To solve this problem, we need to rationalize the denominator of the fraction:

$\frac{4}{\sqrt{5} - \sqrt{3}}$

Step 1: Multiply by the Conjugate

To rationalize the denominator, multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{5} - \sqrt{3}$ is $\sqrt{5} + \sqrt{3}$. So we multiply as follows:

$\frac{4}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{4(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}$

Step 2: Simplify the Denominator

The denominator is now a difference of squares:

$(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2$

Thus, the expression becomes:

$\frac{4(\sqrt{5} + \sqrt{3})}{2}$

Step 3: Simplify the Fraction

Now, divide both the numerator and denominator by 2:

$\frac{4(\sqrt{5} + \sqrt{3})}{2} = 2(\sqrt{5} + \sqrt{3})$

Final Answer:

The simplified expression is $2(\sqrt{5} + \sqrt{3})$, which corresponds to option b in the question.

Would you like any further explanation or details?

Here are 5 related questions to explore:

1. How do you multiply radicals with different bases?
2. What are other examples of conjugates in rationalizing denominators?
3. Can the same method be applied for rationalizing higher-degree roots?
4. How do you simplify the product of two binomials involving square roots?
5. What happens if both the numerator and denominator contain radicals?

Tip: Always check for opportunities to use conjugates when dealing with irrational denominators!