You have the function:

$f(x) = \left( 2x^3 + 5x^2 + 7x + 11 \right) \mod 26 + 1$

This expression involves both polynomial evaluation and modular arithmetic. To evaluate this function for a given value of $x$, you will:

1. Compute the cubic expression $2x^3 + 5x^2 + 7x + 11$.
2. Apply the modulus operation with 26, i.e., reduce the result modulo 26.
3. Add 1 to the result.

For example, if you want to find $f(3)$, you would:

1. Plug $x = 3$ into the cubic expression: $2(3)^3 + 5(3)^2 + 7(3) + 11 = 2(27) + 5(9) + 7(3) + 11 = 54 + 45 + 21 + 11 = 131$

2. Reduce $131 \mod 26$: $131 \div 26 = 5 \quad \text{(quotient)} \quad \text{remainder} = 131 - 5 \times 26 = 131 - 130 = 1$ Thus, $131 \mod 26 = 1$.

3. Finally, add 1: $1 + 1 = 2$

So, $f(3) = 2$.

Would you like to try more values for $x$, or would you like additional steps to be explained?

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Here are 5 related questions you might find interesting:

1. How do we compute polynomial functions under modular arithmetic for higher degree polynomials?
2. What is the significance of modular arithmetic in number theory and cryptography?
3. Can this function $f(x)$ be simplified in some way under modulo 26?
4. How do you handle modular inverses if we needed to solve for values of $x$ in such functions?
5. How do you apply similar functions in encryption algorithms like RSA or AES?

Tip: Modular arithmetic is often used in encryption because it keeps calculations bounded within a certain range, preventing overflow and ensuring manageable numbers.