If x is an integer, which of the following CANNOT be equal to zero?

1. Translate the problem requirements: We need to find which expression among the five choices can NEVER equal zero when x is any integer. This means we're looking for an expression that will always be positive or always be negative, regardless of what integer value we choose for x.
2. Test each expression systematically: For each answer choice, determine if there exists at least one integer value of x that makes the expression equal to zero. If we can find such a value, then that expression CAN be zero.
3. Identify the expression that remains non-zero: The expression that cannot equal zero will be the one where no integer value of x will make it zero. This will be our answer.
4. Verify the mathematical reasoning: Confirm why the chosen expression can never be zero by understanding the mathematical property that prevents it from reaching zero.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this question is really asking us. We have five different expressions, and we need to find which one can NEVER equal zero, no matter what integer we put in for x.

Think of it this way: imagine we're testing each expression like a machine that takes any whole number as input. Most of these expressions will output zero for at least one input, but one of them will stubbornly refuse to ever give us zero.

So our job is to check each expression and see if we can find at least one integer value of x that makes it equal to zero. If we can find such a value, then that expression CAN be zero. The one where we can't find any such value is our answer.

Process Skill: TRANSLATE

2. Test each expression systematically

Let's go through each answer choice and try to make it equal zero:

Choice A: \(\mathrm{x + 1}\)
Can we make \(\mathrm{x + 1 = 0}\)? Sure! If \(\mathrm{x = -1}\), then \(\mathrm{(-1) + 1 = 0}\). So this expression CAN be zero.

Choice B: \(\mathrm{x - 1}\)
Can we make \(\mathrm{x - 1 = 0}\)? Absolutely! If \(\mathrm{x = 1}\), then \(\mathrm{1 - 1 = 0}\). So this expression CAN be zero.

Choice C: \(\mathrm{x^2 + 1}\)
Can we make \(\mathrm{x^2 + 1 = 0}\)? Let's think about this carefully. For this to equal zero, we would need \(\mathrm{x^2 = -1}\). But wait - when we square any real number, we always get a positive result or zero. We can never get a negative number from squaring. So \(\mathrm{x^2}\) can never equal \(\mathrm{-1}\), which means \(\mathrm{x^2 + 1}\) can never equal zero.

Choice D: \(\mathrm{x^2 - 1}\)
Can we make \(\mathrm{x^2 - 1 = 0}\)? Yes! This means \(\mathrm{x^2 = 1}\), which happens when \(\mathrm{x = 1}\) or \(\mathrm{x = -1}\). Let's check: \(\mathrm{1^2 - 1 = 0}\) ✓. So this expression CAN be zero.

Choice E: \(\mathrm{|x + 1|^2}\)
Can we make \(\mathrm{|x + 1|^2 = 0}\)? This equals zero when \(\mathrm{|x + 1| = 0}\), which happens when \(\mathrm{x + 1 = 0}\), so \(\mathrm{x = -1}\). Let's check: \(\mathrm{|-1 + 1|^2 = |0|^2 = 0}\) ✓. So this expression CAN be zero.

Process Skill: CONSIDER ALL CASES

3. Identify the expression that remains non-zero

From our testing above, we found that:

• Choices A, B, D, and E can all equal zero for specific integer values of x
• Choice C (\(\mathrm{x^2 + 1}\)) cannot equal zero for any integer value of x

Therefore, \(\mathrm{x^2 + 1}\) is the expression that CANNOT be equal to zero.

4. Verify the mathematical reasoning

Let's double-check our reasoning for Choice C. The expression \(\mathrm{x^2 + 1}\) represents "a perfect square plus 1."

Since x is an integer, \(\mathrm{x^2}\) (x squared) will always be non-negative:

• If \(\mathrm{x = 0}\), then \(\mathrm{x^2 = 0}\)
• If x is any positive integer, then \(\mathrm{x^2}\) is positive
• If x is any negative integer, then \(\mathrm{x^2}\) is still positive (negative times negative equals positive)

So the smallest possible value of \(\mathrm{x^2}\) is 0. This means the smallest possible value of \(\mathrm{x^2 + 1}\) is \(\mathrm{0 + 1 = 1}\).

Since \(\mathrm{x^2 + 1 \geq 1}\) for all integers x, it can never equal zero.

Final Answer

The answer is C. The expression \(\mathrm{x^2 + 1}\) cannot be equal to zero because \(\mathrm{x^2}\) is always non-negative for any integer x, making \(\mathrm{x^2 + 1}\) always at least 1.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "CANNOT be equal to zero"
Students often misread this as "CAN be equal to zero" and look for expressions that have solutions, rather than the one that has NO solutions. This leads them to pick expressions like \(\mathrm{x + 1}\) or \(\mathrm{x - 1}\) that clearly can equal zero.

2. Not recognizing this as a "test all cases" problem
Some students try to solve this algebraically by setting each expression equal to zero simultaneously, rather than understanding they need to test each expression individually to see which one can never equal zero.

Errors while executing the approach

1. Forgetting that \(\mathrm{x^2}\) is always non-negative
When testing \(\mathrm{x^2 + 1 = 0}\), students may attempt to solve \(\mathrm{x^2 = -1}\) without recognizing that squaring any real number (including integers) always gives a non-negative result. They might think there could be some integer solution.

2. Incorrectly handling absolute value expressions
For choice E, \(\mathrm{|x + 1|^2}\), students might get confused about when absolute values equal zero, or make errors in computing \(\mathrm{|x + 1|^2}\) when \(\mathrm{x = -1}\), potentially thinking this expression cannot be zero.

3. Making sign errors when testing negative integers
When checking if \(\mathrm{x^2 - 1}\) can equal zero, students might make computational errors with \(\mathrm{x = -1}\), getting confused about \(\mathrm{(-1)^2 = 1}\), and incorrectly conclude this expression cannot be zero.

Errors while selecting the answer

1. Selecting an expression that CAN be zero instead of CANNOT
After correctly identifying which expressions can and cannot be zero, students may accidentally select one of the expressions that CAN be zero (like A or B) due to the negative phrasing in the question stem.