Answer
$2$
Restrictions: $x \ne -3, 1$
Work Step by Step
Rewrite the exercise using the division ($\div$) symbol:
$(2x + 6) \div \left[{(x - 1)^{-1}(x^2 + 2x - 3)}\right]$
Move the expression with the negative exponent from the denominator to the numerator:
$=(2x + 6)(x - 1) \div \left[{x^2 + 2x - 3}\right]$
Invert the divisor and change the operation to multiplication:
$=(2x + 6)(x - 1) \cdot \dfrac{1}{x^2 + 2x - 3}$
Rewrite as one rational expression:
$=\dfrac{(2x + 6)(x - 1)}{x^2 + 2x - 3}$
Factor each polynomial completely:
$=\dfrac{2(x + 3)(x - 1)}{(x + 3)(x - 1)}$
Cancel common factors in the numerator and denominator:
$=2$
Restrictions occur when the expression becomes undefined, meaning when the denominator is $0$.
To find the restrictions, set each expression in the denominators of the original rational expressions as well as in the reciprocal equal to zero and solve:
Restriction:
$x + 3 = 0$
$x = -3$
Restriction:
$x - 1 = 0$
$x = 1$
Therefore, the restrictions are $x \ne -3, 1$
