<Question>
A polynomial with integer coefficients is of the form
\[2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 0.\]Find the number of different possible rational roots of this polynomial.
</Question>

<think>
<Plan_1>
Okay, so I need to find the number of different possible rational roots for this polynomial: 2x⁴ + a₃x³ + a₂x² + a₁x + 1, where the coefficients a₃, a₂, a₁ are integers. Hmm, let me remember how to approach this.

First, I recall the Rational Root Theorem. If I have a polynomial with integer coefficients, then any possible rational root, expressed in lowest terms as p/q, must have p as a divisor of the constant term and q as a divisor of the leading coefficient. Right? So in this case, the constant term is 1, and the leading coefficient is 2.
So the possible values for p are the divisors of 1, which are ±1, and the possible values for q are the divisors of 2, which are ±1, ±2. Therefore, the possible rational roots are ±1, ±1/2.

Wait, but the leading coefficient is 2, so q can be 1 or 2. So the possible denominators are 1 and 2. So the possible rational roots are ±1/1 and ±1/2. That simplifies to ±1 and ±1/2. So there are four possible rational roots.

But hold on, the question says "the number of different possible rational roots of this polynomial". But the coefficients a₃, a₂, a₁ can be any integers.
So does that mean that the possible rational roots could vary depending on the coefficients? Wait, no. The Rational Root Theorem says that any rational root must be of the form p/q where p divides the constant term and q divides the leading coefficient, regardless of the other coefficients.

<prompt_1.1> Clarify whether the possible rational roots are fixed or depend on the coefficients. </prompt_1.1>

But let me check again. Let's list all possible p and q.

<prompt_1.2> Re-evaluate the possible rational roots by listing all p and q. </prompt_1.2>

But wait, the problem is stated as "Find the number of different possible rational roots of this polynomial". Since the coefficients a₃, a₂, a₁ can be any integers, does that affect the possible rational roots? Or is it that regardless of the coefficients, the possible rational roots are still only ±1 and ±1/2?
Because even if the coefficients change, the constant term is still 1 and the leading coefficient is 2, so the possible rational roots can't change. So the possible rational roots are fixed as ±1 and ±1/2, regardless of the values of a₃, a₂, a₁. Therefore, there are 4 different possible rational roots.
</Plan_1>

<Execution_1>
<execution_1.1>
Because the theorem is about possible rational roots of polynomials with integer coefficients, and even if the coefficients vary, the leading coefficient and constant term are fixed here. Wait, the leading coefficient is 2, constant term is 1. So regardless of a₃, a₂, a₁, the possible rational roots are always ±1, ±1/2.
</execution_1.1>

<execution_1.2>
The constant term is 1, so the divisors p are ±1. The leading coefficient is 2, so divisors q are ±1, ±2. But when we form p/q, we have to consider all combinations. However, ±1 divided by ±1 gives ±1, and ±1 divided by ±2 gives ±1/2. So the possible rational roots are ±1 and ±1/2.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
Therefore, there are 4 different possible rational roots. So the answer should be 4. But wait, let me make sure I'm not missing something here. The problem says "different possible rational roots of this polynomial". So if the polynomial can have different roots depending on the coefficients, but the question is about the possible rational roots that could exist for any such polynomial.
But according to the Rational Root Theorem, the possible rational roots are only those that divide the constant term over the leading coefficient, so ±1 and ±1/2. So regardless of the coefficients, these are the only possible candidates. Therefore, even if the coefficients vary, the possible rational roots can't be anything else. So the number of different possible rational roots is 4.

Wait, but another thought: maybe some of these possible roots are not actually roots for any choice of coefficients. For example, could 1 be a root for some choice of a₃, a₂, a₁? Let's check.

<prompt_2.1> Check if each possible rational root can be an actual root for some choice of coefficients. </prompt_2.1>

Therefore, each of the four possible rational roots is indeed possible, depending on the coefficients. So the number of different possible rational roots is 4.
Therefore, the answer is 4.

But let me verify again. Suppose we take each possible root and see if there exists a polynomial of the given form for which it is a root. For example, take x=1.

<prompt_2.2> Verify that each possible rational root can be an actual root for some specific choice of coefficients. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
If I plug x=1 into the polynomial, we get 2(1)^4 + a₃(1)^3 + a₂(1)^2 + a₁(1) + 1 = 2 + a₃ + a₂ + a₁ + 1 = a₃ + a₂ + a₁ + 3. For this to be zero, we need a₃ + a₂ + a₁ = -3. Since a₃, a₂, a₁ are integers, this is possible. For example, choose a₃ = 0, a₂ = 0, a₁ = -3. Then the polynomial becomes 2x⁴ + 0x³ + 0x² -3x +1. Then plugging in x=1: 2 + 0 + 0 -3 +1 = 0. So yes, x=1 can be a root. Similarly, x=-1: plug in x=-1: 2(-1)^4 + a₃(-1)^3 + a₂(-1)^2 + a₁(-1) +1 = 2 -a₃ + a₂ -a₁ +1 = (2+1) + (-a₃ + a₂ -a₁) = 3 - (a₃ - a₂ + a₁). To make this zero, we need a₃ - a₂ + a₁ = 3. Again, possible by choosing appropriate integers. For example, a₃=3, a₂=0, a₁=0. Then the polynomial is 2x⁴ +3x³ +0x² +0x +1. Plugging in x=-1: 2 -3 +0 +0 +1=0. So x=-1 is a root in this case.

For x=1/2: plug into the polynomial: 2*(1/2)^4 + a₃*(1/2)^3 + a₂*(1/2)^2 + a₁*(1/2) +1. Let's compute each term:

2*(1/16) = 1/8

a₃*(1/8)

a₂*(1/4)

a₁*(1/2)

Plus 1.

So total: 1/8 + a₃/8 + a₂/4 + a₁/2 +1. Let's combine terms:

Multiply everything by 8 to eliminate denominators:

1 + a₃ + 2a₂ + 4a₁ +8 = 0

So 9 + a₃ + 2a₂ +4a₁ =0

Thus, a₃ +2a₂ +4a₁ = -9. Since a₃, a₂, a₁ are integers, this is possible. For example, choose a₁ = -1, then 4a₁ = -4, so a₃ +2a₂ = -5. Let's choose a₂ = -1, then a₃ = -5 +2 = -3. So coefficients a₃=-3, a₂=-1, a₁=-1. Then the polynomial is 2x⁴ -3x³ -x² -x +1. Let's check x=1/2:

2*(1/2)^4 = 2*(1/16)=1/8

-3*(1/2)^3 = -3*(1/8)=-3/8

-1*(1/2)^2 = -1*(1/4)=-1/4

-1*(1/2) = -1/2

Plus 1.

Adding up: 1/8 -3/8 -1/4 -1/2 +1 = (1 -3)/8 + (-2/8) + (-4/8) +8/8 = (-2/8) + (-2/8) + (-4/8) +8/8 = (-8/8) +8/8=0. So yes, x=1/2 is a root here.

Similarly, x=-1/2: Let's check if that can be a root. Plug in x=-1/2:

2*(-1/2)^4 + a₃*(-1/2)^3 + a₂*(-1/2)^2 + a₁*(-1/2) +1.

Compute each term:

2*(1/16) =1/8

a₃*(-1/8)

a₂*(1/4)

a₁*(-1/2)

Plus 1.

Total: 1/8 -a₃/8 +a₂/4 -a₁/2 +1. Multiply by 8:

1 -a₃ +2a₂ -4a₁ +8 =0

So 9 -a₃ +2a₂ -4a₁=0 => -a₃ +2a₂ -4a₁ = -9 => a₃ =9 +2a₂ -4a₁.

Since a₂ and a₁ are integers, we can choose them such that a₃ is also an integer. For example, let a₂=0, a₁=3. Then a₃=9 +0 -12= -3. So coefficients a₃=-3, a₂=0, a₁=3. The polynomial is 2x⁴ -3x³ +0x² +3x +1. Plugging in x=-1/2:

2*(1/16) -3*(-1/8) +0*(1/4) +3*(-1/2) +1 = 1/8 +3/8 +0 -3/2 +1 = (1+3)/8 -3/2 +1 = 4/8 -12/8 +8/8 = (4 -12 +8)/8=0/8=0. So x=-1/2 is a root here.

Therefore, all four possible rational roots are indeed possible, given appropriate choices of a₃, a₂, a₁. Therefore, the different possible rational roots are ±1, ±1/2, which are four distinct values.
</execution_2.1>

<execution_2.2>
If x=1 is a root, then substituting x=1 into the polynomial gives 2(1)^4 + a₃(1)^3 + a₂(1)^2 + a₁(1) +1 = 2 + a₃ + a₂ + a₁ +1 = a₃ + a₂ + a₁ +3 =0. So we can choose a₃, a₂, a₁ such that a₃ + a₂ + a₁ = -3. For example, set a₃ = -1, a₂ = -1, a₁ = -1. Then their sum is -3, so 1 is a root.

Similarly, for x=-1. Substitute x=-1: 2(-1)^4 + a₃(-1)^3 + a₂(-1)^2 + a₁(-1) +1 = 2(1) + a₃(-1) + a₂(1) + a₁(-1) +1 = 2 -a₃ + a₂ -a₁ +1 = (2+1) + (-a₃ + a₂ -a₁) = 3 + (-a₃ + a₂ -a₁). Setting this equal to zero: -a₃ + a₂ -a₁ = -3. For example, choose a₃=1, a₂=1, a₁=1. Then -1 +1 -1 = -1, which is not -3. Hmm, maybe another choice. Let's solve -a₃ + a₂ -a₁ = -3. Let me set a₃=3, a₂=0, a₁=0. Then -3 +0 -0 = -3. So yes, with a₃=3, a₂=0, a₁=0, x=-1 is a root.

For x=1/2. Let's check if 1/2 can be a root. Substitute x=1/2: 2*(1/2)^4 + a₃*(1/2)^3 + a₂*(1/2)^2 + a₁*(1/2) +1 = 2*(1/16) + a₃*(1/8) + a₂*(1/4) + a₁*(1/2) +1 = (2/16) + (a₃)/8 + (a₂)/4 + (a₁)/2 +1 = (1/8) + (a₃)/8 + (a₂)/4 + (a₁)/2 +1. Let's combine terms: (1/8 +1) + (a₃)/8 + (a₂)/4 + (a₁)/2 = (9/8) + (a₃ + 2a₂ + 4a₁)/8. So total is (9 + a₃ + 2a₂ + 4a₁)/8. Setting this equal to zero: 9 + a₃ + 2a₂ + 4a₁ =0. So we need a₃ + 2a₂ +4a₁ = -9. For example, choose a₁= -2, then 4a₁= -8, so a₃ +2a₂ = -1. Let a₂=0, then a₃= -1. So coefficients a₃=-1, a₂=0, a₁=-2. Then the polynomial is 2x⁴ -x³ +0x² -2x +1. Let's test x=1/2: 2*(1/2)^4 - (1/2)^3 +0 -2*(1/2) +1 = 2*(1/16) - (1/8) -1 +1 = (1/8) - (1/8) -1 +1 = 0. Yes, works.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
Therefore, all four possible rational roots exist for some choices of coefficients. Hence, the number of different possible rational roots is 4.
**Final Answer**
The number of different possible rational roots is \boxed{4}.
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the number of different possible rational roots of this polynomial is $\boxed{4}$.
</Final_answer>
</think>