Product Property of Square Roots: Number of terms

Notice that in the given problem, a multiplication is performed between three terms, one of which is:

$\sqrt{0}$and let's remember that the root (of any order) of the number 0 is 0, meaning that:

$\sqrt{0}=0$and since multiplying any number by 0 will always yield the result 0,

therefore the result of the multiplication in the problem is 0, meaning:

$\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{0}=\\ \sqrt{2}\cdot\sqrt{2}\cdot0=\\ \boxed{0}$and thus the correct answer is answer C.

In order to simplify the given expression we use two laws of exponents:

A. Defining the root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$B. The law of exponents for a product of numbers with the same base (in the opposite direction):

$x^n\cdot y^n =(x\cdot y)^n$

Let's start by definging the roots as exponents using the law of exponents shown in A:

$\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$Since we are multiplying between four numbers with the same exponents we can use the law of exponents shown in B (which also applies to a product of numbers with the same base) and combine them together in a product wit the same base which is raised to the same exponent:

$2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (2\cdot5\cdot2\cdot2)^{\frac{1}{2}}=\\ 40^{\frac{1}{2}}=\\ \boxed{\sqrt{40}}$In the last step we performed the product which is in the base, then we used again the definition of the root as an exponent shown earlier in A (in the opposite direction) to return to writing the root.

Therefore, note that the correct answer is answer C.