Answer
$\dfrac{x + 1}{x - 1}$
Restriction: $x \ne -2, -\frac{1}{2}, \frac{1}{2}, 1$
Work Step by Step
Let's factor out common factors first:
$\dfrac{2x^2 + 5x + 2)}{4x^2 - 1} \cdot \dfrac{2x^2 + x - 1)}{x^2 + x - 2}$
Now we can factor out polynomials:
$2x^2 + 5x + 2$ = $(2x + 1)(x + 2)$
$4x^2 - 1$ = $(2x + 1)(2x - 1)$
$2x^2 + x - 1$ = $(2x - 1)(x + 1)$
$x^2 + x - 2$ = $(x + 2)(x - 1)$
Multiply the two expressions using the factors instead of the polynomials:
$\dfrac{(2x + 1)(x + 2)(2x - 1)(x + 1)}{(2x + 1)(2x - 1)(x + 2)(x - 1)}$
Factor out $2x + 1$, $2x - 1$, $x + 2$:
$\dfrac{x + 1}{x - 1}$
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:
$(2x + 1)(2x - 1) = 0$
Set each factor equal to zero:
First factor:
$2x + 1 = 0$
Subtract $1$ from each side:
$2x = -1$
Divide each side by $2$:
$x = -\frac{1}{2}$
Second factor:
$2x - 1 = 0$
Add $1$ to each side:
$2x = 1$
Divide each side by $2$:
$x = \frac{1}{2}$
Second denominator:
$(x + 2)(x - 1) = 0$
Set each factor equal to zero:
First factor:
$x + 2 = 0$
Subtract $2$ from each side:
$x = -2$
Second factor:
$x - 1 = 0$
Add $1$ to each side of the equation:
$x = 1$
Restrictions: $x \ne -2, -\frac{1}{2}, \frac{1}{2}, 1$
