1. What is the quotient in lowest terms?

We are given the division:

$\dfrac{\,\,\,\dfrac{x^2-3x-10}{x-3}\,\,\,}{\dfrac{x+2}{3x-9}}$

First, we rewrite the division as multiplication by the reciprocal:

$\dfrac{x^2-3x-10}{x-3} \times \dfrac{3x-9}{x+2}$

Now, we simplify the terms:

• Factor $x^2 - 3x - 10$. This factors as $(x - 5)(x + 2)$.
• Recognize that $3x - 9 = 3(x - 3)$.

Now substitute these factored forms:

$\dfrac{(x - 5)(x + 2)}{x - 3} \times \dfrac{3(x - 3)}{x + 2}$

Next, cancel out the common factors $(x - 3)$ and $(x + 2)$ from the numerator and denominator:

$3(x - 5)$

Thus, the quotient in its lowest terms is:

$\boxed{3(x - 5) = 3x - 15}$

2. What values of $x$ must we exclude from the domains of the expressions?

To find the excluded values, we check the factors that were canceled or could make any denominator zero:

• In $\dfrac{x^2 - 3x - 10}{x - 3}$, we have a denominator of $x - 3$, so $x \neq 3$.
• In $\dfrac{x+2}{3x-9}$, the denominator $3x - 9 = 3(x - 3)$ again gives $x \neq 3$.
• Additionally, the term $x + 2$ was in the denominator before canceling, so $x \neq -2$.

Thus, the values of $x$ to exclude are:

$\boxed{x = 3 \text{ and } x = -2}$

Correct answers:

• Choice A: $x = -2$
• Choice C: $x = 3$

Questions:

1. Why do we factor the expressions before simplifying the rational division?
2. How does multiplying by the reciprocal help in simplifying rational expressions?
3. What would the quotient be if we didn't cancel common factors?
4. Why must values that make any denominator zero be excluded?
5. How would the answer change if one of the factors were non-cancelable?

Tip:

Always check for restrictions on the variable from all denominators before and after simplifying!