Okay, let's try to figure out how to solve this problem. So, we have a monic polynomial p(x) of degree 4, and three of its roots are 1, 2, and 3. We need to find p(0) + p(4). Hmm, monic means the leading coefficient is 1, right? So, a monic polynomial of degree 4 would look like x⁴ + something x³ + something x² + something x + something. Since it's degree 4, there are four roots, but three of them are given: 1, 2, and 3. So, there must be a fourth root, let's call it r. Then, the polynomial can be written as (x - 1)(x - 2)(x - 3)(x - r). Since it's monic, that's the expanded form. 

Wait, so p(x) = (x - 1)(x - 2)(x - 3)(x - r). Then, to find p(0) and p(4), we can plug in x = 0 and x = 4 into this polynomial. Let's see. 

First, let's expand the polynomial. Maybe expanding it will help us find the coefficients, but I'm not sure if we need that. Alternatively, since we need p(0) and p(4), perhaps there's a smarter way without expanding everything. Let me think. 

P(0) is the product of (-1)(-2)(-3)(-r). Because when x=0, each term becomes (0 - root) = -root. So, p(0) = (-1)(-2)(-3)(-r) = (-1)^4 * (1*2*3*r) = 1*6*r = 6r. Wait, is that right? Let me check. 

Each (0 - root) is -root, so multiplying four terms: (-1)(-2)(-3)(-r) = (-1)^4 * (1*2*3*r) = 1 * 6r = 6r. Yes, that seems correct. So p(0) = 6r. 

Similarly, p(4) is (4 - 1)(4 - 2)(4 - 3)(4 - r). Let's compute each term: (4 -1)=3, (4 -2)=2, (4 -3)=1, and (4 - r). So, p(4) = 3*2*1*(4 - r) = 6*(4 - r) = 24 - 6r. 

Therefore, p(0) + p(4) = 6r + (24 - 6r) = 24. Wait, that's interesting! The r terms cancel out. So, regardless of the value of r, the sum p(0) + p(4) is 24. Is that possible? Let me verify again. 

For p(0): (-1)(-2)(-3)(-r) = 6r. Correct. For p(4): (3)(2)(1)(4 - r) = 6*(4 - r) = 24 - 6r. So adding them: 6r + 24 - 6r = 24. Yes, that's right. So, the answer is 24. 

But wait, why is the fourth root r not affecting the sum? That seems counterintuitive at first, but when we look at the terms, the p(0) has +6r and p(4) has -6r, so they cancel each other. Therefore, the sum is independent of r. That's a neat result. 

So, even though we don't know the fourth root, the sum p(0) + p(4) is always 24. Therefore, the answer is 24. 

I think that's the solution. Let me check if there's another way to approach this. Maybe using Vieta's formula? Let's see. 

Vieta's formula relates the sum and products of roots to the coefficients of the polynomial. For a quartic polynomial x⁴ + a x³ + b x² + c x + d, the sum of roots is -a, the sum of products of roots two at a time is b, the sum of products three at a time is -c, and the product of roots is d. 

Since p(x) is monic and degree 4, the polynomial is x⁴ + a x³ + b x² + c x + d. The roots are 1, 2, 3, and r. Then, sum of roots is 1 + 2 + 3 + r = 6 + r = -a. 

Sum of products two at a time: (1*2) + (1*3) + (1*r) + (2*3) + (2*r) + (3*r) = 2 + 3 + r + 6 + 2r + 3r = 11 + 6r = b. 

Sum of products three at a time: (1*2*3) + (1*2*r) + (1*3*r) + (2*3*r) = 6 + 2r + 3r + 6r = 6 + 11r = -c. 

Product of roots: 1*2*3*r = 6r = d. 

But how does this help us? Well, we need to compute p(0) + p(4). 

Wait, p(0) is the constant term of the polynomial, which is d = 6r. 

p(4) is 4⁴ + a*4³ + b*4² + c*4 + d. Let me compute that: 

4⁴ = 256, 4³ = 64, 4² = 16, so:

p(4) = 256 + a*64 + b*16 + c*4 + d. 

But we know d = 6r. 

Now, let's express everything in terms of r. 

From Vieta's formula, we have a = -(sum of roots) = -(6 + r). So, a = -6 - r. 

Similarly, b = 11 + 6r. 

c = -(sum of products three at a time) = -(6 + 11r). So, c = -6 -11r. 

Therefore, p(4) = 256 + (-6 - r)*64 + (11 + 6r)*16 + (-6 -11r)*4 + 6r. 

Let me compute each term step by step:

First term: 256. 

Second term: (-6 - r)*64 = (-6*64) + (-r*64) = -384 -64r. 

Third term: (11 + 6r)*16 = 11*16 + 6r*16 = 176 + 96r. 

Fourth term: (-6 -11r)*4 = -6*4 + (-11r)*4 = -24 -44r. 

Fifth term: 6r. 

Now, adding all these up:

256 -384 -64r +176 +96r -24 -44r +6r. 

Combine like terms:

Constants: 256 -384 +176 -24. 

Let's compute that:

256 -384 = -128

-128 +176 = 48

48 -24 = 24

Variables: -64r +96r -44r +6r. 

Compute coefficients:

-64 +96 = 32

32 -44 = -12

-12 +6 = -6

So, variables term is -6r.

Therefore, p(4) = 24 -6r.

Therefore, p(4) = 24 -6r.

Earlier, we found p(0) =6r. Therefore, p(0) + p(4) =6r +24 -6r =24. 

So, same result. Therefore, regardless of the value of r, the sum is 24. 

So, the answer is 24. 

Therefore, the final answer is \boxed{24}. 

**Final Answer**
\boxed{24}
Therefore, the value of $p(0) + p(4)$ is $\boxed{24}$.
</think>