Answer
$\dfrac{3(x + 5)}{x + 3}$
Because we cannot have the denominator equal zero, $x$ cannot be $-6, 2, -3$.
Work Step by Step
First, we want to factor all polynomials to see if we can cancel out anything.
$x^2 + 3x - 10$ factors out to $(x + 5)(x - 2)$
$x^2 + 4x - 13$ factors out to $(x + 6)(x - 2)$
$3x + 18$ factors out to $3(x + 6)$
Let's go ahead and rewrite the problem using factorization:
$\dfrac{(x + 5)(x - 2)}{(x + 6)(x - 2)} \cdot \dfrac{3(x + 6)}{x + 3}$
Now, let's see what factors we can cancel out. We can take out $x - 2$ and $x + 6$. Let's see what we have left:
$\dfrac{x + 5}{1} \cdot \dfrac{3}{x + 3}$
Let's combine the two fractions:
$\dfrac{3(x + 5)}{x + 3}$
To check what restrictions we have for $x$, we need to find which values of $x$ will make the denominator $0$, which would make the fraction undefined.
Let's set each of the factors in the denominators of the original fractions equal to $0$:
First factor:
$x + 6 = 0$
Subtract $6$ from each side of the equation:
$x = -6$
Second factor:
$x - 2 = 0$
Add $2$ to each side of the equation:
$x = 2$
Third factor:
$x + 3 = 0$
Subtract $3$ from each side of the equation:
$x = -3$
Because we cannot have the denominator equal zero, $x$ cannot be $-6, 2, -3$.
