The problem asks to use the formulas of factorization to express the given polynomial as a product of factors. Let's analyze the expression:

$(7z^2)^2 - \left(6z^2 + 11^2\right)^2$

This is a difference of squares, which can be factored using the formula:

$a^2 - b^2 = (a - b)(a + b)$

Here:

• $a = 7z^2$
• $b = 6z^2 + 11^2 = 6z^2 + 121$

Substitute these values into the formula:

$(7z^2)^2 - (6z^2 + 121)^2 = \Big(7z^2 - (6z^2 + 121)\Big)\Big(7z^2 + (6z^2 + 121)\Big)$

Simplify Each Factor:

1. First factor: $7z^2 - (6z^2 + 121) = 7z^2 - 6z^2 - 121 = z^2 - 121$

2. Second factor: $7z^2 + (6z^2 + 121) = 7z^2 + 6z^2 + 121 = 13z^2 + 121$

Final Expression:

The fully factored form is:

$(7z^2)^2 - \left(6z^2 + 11^2\right)^2 = (z^2 - 121)(13z^2 + 121)$

Would you like me to verify or break down any part of this further? Here are 5 related questions to expand your understanding:

1. How does the difference of squares formula simplify expressions with squared terms?
2. Can the term $z^2 - 121$ be factored further (e.g., into smaller factors)?
3. Why is it important to simplify terms inside parentheses before applying formulas?
4. How can similar techniques be applied to cube-based expressions, like $a^3 - b^3$?
5. What would happen if $a^2 + b^2$ (sum of squares) was used instead? Why can't it be factored over real numbers?

Tip: Always look for standard algebraic identities, such as $(a+b)^2, (a-b)^2,$ or $a^2 - b^2$, when dealing with polynomials. These shortcuts save time!