If x = 11, evaluate the polynomial expression x⁵ − 12x⁴ + 12x³ − 12x² + 12x − 1 and determine its numerical value.

Introduction / Context:
This question involves evaluating a higher degree polynomial at a specific value of x. Such expressions often have hidden patterns, but even without factoring, careful substitution and stepwise calculation lead to the correct value. This type of problem is useful for checking accuracy in polynomial arithmetic.

Given Data / Assumptions:

• The polynomial is x⁵ − 12x⁴ + 12x³ − 12x² + 12x − 1.
• We are given x = 11.
• We must compute the numerical value when x is substituted.
• All calculations use standard integer arithmetic.

Concept / Approach:
The direct approach is to substitute x = 11 into each term, compute powers of 11, multiply by coefficients, and then add or subtract the results. To reduce errors, we evaluate stepwise, grouping terms if possible. Alternatively, one may look for a possible factorisation pattern, but here substitution is straightforward and reliable.

Step-by-Step Solution:
Substitute x = 11 into x⁵: 11⁵ = 161051.Compute 12x⁴: 11⁴ = 14641, so 12 × 14641 = 175692.Compute 12x³: 11³ = 1331, so 12 × 1331 = 15972.Compute 12x²: 11² = 121, so 12 × 121 = 1452.Compute 12x: 12 × 11 = 132.Now assemble the expression: x⁵ − 12x⁴ + 12x³ − 12x² + 12x − 1.This becomes 161051 − 175692 + 15972 − 1452 + 132 − 1.Calculate stepwise: 161051 − 175692 = −14641.Then −14641 + 15972 = 1331.Next 1331 − 1452 = −121.Then −121 + 132 = 11.Finally 11 − 1 = 10.

Verification / Alternative check:
Notice that during computation the intermediate values 14641, 1331, and 121 appear, which are 11⁴, 11³, and 11² respectively. The pattern in the coefficients suggests a telescoping effect where many terms cancel, leaving a small final number. Recalculating the additions and subtractions confirms that the final result is 10. A quick calculator check would provide the same result, verifying accuracy.

Why Other Options Are Wrong:
11, 12, −10, and 0 would require different cancellation patterns among the terms. A small slip in adding or subtracting intermediate results could lead to one of these values, but careful recalculation shows that the correct final value is 10. Only option 10 matches the accurate evaluation of the polynomial at x = 11.

Common Pitfalls:
Common mistakes include miscomputing powers of 11, for example confusing 11³ and 11⁴, and mishandling negative signs in the successive subtractions and additions. To avoid errors, it is helpful to write each intermediate result clearly and to recalculate any suspicious step. Recognising that many terms are designed to cancel also motivates a more structured approach.

Final Answer:
The value of the polynomial when x = 11 is 10.