Write a factor that you can use to rationalize the denominator of $\frac{\sqrt{2}}{\sqrt{5}-8}$

Answer:
[tex]\sqrt{5}+8[/tex]
Step-by-step explanation:
To rationalize the denominator, you would need to multiply the numerator and denominator by its conjugate. The conjugate of [tex]\sqrt{5}-8[/tex] is [tex]\sqrt{5}+8[/tex], so, if you wanted to do the actual math:
[tex]\displaystyle \frac{\sqrt{2}}{\sqrt{5}-8}\\ \\\frac{\sqrt{2}}{\sqrt{5}-8}\cdot\frac{\sqrt{5}+8}{\sqrt{5}+8}\\ \\\frac{\sqrt{10}+8\sqrt{2}}{5-64}\\ \\\frac{\sqrt{10}+8\sqrt{2}}{-59}\\ \\\frac{\sqrt{10}-8\sqrt{2}}{59}[/tex]
This method works because [tex](\sqrt{5}-8)(\sqrt{5}+8)=(\sqrt{5}^2-8^2)[/tex], which is a difference of squares, and helps to eliminate the radicals in the denominator.
Hope this helps!