Let's work through each expression one by one and simplify them.

a) $\sqrt{150}$

1. First, find the prime factorization of 150: $150 = 2 \times 3 \times 5^2$

2. Break the square root into factors: $\sqrt{150} = \sqrt{2 \times 3 \times 5^2}$

3. Apply the square root property: $\sqrt{150} = \sqrt{2} \times \sqrt{3} \times 5$

4. Simplify: $\sqrt{150} = 5\sqrt{6}$

b) $(2\sqrt{10})(5\sqrt{2})$

1. Multiply the constants and the square roots separately: $(2 \times 5) \times (\sqrt{10} \times \sqrt{2})$

2. Simplify: $10 \times \sqrt{20}$

3. Now, simplify $\sqrt{20}$: $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$

4. Final simplification: $10 \times 2\sqrt{5} = 20\sqrt{5}$

c) $3\sqrt{48} - 5\sqrt{27} + 2\sqrt{3} - \sqrt{32}$

1. Simplify each square root term:

   • $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
   • $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$
   • $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

2. Substitute back into the expression: $3\sqrt{48} - 5\sqrt{27} + 2\sqrt{3} - \sqrt{32} = 3(4\sqrt{3}) - 5(3\sqrt{3}) + 2\sqrt{3} - 4\sqrt{2}$

3. Simplify each term: $= 12\sqrt{3} - 15\sqrt{3} + 2\sqrt{3} - 4\sqrt{2}$

4. Combine like terms for $\sqrt{3}$: $(12\sqrt{3} - 15\sqrt{3} + 2\sqrt{3}) - 4\sqrt{2}$

   $= -1\sqrt{3} - 4\sqrt{2}$

   Final answer: $-\sqrt{3} - 4\sqrt{2}$

Final Results:

• a) $\sqrt{150} = 5\sqrt{6}$
• b) $(2\sqrt{10})(5\sqrt{2}) = 20\sqrt{5}$
• c) $3\sqrt{48} - 5\sqrt{27} + 2\sqrt{3} - \sqrt{32} = -\sqrt{3} - 4\sqrt{2}$

Would you like more detailed steps for any part of this? Or have any questions?

Here are some related questions:

1. How do you simplify square roots with perfect squares in them?
2. What is the square root property, and how is it used in simplification?
3. How do you handle terms with different square roots in an expression?
4. What is the significance of prime factorization in simplifying square roots?
5. Can you simplify expressions with cube roots in a similar way to square roots?

Tip: When simplifying square roots, always look for perfect square factors and pull them out of the square root to simplify the expression.