To simplify $\sqrt{56z^7}$, we can break down the expression under the square root and remove any perfect squares.

Step-by-Step Solution:

1. Factor 56 into prime factors: $56 = 4 \times 14 = 4 \times 2 \times 7 = 2^2 \times 14$ So we can rewrite $\sqrt{56z^7}$ as: $\sqrt{4 \times 14 \times z^7}$

2. Separate perfect squares: Since $4$ is a perfect square ($2^2$), we can take it out of the square root: $\sqrt{4 \times 14 \times z^7} = \sqrt{4} \cdot \sqrt{14} \cdot \sqrt{z^7}$ $= 2 \cdot \sqrt{14} \cdot \sqrt{z^7}$

3. Simplify $\sqrt{z^7}$: We can split $z^7$ as $z^6 \cdot z$ because $z^6$ is a perfect square: $\sqrt{z^7} = \sqrt{z^6 \cdot z} = \sqrt{z^6} \cdot \sqrt{z} = z^3 \cdot \sqrt{z}$

4. Combine everything: Substitute back to get: $\sqrt{56z^7} = 2 \cdot z^3 \cdot \sqrt{14z}$

Final Answer:

$\sqrt{56z^7} = 2z^3\sqrt{14z}$

Would you like further details on any step?

Here are five related questions for further practice:

1. Simplify $\sqrt{72x^8}$ by removing all perfect squares.
2. How would you simplify $\sqrt{98y^{10}}$?
3. What is the simplified form of $\sqrt{200k^5}$?
4. If given $\sqrt{45m^6}$, how would you remove the perfect squares?
5. How would you approach $\sqrt{64n^9}$ to simplify it?

Tip: When simplifying square roots, always look for the largest perfect square factor to make the process quicker.