Okay, so I need to factor the polynomial \(x^8 + 3x^4 - 4\) into monic polynomials with integer coefficients that can't be factored further over the integers. Then, evaluate each of these polynomials at 1 and sum them up. Let me start by understanding the problem step by step.

First, let me write down the polynomial:

\(f(x) = x^8 + 3x^4 - 4\)

I need to factor this into irreducible monic polynomials over the integers. Then, sum their values at x=1. So, the first step is to factor f(x).

Looking at the polynomial, I notice that it's a quadratic in terms of \(x^4\). Let me set \(y = x^4\), so the equation becomes:

\(y^2 + 3y - 4 = 0\)

Solving this quadratic equation:

\(y = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2}\)

So, the roots are:

\(y = \frac{-3 + 5}{2} = 1\) and \(y = \frac{-3 - 5}{2} = -4\)

Therefore, the original polynomial factors as:

\((x^4 - 1)(x^4 + 4)\)

So, \(x^8 + 3x^4 -4 = (x^4 -1)(x^4 +4)\)

Now, I need to check if these factors can be factored further over the integers. Let's start with \(x^4 -1\). That's a difference of squares:

\(x^4 -1 = (x^2)^2 -1^2 = (x^2 -1)(x^2 +1)\)

Then, \(x^2 -1\) is also a difference of squares:

\(x^2 -1 = (x -1)(x +1)\)

So, \(x^4 -1\) factors into \((x -1)(x +1)(x^2 +1)\). Are these factors irreducible over the integers?

Yes, because each of these polynomials is either linear (degree 1) or quadratic with discriminant \(0^2 - 4*1*1 = -4\), which is negative, so they can't be factored further into real polynomials, let alone integer polynomials.

Now, the other factor is \(x^4 +4\). Let's see if this can be factored. It's similar to the sum of squares, but since 4 is a square, perhaps it's a biquadratic that can be factored into quadratics.

Recall that \(a^4 + 4b^4\) can be factored as \((a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)\). Let me check if this applies here.

In our case, \(x^4 +4 = x^4 + 4*1^4\). So, applying the formula with b=1:

\(x^4 +4 = (x^2 + 2*1*x + 2*1^2)(x^2 - 2*1*x + 2*1^2) = (x^2 + 2x + 2)(x^2 - 2x + 2)\)

Let me verify this multiplication:

\((x^2 + 2x + 2)(x^2 - 2x + 2) = x^4 -2x^3 +2x^2 +2x^3 -4x^2 +4x +2x^2 -4x +4\)

Combine like terms:

x^4 + ( -2x^3 +2x^3 ) + (2x^2 -4x^2 +2x^2 ) + (4x -4x ) +4

Simplifies to:

x^4 + 0x^3 + 0x^2 + 0x +4 = x^4 +4. Perfect, that works.

So, \(x^4 +4\) factors into \((x^2 + 2x +2)(x^2 -2x +2)\). Now, are these quadratic factors irreducible over the integers?

Check if they can be factored further. For a quadratic \(ax^2 +bx +c\), it factors over integers if discriminant \(b^2 -4ac\) is a perfect square.

For \(x^2 +2x +2\), discriminant is \(4 -8 = -4\), which is not a perfect square, so it's irreducible over integers. Similarly, \(x^2 -2x +2\) has discriminant \(4 -8 = -4\), also irreducible.

Therefore, the complete factorization of the original polynomial is:

\(x^8 +3x^4 -4 = (x -1)(x +1)(x^2 +1)(x^2 +2x +2)(x^2 -2x +2)\)

So, these are all monic polynomials with integer coefficients, and each is irreducible over the integers. Therefore, the factors are:

\(p_1(x) = x -1\), \(p_2(x) = x +1\), \(p_3(x) = x^2 +1\), \(p_4(x) = x^2 +2x +2\), \(p_5(x) = x^2 -2x +2\)

Now, compute \(p_1(1) + p_2(1) + p_3(1) + p_4(1) + p_5(1)\).

Let's compute each term:

1. \(p_1(1) = 1 -1 = 0\)
2. \(p_2(1) = 1 +1 = 2\)
3. \(p_3(1) = 1^2 +1 = 1 +1 = 2\)
4. \(p_4(1) = 1^2 +2*1 +2 = 1 +2 +2 =5\)
5. \(p_5(1) = 1^2 -2*1 +2 =1 -2 +2 =1\)

Now, sum them up:

0 + 2 + 2 +5 +1 = 10

So, the final answer is 10. Let me just check if I made any mistakes in the factorization steps.

Original polynomial: \(x^8 +3x^4 -4\). Factored into \((x^4 -1)(x^4 +4)\). Then \(x^4 -1\) factors into \((x -1)(x +1)(x^2 +1)\), and \(x^4 +4\) into \((x^2 +2x +2)(x^2 -2x +2)\). All factors are monic and irreducible over integers. Evaluating each at 1 gives 0,2,2,5,1. Sum is 10. Seems correct.

**Final Answer**
\boxed{10}
Therefore, the sum of the values of all the irreducible monic polynomial factors evaluated at \(1\) is \(\boxed{10}\).
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