Okay, let's tackle this problem step by step. The question is about finding the number of different possible rational roots of a polynomial with integer coefficients. The polynomial is given as 2x⁴ + a₃x³ + a₂x² + a₁x + 1 = 0, where a₃, a₂, and a₁ are integers. We need to determine how many distinct rational roots are possible for this polynomial. 

First, I remember that the Rational Root Theorem is a tool that helps us find possible rational roots of a polynomial with integer coefficients. The theorem states that any possible rational root, expressed in its lowest terms p/q, must have p as a factor of the constant term and q as a factor of the leading coefficient. 

In this polynomial, the constant term is 1, and the leading coefficient is 2. So, according to the Rational Root Theorem, the possible values for p (the numerator) are the factors of 1, which are ±1. The possible values for q (the denominator) are the factors of 2, which are ±1 and ±2. 

Therefore, the possible rational roots are the combinations of p/q where p is ±1 and q is ±1 or ±2. Let me list them out:

1. p = 1, q = 1 → 1/1 = 1
2. p = 1, q = -1 → 1/-1 = -1
3. p = 1, q = 2 → 1/2
4. p = 1, q = -2 → 1/-2 = -1/2
5. p = -1, q = 1 → -1/1 = -1
6. p = -1, q = -1 → -1/-1 = 1
7. p = -1, q = 2 → -1/2
8. p = -1, q = -2 → -1/-2 = 1/2

Now, let's look at these possibilities and see which are distinct. 

Looking at the list:

- 1 appears twice (from cases 1 and 6)
- -1 appears twice (from cases 2 and 5)
- 1/2, -1/2, 1/2, -1/2 (cases 3,4,7,8)

Wait, actually, let's list them all:

1, -1, 1/2, -1/2. So, even though there are multiple ways to get these values, they are all duplicates in terms of distinctness. 

So, the distinct possible rational roots are 1, -1, 1/2, and -1/2. That gives us four distinct rational roots. 

But wait, the question is asking for the number of different possible rational roots of this polynomial. The polynomial is of degree 4, so it can have up to 4 roots. However, since the coefficients a₃, a₂, a₁ are variables (they can be any integers), the actual roots can vary. But the question is about the possible rational roots that could exist for any such polynomial. 

So, regardless of the coefficients a₃, a₂, a₁, the possible rational roots are constrained by the Rational Root Theorem to be ±1, ±1/2. Therefore, the number of different possible rational roots is 4. 

But let me check if there are any other possible factors. The constant term is 1, so the factors of 1 are indeed ±1. The leading coefficient is 2, so factors are ±1, ±2. Therefore, the possible rational roots are ±1, ±1/2. That's four distinct possible rational roots. 

I don't see any other possibilities here. So, the answer should be 4. 

Wait, but the question says "different possible rational roots". Since the polynomial can have different coefficients, the actual roots can vary, but the possible candidates for rational roots are limited. So, the number of possible distinct rational roots is 4. 

Therefore, the answer is 4. 

But let me just verify once again. The Rational Root Theorem says that any rational root p/q must have p dividing the constant term and q dividing the leading coefficient. Here, constant term is 1, leading coefficient is 2. So p can be ±1, q can be ±1, ±2. Therefore, possible roots are ±1, ±1/2. That's four. 

Yes, that seems correct. So, the number of different possible rational roots is 4. 

**Final Answer**
The number of different possible rational roots is \boxed{4}.
Therefore, the number of different possible rational roots of this polynomial is $\boxed{4}$.
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