem to reflect this:
$\dfrac{x^2 - x - 2}{2x^2 - 5x + 2} \cdot \dfrac{2x^2 + 5x - 3}{x^2 - x - 12}$
Let's factor the polynomials:
$x^2 - x - 2$ = $(x - 2)(x + 1)$
$2x^2 - 5x + 2$ = $(2x - 1)(x - 2)$
$2x^2 + 5x - 3$ = $(2x - 1)(x + 3)$
$x^2 - x - 12$ = $(x - 4)(x + 3)$
Multiply the two expressions using the factors instead of the polynomials:
$\dfrac{(x - 2)(x + 1)(2x - 1)(x + 3)}{(2x - 1)(x - 2)(x - 4)(x + 3)}$
Factor out $x - 2$, $2x - 1$, $x + 3$:
$\dfrac{x + 1}{x - 4}$
To check what restrictions we have for the variables, we need to find which values of the variables will make the denominators of the original expressions equal $0$, which would make the fraction undefined. Let's set the denominators equal to $0$, and then solve:
First denominator:
$(2x - 1)(x - 2) = 0$
Set each factor equal to zero:
First factor:
$2x - 1 = 0$
Add $1$ to both sides:
$2x = 1$
Divide both sides by $2$:
$x = \frac{1}{2}$
Second factor:
$x - 2 = 0$
Add $2$ to each side:
$x = 2$
Second denominator:
$(x - 4)(x + 3) = 0$
Set each factor equal to zero:
First factor:
$x - 4 = 0$
Add $4$ to each side:
$x = 4$
Second factor:
$x + 3 = 0$
Subtract $3$ from each side of the equation:
$x = -3$
Restrictions: $x \ne -3, \frac{1}{2}, 2, 4$
