The problem asks which expression is needed to multiply both the numerator and denominator of the expression:

$\frac{\sqrt{x + 1} - 3}{8 - x}$

to obtain:

$-\frac{1}{3 + \sqrt{x + 1}}$

Analysis and Solution:

To simplify this, notice the denominator in the target expression is $3 + \sqrt{x + 1}$, while the original expression has a denominator of $8 - x$. If we multiply both the numerator and the denominator of the original fraction by the conjugate of $\sqrt{x + 1} - 3$, we should be able to simplify it into the required form.

The conjugate of $\sqrt{x + 1} - 3$ is $\sqrt{x + 1} + 3$. When we multiply these, we use the difference of squares formula:

$(\sqrt{x + 1} - 3)(\sqrt{x + 1} + 3) = (\sqrt{x + 1})^2 - 3^2 = (x + 1) - 9 = x - 8.$

This will allow the denominator $8 - x$ to simplify nicely, since:

$8 - x = -(x - 8).$

Thus, multiplying by $\sqrt{x + 1} + 3$ will result in the desired denominator of $3 + \sqrt{x + 1}$.

Conclusion:

The correct expression to multiply by is $\sqrt{x + 1} + 3$, which corresponds to the second option in the list:

$\boxed{\sqrt{x + 1} + 3}.$

Would you like more details or have any questions?

Here are five related questions:

1. How do you multiply expressions with radicals?
2. What is the difference of squares, and how does it apply to this problem?
3. How do you rationalize the denominator of a fraction involving square roots?
4. Can you explain how the conjugate of a binomial with a square root works?
5. Why is multiplying by the conjugate effective in simplifying expressions like this?

Tip: When simplifying expressions involving square roots, always consider using the conjugate to eliminate the square root in the denominator.