Answer
$1$
Restrictions: $x \ne -2, 2, 3, -1$
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:
$\dfrac{x^2 - 5x + 6}{x^2 - 4} \cdot \dfrac{x^2 + 3x + 2}{x^2 - 2x - 3}$
Let's take each polynomial separately and factor them out:
$x^2 - 5x + 6$ = $(x - 3)(x - 1)$
$x^2 - 4$ = $(x - 2)(x + 2)$
$x^2 + 3x + 2$ = $(x + 2)(x + 1)$
$x^2 - 2x - 3$ = $(x - 3)(x + 1)$
Let's replace the polynomials with the factors in the expressions:
$\dfrac{(x - 3)(x - 2)}{(x - 2)(x + 2)} \cdot \dfrac{(x + 2)(x + 1)}{(x - 3)(x + 1)}$
Let's multiply the two expressions:
$\dfrac{(x - 3)(x - 2)(x + 2)(x + 1)}{(x - 2)(x + 2)(x - 3)(x + 1)}$
We can cancel out $x - 3$, $x - 2$, $x + 1$, $x + 2$. Everything cancels out, so we are left with $1$:
$1$
To find out if there are any restrictions on the variables, we need to find which values of $x$ will cause the denominator to equal zero, which would make the whole expression undefined. Let's set the denominator equal to zero and solve for $x$:
Denominator in first expression:
$(x - 2)(x + 2) = 0$
Set each factor equal to zero and solve:
$x - 2 = 0$ or $x + 2 = 0$
$x = 2$ or $x = -2$
Denominator in second expression:
$(x - 3)(x + 1) = 0$
Set each factor equal to zero and solve:
$x - 3 = 0$ or $x + 1 = 0$
$x = 3$ or $x = -1$
Restriction: $x \ne -2, 2, 3, -1$
