Explanation:
Okay, so I need to find the smallest positive real number C such that the norm of the matrix multiplied by any vector v is less than or equal to C times the norm of v. The matrix given is [[2, 3], [0, -2]]. Let me think about how to approach this.

First, I remember that for a matrix A, the operator norm (which is the maximum value of ||Av|| / ||v|| for all non-zero vectors v) is the largest singular value of A. So, if I can find the largest singular value of this matrix, that should be the smallest such C. But maybe I need to verify this.

Alternatively, maybe I can think about it in terms of eigenvalues? Wait, but the operator norm isn't directly the eigenvalues unless the matrix is symmetric. This matrix isn't symmetric, though. The transpose of the matrix is [[2, 0], [3, -2]], which isn't the same as the original matrix. So maybe singular values are the way to go.

Singular values are the square roots of the eigenvalues of A*A, right? So let me compute A*A first. Let me write down matrix A:

A = [[2, 3],
     [0, -2]]

So A transpose is [[2, 0], [3, -2]]. Then A*A would be:

First row first column: (2)(2) + (3)(3) = 4 + 9 = 13

First row second column: (2)(0) + (3)(-2) = 0 - 6 = -6

Second row first column: (0)(2) + (-2)(3) = 0 - 6 = -6

Second row second column: (0)(0) + (-2)(-2) = 0 + 4 = 4

So A*A is [[13, -6], [-6, 4]]. Now, to find the singular values, I need to find the eigenvalues of this matrix. Then take their square roots.

So let me find the eigenvalues of [[13, -6], [-6, 4]]. The characteristic equation is det(A*A - λI) = 0.

So determinant of [[13 - λ, -6], [-6, 4 - λ]] is (13 - λ)(4 - λ) - (-6)(-6) = 0.

Calculating that:

(13 - λ)(4 - λ) - 36 = 0

First expand (13 - λ)(4 - λ):

13*4 - 13λ - 4λ + λ² = 52 - 17λ + λ²

Then subtract 36: 52 - 17λ + λ² - 36 = 16 - 17λ + λ² = 0

So the equation is λ² -17λ +16 = 0

Wait, let me check that again. Wait, 52 - 36 is 16, yes. So the quadratic equation is λ² -17λ +16 = 0.

Wait, solving this quadratic equation: λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2.

So sqrt(225) is 15. Therefore, the eigenvalues are (17 + 15)/2 = 32/2 = 16, and (17 -15)/2 = 2/2 = 1.

Therefore, the eigenvalues of A*A are 16 and 1. Therefore, the singular values of A are sqrt(16) = 4 and sqrt(1) = 1. Therefore, the largest singular value is 4, so the operator norm is 4. Therefore, the smallest C is 4?

Wait, but let me check if this is correct. Let me think again.

Alternatively, maybe I can approach this problem by considering the norm of Av for any vector v. Let me write v as [x, y]^T. Then Av is [2x + 3y, -2y]^T. Then the norm squared of Av is (2x + 3y)^2 + (-2y)^2. Let me compute that:

(2x + 3y)^2 + ( -2y)^2 = 4x² + 12xy + 9y² + 4y² = 4x² + 12xy + 13y².

So the norm squared of Av is 4x² + 12xy + 13y². The norm squared of v is x² + y². Therefore, the ratio squared is (4x² + 12xy + 13y²)/(x² + y²). Therefore, the maximum value of this ratio is the square of the operator norm, so the square root of that maximum is the operator norm. Therefore, we need to find the maximum of (4x² + 12xy + 13y²)/(x² + y²) over all (x, y) ≠ (0, 0). Then take the square root of that maximum to get C.

Alternatively, since this is a homogeneous function, we can set x² + y² = 1, and maximize 4x² + 12xy + 13y². So, let me consider the function f(x, y) = 4x² + 12xy + 13y² with the constraint x² + y² = 1. Then, using Lagrange multipliers, we can find the maximum.

Let me set up the Lagrangian: L = 4x² + 12xy + 13y² - λ(x² + y² - 1). Taking partial derivatives:

∂L/∂x = 8x + 12y - 2λx = 0

∂L/∂y = 12x + 26y - 2λy = 0

∂L/∂λ = -(x² + y² - 1) = 0

So the equations are:

8x + 12y = 2λx --> 8x + 12y - 2λx = 0 --> (8 - 2λ)x + 12y = 0

12x + 26y = 2λy --> 12x + 26y - 2λy = 0 --> 12x + (26 - 2λ)y = 0

And x² + y² = 1.

So we have a system of equations:

(8 - 2λ)x + 12y = 0

12x + (26 - 2λ)y = 0

And x² + y² = 1.

This is a homogeneous system for x and y. For non-trivial solutions, the determinant of the coefficients must be zero.

So determinant:

| (8 - 2λ)    12        |
| 12        (26 - 2λ) |  = 0

Calculating determinant:

(8 - 2λ)(26 - 2λ) - (12)(12) = 0

Let me compute that:

First expand (8 - 2λ)(26 - 2λ):

Multiply 8*26 = 208

8*(-2λ) = -16λ

-2λ*26 = -52λ

-2λ*(-2λ) = 4λ²

So altogether: 208 -16λ -52λ +4λ² = 208 -68λ +4λ²

Then subtract 144 (since 12*12=144):

208 -68λ +4λ² -144 = 64 -68λ +4λ² = 0

So equation is 4λ² -68λ +64 = 0. Let me divide both sides by 4:

λ² -17λ +16 = 0

Wait, this is the same equation as before! Which gives λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2, so λ = 16 or 1. Therefore, the maximum value of f(x, y) is 16, and the minimum is 1. Therefore, the maximum ratio is 16, so the operator norm is sqrt(16) = 4. Therefore, the smallest C is 4. So that seems to confirm the previous result.

Alternatively, maybe I can think of the matrix A as a linear transformation and find its maximum stretching factor. Since the singular values are 4 and 1, then the maximum stretching factor is 4, so C is 4.

But let me check with specific vectors. For example, if I take a vector that is an eigenvector of A*A corresponding to the eigenvalue 16, then Av would have norm 4 times the norm of v. Let me find such a vector.

From the previous equations, when λ = 16, we can find the corresponding eigenvectors. Let me take the first equation:

(8 - 2λ)x + 12y = 0. If λ = 16, then:

(8 - 32)x + 12y = 0 => (-24)x + 12y = 0 => -24x +12y = 0 => Simplify: -2x + y = 0 => y = 2x.

So the eigenvectors corresponding to λ = 16 are vectors where y = 2x. Let me take x = 1, y = 2. Then the vector v = [1, 2]^T. Let me compute Av:

Av = [[2, 3], [0, -2]] * [1, 2]^T = [2*1 + 3*2, 0*1 + (-2)*2]^T = [2 + 6, 0 - 4]^T = [8, -4]^T. The norm of Av is sqrt(8² + (-4)^2) = sqrt(64 + 16) = sqrt(80) = 4*sqrt(5). The norm of v is sqrt(1² + 2²) = sqrt(5). Therefore, the ratio is 4*sqrt(5)/sqrt(5) = 4. So that checks out. Therefore, C must be at least 4. And since we found that the maximum ratio is 4, then C=4 is indeed the smallest such constant.

Alternatively, let me check another vector. For example, take the vector [1, 0]^T. Then Av = [2, 0]^T, norm is 2, and ||v|| is 1, so ratio is 2. For vector [0, 1]^T, Av = [3, -2]^T, norm is sqrt(9 + 4) = sqrt(13) ≈ 3.605, which is less than 4. For the vector [1, 2]^T as above, ratio is 4. So seems like 4 is the maximum.

Alternatively, maybe check another vector. Suppose I take a vector in the direction of the other singular value. For λ = 1, the eigenvectors would satisfy:

From the first equation: (8 - 2λ)x + 12y = 0. If λ = 1, then:

(8 - 2)x + 12y = 0 => 6x + 12y = 0 => x + 2y = 0 => x = -2y. So take y = 1, x = -2. Then vector v = [-2, 1]^T. Compute Av:

Av = [[2, 3], [0, -2]] * [-2, 1]^T = [2*(-2) + 3*1, 0*(-2) + (-2)*1]^T = [-4 + 3, 0 - 2]^T = [-1, -2]^T. The norm of Av is sqrt(1 + 4) = sqrt(5). The norm of v is sqrt(4 + 1) = sqrt(5). Therefore, ratio is 1. Which matches the singular value sqrt(1) = 1. So that checks out.

Therefore, all seems consistent. Therefore, the minimal C is 4.

But just to make sure, let me think if there's another approach. For example, using the fact that for any vector v, ||Av|| ≤ ||A|| ||v||, where ||A|| is the operator norm. And we need the minimal C such that this holds. So since we found that the operator norm is 4, then C=4 is the minimal.

Alternatively, maybe think in terms of the maximum absolute value of the eigenvalues? Wait, but for non-symmetric matrices, the operator norm isn't directly related to eigenvalues. However, for symmetric matrices, the operator norm is the maximum absolute value of the eigenvalues. But this matrix isn't symmetric. However, the singular values are related to the eigenvalues of A*A, which we already calculated.

Alternatively, maybe think of the matrix as a combination of scaling and shearing? The matrix [[2, 3], [0, -2]] can be thought of as scaling the x-axis by 2, scaling the y-axis by -2, and shearing in the x-direction by 3. But I don't know if that helps directly.

Alternatively, maybe parametrize the vector v. Let me suppose that v is a vector with norm 1, then find the maximum of ||Av||. Which is exactly what I did earlier with Lagrange multipliers. Which led me to the maximum of 4. So that seems solid.

Alternatively, think of the matrix as a linear transformation. The maximum stretching factor is the operator norm. Since we found that there exists a vector which gets stretched by 4, and no vector gets stretched more than that, then 4 is the minimal C.

Alternatively, maybe think of the matrix in terms of its action on vectors. For example, if I write the vector v as (x, y), then Av = (2x + 3y, -2y). The norm squared is (2x + 3y)^2 + ( -2y)^2 = 4x² + 12xy + 9y² + 4y² = 4x² + 12xy + 13y². To find the maximum of this expression under the constraint x² + y² = 1. Which is equivalent to finding the maximum eigenvalue of the matrix [[4, 6], [6, 13]]? Wait, because if you think of the quadratic form 4x² + 12xy + 13y², that can be represented as [x y] * [[4, 6], [6, 13]] * [x; y]. Wait, because the coefficient of x² is 4, the coefficient of y² is 13, and the coefficient of xy is 12, which is twice the off-diagonal term. Therefore, the matrix is [[4, 6], [6, 13]]. Then the maximum eigenvalue of this matrix is the maximum value of the quadratic form, which is the same as the maximum of the ratio we had earlier. Wait, but earlier we had that the maximum of (4x² + 12xy + 13y²)/(x² + y²) is 16, so the maximum eigenvalue of [[4, 6], [6, 13]] would be 16? Wait, but if we have the quadratic form [x y] * [[4, 6], [6, 13]] * [x; y], then the maximum value of this quadratic form over unit vectors is the maximum eigenvalue of the matrix. But earlier, we found that the maximum value of (4x² + 12xy + 13y²) is 16 when x² + y² = 1. Therefore, the maximum eigenvalue of [[4, 6], [6, 13]] is 16? Let me check.

Let me compute the eigenvalues of [[4, 6], [6, 13]]. The characteristic equation is det([[4 - λ, 6], [6, 13 - λ]]) = (4 - λ)(13 - λ) - 36 = 0.

Calculating that:

(4 - λ)(13 - λ) - 36 = 52 - 4λ -13λ + λ² - 36 = λ² -17λ + 16 = 0. Which is the same equation as before! Therefore, the eigenvalues are 16 and 1. Therefore, the maximum eigenvalue is 16, so the maximum value of the quadratic form is 16. Therefore, the maximum of ||Av||² is 16 when ||v||² = 1, so ||Av|| is 4. Therefore, again, confirming that the operator norm is 4.

Therefore, all methods lead to the conclusion that the minimal C is 4. Therefore, I think that is the answer.

But just to be thorough, let me check if there's a possibility that the maximum occurs somewhere else. For example, suppose I take a vector that's not aligned with the eigenvectors. Let me pick a random vector, say [1, 1]^T. Then Av = [2*1 + 3*1, -2*1]^T = [5, -2]^T. The norm is sqrt(25 + 4) = sqrt(29) ≈ 5.385. The norm of v is sqrt(2) ≈ 1.414. The ratio is sqrt(29)/sqrt(2) ≈ 5.385 / 1.414 ≈ 3.807, which is less than 4. So that's still less than 4.

Another vector, say [1, -1]^T. Then Av = [2*1 + 3*(-1), -2*(-1)] = [2 - 3, 2] = [-1, 2]^T. The norm is sqrt(1 + 4) = sqrt(5) ≈ 2.236. The norm of v is sqrt(2) ≈ 1.414. The ratio is sqrt(5)/sqrt(2) ≈ 1.58, which is less than 4.

Another vector, say [2, 1]^T. Then Av = [2*2 + 3*1, -2*1] = [4 + 3, -2] = [7, -2]^T. Norm squared is 49 + 4 = 53. ||v|| squared is 4 + 1 = 5. Ratio squared is 53/5 = 10.6, so ratio is sqrt(10.6) ≈ 3.26, still less than 4.

Alternatively, take the vector [1, 2]^T as before, which gives ratio 4. So that's the maximum. Therefore, seems like 4 is indeed the maximum.

Alternatively, think of the matrix as a combination of scaling and rotation? Wait, the matrix isn't orthogonal, so it's not just rotation and scaling. But maybe decompose it? Not sure if that helps.

Alternatively, think of the matrix in terms of its columns. The columns are [2, 0] and [3, -2]. The norm of the first column is 2, the norm of the second column is sqrt(9 + 4) = sqrt(13). But the operator norm isn't directly the maximum column norm, because the operator norm is about the maximum stretching of any vector, not just the standard basis vectors. For example, the maximum column norm is sqrt(13) ≈ 3.605, which is less than 4, so that's not the operator norm. So that approach doesn't give the answer.

Alternatively, think of the matrix as a linear transformation. The maximum stretching factor is the operator norm. Since we have found that there exists a vector which gets stretched by 4, and that the maximum is 4, then that's the answer.

Therefore, after all these checks, I think the minimal C is 4.

**Final Answer**
The smallest positive real number $ C $ is $\boxed{4}$.

To find the smallest positive real number $ C $ such that the inequality

$$
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| \le C \|\bold{v}\|
$$

holds for all two-dimensional vectors $ \bold{v} $, we need to determine the **operator norm** of the matrix $ A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} $.

---

### Step 1: Understanding the Operator Norm

The **operator norm** of a matrix $ A $ is defined as:

$$
\|A\| = \sup_{\|\bold{v}\| = 1} \|A\bold{v}\|
$$

This is the **maximum stretching factor** of the matrix $ A $ when applied to unit vectors. The smallest such $ C $ that satisfies the inequality for all $ \bold{v} $ is exactly this operator norm.

---

### Step 2: Using Singular Values

For any matrix $ A $, the operator norm is equal to the **largest singular value** of $ A $. The singular values of $ A $ are the square roots of the eigenvalues of $ A^T A $.

Let’s compute $ A^T A $:

$$
A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}, \quad A^T = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix}
$$

$$
A^T A = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 4 + 0 & 6 - 6 \\ 6 - 6 & 9 + 4 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 13 \end{pmatrix}
$$

Wait, that’s not correct. Let's recompute:

$$
A^T A = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}
= \begin{pmatrix}
(2)(2) + (0)(0) & (2)(3) + (0)(-2) \\
(3)(2) + (-2)(0) & (3)(3) + (-2)(-2)
\end{pmatrix}
= \begin{pmatrix}
4 & 6 \\
6 & 13
\end{pmatrix}
$$

Now, find the eigenvalues of $ A^T A $:

$$
\det(A^T A - \lambda I) = \det\left( \begin{pmatrix} 4 - \lambda & 6 \\ 6 & 13 - \lambda \end{pmatrix} \right)
= (4 - \lambda)(13 - \lambda) - 36 = \lambda^2 - 17\lambda + 16
$$

Solving the quadratic:

$$
\lambda = \frac{17 \pm \sqrt{289 - 64}}{2} = \frac{17 \pm \sqrt{225}}{2} = \frac{17 \pm 15}{2}
\Rightarrow \lambda = 16 \text{ or } 1
$$

So the singular values of $ A $ are $ \sqrt{16} = 4 $ and $ \sqrt{1} = 1 $. Therefore, the **largest singular value** is $ 4 $, and hence the **operator norm** of $ A $ is $ 4 $.

---

### Step 3: Verifying with a Specific Vector

To confirm, consider the vector $ \bold{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} $, which is an eigenvector of $ A^T A $ corresponding to the eigenvalue $ 16 $.

Compute $ A\bold{v} $:

$$
A \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2(1) + 3(2) \\ 0(1) - 2(2) \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \end{pmatrix}
$$

$$
\|A\bold{v}\| = \sqrt{8^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}
$$

$$
\|\bold{v}\| = \sqrt{1^2 + 2^2} = \sqrt{5}
$$

$$
\frac{\|A\bold{v}\|}{\|\bold{v}\|} = \frac{4\sqrt{5}}{\sqrt{5}} = 4
$$

This confirms that the maximum ratio is indeed $ 4 $, and thus the smallest $ C $ satisfying the inequality for all $ \bold{v} $ is:

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest positive real number C such that the norm of the matrix multiplied by any vector v is less than or equal to C times the norm of v. The matrix given is [[2, 3], [0, -2]]. Let me think about how to approach this.

First, I remember that for a matrix A, the operator norm (which is the maximum value of ||Av|| / ||v|| for all non-zero vectors v) is the largest singular value of A. So, if I can find the largest singular value of this matrix, that should be the smallest such C. But maybe I need to verify this.

Alternatively, maybe I can think about it in terms of eigenvalues? Wait, but the operator norm isn't directly the eigenvalues unless the matrix is symmetric. This matrix isn't symmetric, though. The transpose of the matrix is [[2, 0], [3, -2]], which isn't the same as the original matrix. So maybe singular values are the way to go.

Singular values are the square roots of the eigenvalues of A*A, right? So let me compute A*A first. Let me write down matrix A:

A = [[2, 3],
     [0, -2]]

So A transpose is [[2, 0], [3, -2]]. Then A*A would be:

First row first column: (2)(2) + (3)(3) = 4 + 9 = 13

First row second column: (2)(0) + (3)(-2) = 0 - 6 = -6

Second row first column: (0)(2) + (-2)(3) = 0 - 6 = -6

Second row second column: (0)(0) + (-2)(-2) = 0 + 4 = 4

So A*A is [[13, -6], [-6, 4]]. Now, to find the singular values, I need to find the eigenvalues of this matrix. Then take their square roots.

So let me find the eigenvalues of [[13, -6], [-6, 4]]. The characteristic equation is det(A*A - λI) = 0.

So determinant of [[13 - λ, -6], [-6, 4 - λ]] is (13 - λ)(4 - λ) - (-6)(-6) = 0.

Calculating that:

(13 - λ)(4 - λ) - 36 = 0

First expand (13 - λ)(4 - λ):

13*4 - 13λ - 4λ + λ² = 52 - 17λ + λ²

Then subtract 36: 52 - 17λ + λ² - 36 = 16 - 17λ + λ² = 0

So the equation is λ² -17λ +16 = 0

Wait, let me check that again. Wait, 52 - 36 is 16, yes. So the quadratic equation is λ² -17λ +16 = 0.

Wait, solving this quadratic equation: λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2.

So sqrt(225) is 15. Therefore, the eigenvalues are (17 + 15)/2 = 32/2 = 16, and (17 -15)/2 = 2/2 = 1.

Therefore, the eigenvalues of A*A are 16 and 1. Therefore, the singular values of A are sqrt(16) = 4 and sqrt(1) = 1. Therefore, the largest singular value is 4, so the operator norm is 4. Therefore, the smallest C is 4?

Wait, but let me check if this is correct. Let me think again.

Alternatively, maybe I can approach this problem by considering the norm of Av for any vector v. Let me write v as [x, y]^T. Then Av is [2x + 3y, -2y]^T. Then the norm squared of Av is (2x + 3y)^2 + (-2y)^2. Let me compute that:

(2x + 3y)^2 + ( -2y)^2 = 4x² + 12xy + 9y² + 4y² = 4x² + 12xy + 13y².

So the norm squared of Av is 4x² + 12xy + 13y². The norm squared of v is x² + y². Therefore, the ratio squared is (4x² + 12xy + 13y²)/(x² + y²). Therefore, the maximum value of this ratio is the square of the operator norm, so the square root of that maximum is the operator norm. Therefore, we need to find the maximum of (4x² + 12xy + 13y²)/(x² + y²) over all (x, y) ≠ (0, 0). Then take the square root of that maximum to get C.

Alternatively, since this is a homogeneous function, we can set x² + y² = 1, and maximize 4x² + 12xy + 13y². So, let me consider the function f(x, y) = 4x² + 12xy + 13y² with the constraint x² + y² = 1. Then, using Lagrange multipliers, we can find the maximum.

Let me set up the Lagrangian: L = 4x² + 12xy + 13y² - λ(x² + y² - 1). Taking partial derivatives:

∂L/∂x = 8x + 12y - 2λx = 0

∂L/∂y = 12x + 26y - 2λy = 0

∂L/∂λ = -(x² + y² - 1) = 0

So the equations are:

8x + 12y = 2λx --> 8x + 12y - 2λx = 0 --> (8 - 2λ)x + 12y = 0

12x + 26y = 2λy --> 12x + 26y - 2λy = 0 --> 12x + (26 - 2λ)y = 0

And x² + y² = 1.

So we have a system of equations:

(8 - 2λ)x + 12y = 0

12x + (26 - 2λ)y = 0

And x² + y² = 1.

This is a homogeneous system for x and y. For non-trivial solutions, the determinant of the coefficients must be zero.

So determinant:

| (8 - 2λ)    12        |
| 12        (26 - 2λ) |  = 0

Calculating determinant:

(8 - 2λ)(26 - 2λ) - (12)(12) = 0

Let me compute that:

First expand (8 - 2λ)(26 - 2λ):

Multiply 8*26 = 208

8*(-2λ) = -16λ

-2λ*26 = -52λ

-2λ*(-2λ) = 4λ²

So altogether: 208 -16λ -52λ +4λ² = 208 -68λ +4λ²

Then subtract 144 (since 12*12=144):

208 -68λ +4λ² -144 = 64 -68λ +4λ² = 0

So equation is 4λ² -68λ +64 = 0. Let me divide both sides by 4:

λ² -17λ +16 = 0

Wait, this is the same equation as before! Which gives λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± sqrt(225)] / 2 = [17 ± 15]/2, so λ = 16 or 1. Therefore, the maximum value of f(x, y) is 16, and the minimum is 1. Therefore, the maximum ratio is 16, so the operator norm is sqrt(16) = 4. Therefore, the smallest C is 4. So that seems to confirm the previous result.

Alternatively, maybe I can think of the matrix A as a linear transformation and find its maximum stretching factor. Since the singular values are 4 and 1, then the maximum stretching factor is 4, so C is 4.

But let me check with specific vectors. For example, if I take a vector that is an eigenvector of A*A corresponding to the eigenvalue 16, then Av would have norm 4 times the norm of v. Let me find such a vector.

From the previous equations, when λ = 16, we can find the corresponding eigenvectors. Let me take the first equation:

(8 - 2λ)x + 12y = 0. If λ = 16, then:

(8 - 32)x + 12y = 0 => (-24)x + 12y = 0 => -24x +12y = 0 => Simplify: -2x + y = 0 => y = 2x.

So the eigenvectors corresponding to λ = 16 are vectors where y = 2x. Let me take x = 1, y = 2. Then the vector v = [1, 2]^T. Let me compute Av:

Av = [[2, 3], [0, -2]] * [1, 2]^T = [2*1 + 3*2, 0*1 + (-2)*2]^T = [2 + 6, 0 - 4]^T = [8, -4]^T. The norm of Av is sqrt(8² + (-4)^2) = sqrt(64 + 16) = sqrt(80) = 4*sqrt(5). The norm of v is sqrt(1² + 2²) = sqrt(5). Therefore, the ratio is 4*sqrt(5)/sqrt(5) = 4. So that checks out. Therefore, C must be at least 4. And since we found that the maximum ratio is 4, then C=4 is indeed the smallest such constant.

Alternatively, let me check another vector. For example, take the vector [1, 0]^T. Then Av = [2, 0]^T, norm is 2, and ||v|| is 1, so ratio is 2. For vector [0, 1]^T, Av = [3, -2]^T, norm is sqrt(9 + 4) = sqrt(13) ≈ 3.605, which is less than 4. For the vector [1, 2]^T as above, ratio is 4. So seems like 4 is the maximum.

Alternatively, maybe check another vector. Suppose I take a vector in the direction of the other singular value. For λ = 1, the eigenvectors would satisfy:

From the first equation: (8 - 2λ)x + 12y = 0. If λ = 1, then:

(8 - 2)x + 12y = 0 => 6x + 12y = 0 => x + 2y = 0 => x = -2y. So take y = 1, x = -2. Then vector v = [-2, 1]^T. Compute Av:

Av = [[2, 3], [0, -2]] * [-2, 1]^T = [2*(-2) + 3*1, 0*(-2) + (-2)*1]^T = [-4 + 3, 0 - 2]^T = [-1, -2]^T. The norm of Av is sqrt(1 + 4) = sqrt(5). The norm of v is sqrt(4 + 1) = sqrt(5). Therefore, ratio is 1. Which matches the singular value sqrt(1) = 1. So that checks out.

Therefore, all seems consistent. Therefore, the minimal C is 4.

But just to make sure, let me think if there's another approach. For example, using the fact that for any vector v, ||Av|| ≤ ||A|| ||v||, where ||A|| is the operator norm. And we need the minimal C such that this holds. So since we found that the operator norm is 4, then C=4 is the minimal.

Alternatively, maybe think in terms of the maximum absolute value of the eigenvalues? Wait, but for non-symmetric matrices, the operator norm isn't directly related to eigenvalues. However, for symmetric matrices, the operator norm is the maximum absolute value of the eigenvalues. But this matrix isn't symmetric. However, the singular values are related to the eigenvalues of A*A, which we already calculated.

Alternatively, maybe think of the matrix as a combination of scaling and shearing? The matrix [[2, 3], [0, -2]] can be thought of as scaling the x-axis by 2, scaling the y-axis by -2, and shearing in the x-direction by 3. But I don't know if that helps directly.

Alternatively, maybe parametrize the vector v. Let me suppose that v is a vector with norm 1, then find the maximum of ||Av||. Which is exactly what I did earlier with Lagrange multipliers. Which led me to the maximum of 4. So that seems solid.

Alternatively, think of the matrix as a linear transformation. The maximum stretching factor is the operator norm. Since we found that there exists a vector which gets stretched by 4, and no vector gets stretched more than that, then 4 is the minimal C.

Alternatively, maybe think of the matrix in terms of its action on vectors. For example, if I write the vector v as (x, y), then Av = (2x + 3y, -2y). The norm squared is (2x + 3y)^2 + ( -2y)^2 = 4x² + 12xy + 9y² + 4y² = 4x² + 12xy + 13y². To find the maximum of this expression under the constraint x² + y² = 1. Which is equivalent to finding the maximum eigenvalue of the matrix [[4, 6], [6, 13]]? Wait, because if you think of the quadratic form 4x² + 12xy + 13y², that can be represented as [x y] * [[4, 6], [6, 13]] * [x; y]. Wait, because the coefficient of x² is 4, the coefficient of y² is 13, and the coefficient of xy is 12, which is twice the off-diagonal term. Therefore, the matrix is [[4, 6], [6, 13]]. Then the maximum eigenvalue of this matrix is the maximum value of the quadratic form, which is the same as the maximum of the ratio we had earlier. Wait, but earlier we had that the maximum of (4x² + 12xy + 13y²)/(x² + y²) is 16, so the maximum eigenvalue of [[4, 6], [6, 13]] would be 16? Wait, but if we have the quadratic form [x y] * [[4, 6], [6, 13]] * [x; y], then the maximum value of this quadratic form over unit vectors is the maximum eigenvalue of the matrix. But earlier, we found that the maximum value of (4x² + 12xy + 13y²) is 16 when x² + y² = 1. Therefore, the maximum eigenvalue of [[4, 6], [6, 13]] is 16? Let me check.

Let me compute the eigenvalues of [[4, 6], [6, 13]]. The characteristic equation is det([[4 - λ, 6], [6, 13 - λ]]) = (4 - λ)(13 - λ) - 36 = 0.

Calculating that:

(4 - λ)(13 - λ) - 36 = 52 - 4λ -13λ + λ² - 36 = λ² -17λ + 16 = 0. Which is the same equation as before! Therefore, the eigenvalues are 16 and 1. Therefore, the maximum eigenvalue is 16, so the maximum value of the quadratic form is 16. Therefore, the maximum of ||Av||² is 16 when ||v||² = 1, so ||Av|| is 4. Therefore, again, confirming that the operator norm is 4.

Therefore, all methods lead to the conclusion that the minimal C is 4. Therefore, I think that is the answer.

But just to be thorough, let me check if there's a possibility that the maximum occurs somewhere else. For example, suppose I take a vector that's not aligned with the eigenvectors. Let me pick a random vector, say [1, 1]^T. Then Av = [2*1 + 3*1, -2*1]^T = [5, -2]^T. The norm is sqrt(25 + 4) = sqrt(29) ≈ 5.385. The norm of v is sqrt(2) ≈ 1.414. The ratio is sqrt(29)/sqrt(2) ≈ 5.385 / 1.414 ≈ 3.807, which is less than 4. So that's still less than 4.

Another vector, say [1, -1]^T. Then Av = [2*1 + 3*(-1), -2*(-1)] = [2 - 3, 2] = [-1, 2]^T. The norm is sqrt(1 + 4) = sqrt(5) ≈ 2.236. The norm of v is sqrt(2) ≈ 1.414. The ratio is sqrt(5)/sqrt(2) ≈ 1.58, which is less than 4.

Another vector, say [2, 1]^T. Then Av = [2*2 + 3*1, -2*1] = [4 + 3, -2] = [7, -2]^T. Norm squared is 49 + 4 = 53. ||v|| squared is 4 + 1 = 5. Ratio squared is 53/5 = 10.6, so ratio is sqrt(10.6) ≈ 3.26, still less than 4.

Alternatively, take the vector [1, 2]^T as before, which gives ratio 4. So that's the maximum. Therefore, seems like 4 is indeed the maximum.

Alternatively, think of the matrix as a combination of scaling and rotation? Wait, the matrix isn't orthogonal, so it's not just rotation and scaling. But maybe decompose it? Not sure if that helps.

Alternatively, think of the matrix in terms of its columns. The columns are [2, 0] and [3, -2]. The norm of the first column is 2, the norm of the second column is sqrt(9 + 4) = sqrt(13). But the operator norm isn't directly the maximum column norm, because the operator norm is about the maximum stretching of any vector, not just the standard basis vectors. For example, the maximum column norm is sqrt(13) ≈ 3.605, which is less than 4, so that's not the operator norm. So that approach doesn't give the answer.

Alternatively, think of the matrix as a linear transformation. The maximum stretching factor is the operator norm. Since we have found that there exists a vector which gets stretched by 4, and that the maximum is 4, then that's the answer.

Therefore, after all these checks, I think the minimal C is 4.

**Final Answer**
The smallest positive real number $ C $ is $\boxed{4}$.

To find the smallest positive real number $ C $ such that the inequality

$$
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| \le C \|\bold{v}\|
$$

holds for all two-dimensional vectors $ \bold{v} $, we need to determine the **operator norm** of the matrix $ A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} $.

---

### Step 1: Understanding the Operator Norm

The **operator norm** of a matrix $ A $ is defined as:

$$
\|A\| = \sup_{\|\bold{v}\| = 1} \|A\bold{v}\|
$$

This is the **maximum stretching factor** of the matrix $ A $ when applied to unit vectors. The smallest such $ C $ that satisfies the inequality for all $ \bold{v} $ is exactly this operator norm.

---

### Step 2: Using Singular Values

For any matrix $ A $, the operator norm is equal to the **largest singular value** of $ A $. The singular values of $ A $ are the square roots of the eigenvalues of $ A^T A $.

Let’s compute $ A^T A $:

$$
A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}, \quad A^T = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix}
$$

$$
A^T A = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 4 + 0 & 6 - 6 \\ 6 - 6 & 9 + 4 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 13 \end{pmatrix}
$$

Wait, that’s not correct. Let's recompute:

$$
A^T A = \begin{pmatrix} 2 & 0 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}
= \begin{pmatrix}
(2)(2) + (0)(0) & (2)(3) + (0)(-2) \\
(3)(2) + (-2)(0) & (3)(3) + (-2)(-2)
\end{pmatrix}
= \begin{pmatrix}
4 & 6 \\
6 & 13
\end{pmatrix}
$$

Now, find the eigenvalues of $ A^T A $:

$$
\det(A^T A - \lambda I) = \det\left( \begin{pmatrix} 4 - \lambda & 6 \\ 6 & 13 - \lambda \end{pmatrix} \right)
= (4 - \lambda)(13 - \lambda) - 36 = \lambda^2 - 17\lambda + 16
$$

Solving the quadratic:

$$
\lambda = \frac{17 \pm \sqrt{289 - 64}}{2} = \frac{17 \pm \sqrt{225}}{2} = \frac{17 \pm 15}{2}
\Rightarrow \lambda = 16 \text{ or } 1
$$

So the singular values of $ A $ are $ \sqrt{16} = 4 $ and $ \sqrt{1} = 1 $. Therefore, the **largest singular value** is $ 4 $, and hence the **operator norm** of $ A $ is $ 4 $.

---

### Step 3: Verifying with a Specific Vector

To confirm, consider the vector $ \bold{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} $, which is an eigenvector of $ A^T A $ corresponding to the eigenvalue $ 16 $.

Compute $ A\bold{v} $:

$$
A \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2(1) + 3(2) \\ 0(1) - 2(2) \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \end{pmatrix}
$$

$$
\|A\bold{v}\| = \sqrt{8^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}
$$

$$
\|\bold{v}\| = \sqrt{1^2 + 2^2} = \sqrt{5}
$$

$$
\frac{\|A\bold{v}\|}{\|\bold{v}\|} = \frac{4\sqrt{5}}{\sqrt{5}} = 4
$$

This confirms that the maximum ratio is indeed $ 4 $, and thus the smallest $ C $ satisfying the inequality for all $ \bold{v} $ is:

$$
\boxed{4}
$$

Solution:
Let $\bold{v} = \begin{pmatrix} x \\ y \end{pmatrix}$.  Then
\[\|\bold{v}\| = \left\| \begin{pmatrix} x \\ y \end{pmatrix} \right\| = \sqrt{x^2 + y^2},\]and
\begin{align*}
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| &= \left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \right\| \\
&= \left\| \begin{pmatrix} 2x + 3y \\ -2y \end{pmatrix} \right\| \\
&= \sqrt{(2x + 3y)^2 + (-2y)^2} \\
&= \sqrt{4x^2 + 12xy + 13y^2},
\end{align*}so the given inequality becomes
\[\sqrt{4x^2 + 12xy + 13y^2} \le C \sqrt{x^2 + y^2},\]or
\[\sqrt{\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2}} \le C.\]Thus, we can think of $C$ as the maximum value of the expression in the left-hand side.

Maximizing the expression in the left-hand side is equivalent to maximizing its square, namely
\[\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2}.\]Let $k$ be a possible value of this expression, which means the equation
\[\frac{4x^2 + 12xy + 13y^2}{x^2 + y^2} = k\]has a solution in $x$ and $y$.  We can re-write this equation as
\[(4 - k) x^2 + 12xy + (13 - k) y^2 = 0.\]For this quadratic expression to have a solution in $x$ and $y$, its discriminant must be nonnegative.  In other words,
\[12^2 - 4 (4 - k)(13 - k) \ge 0,\]or $4k^2 - 68k + 64 \le 0$.  This inequality factors as $4(k - 1)(k - 16) \le 0$.  The largest value of $k$ that satisfies this inequality is 16, so the value of $C$ we seek is $\sqrt{16} = \boxed{4}$.  Note that equality occurs for
\[\bold{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}.\]


Explanation:
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, I have to compute the sum of each of these polynomials evaluated at 1. Let me start by trying to factor the given polynomial.

First, I notice that the polynomial is $ x^8 + 3x^4 - 4 $. It looks like a quadratic in terms of $ x^4 $. Maybe I can make a substitution to simplify it. Let me set $ y = x^4 $. Then the polynomial becomes $ y^2 + 3y - 4 $. Let me try to factor this quadratic.

Looking for two numbers that multiply to -4 and add up to 3. Let me think... 4 and -1? Yes, 4 * (-1) = -4 and 4 + (-1) = 3. So, the quadratic factors as $ (y + 4)(y - 1) $. Therefore, substituting back $ y = x^4 $, we get:

$ (x^4 + 4)(x^4 - 1) $.

So, the original polynomial factors into $ (x^4 + 4)(x^4 - 1) $. Now, I need to check if these factors can be factored further over the integers.

Starting with $ x^4 - 1 $. This is 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 - 1)(x + 1) $. So, putting it all together:

$ x^4 - 1 = (x - 1)(x + 1)(x^2 + 1) $.

Now, check if $ x^4 + 4 $ can be factored further. Hmm, I remember that $ x^4 + 4 $ is a special case. It's a biquadratic polynomial. Maybe it can be factored into quadratics? Let me try to factor it.

I recall that $ x^4 + 4 $ can be factored using the Sophie Germain identity. The identity is $ a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) $. Let me check if that applies here. If I set $ a = x $ and $ b = 1 $, then:

$ x^4 + 4(1)^4 = x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2) $.

Yes! So, $ x^4 + 4 $ factors into $ (x^2 + 2x + 2)(x^2 - 2x + 2) $. Let me verify this multiplication:

Multiply $ (x^2 + 2x + 2)(x^2 - 2x + 2) $:

First, multiply $ x^2 $ by each term in the second polynomial: $ x^2 * x^2 = x^4 $, $ x^2 * (-2x) = -2x^3 $, $ x^2 * 2 = 2x^2 $.

Then, multiply 2x by each term: $ 2x * x^2 = 2x^3 $, $ 2x * (-2x) = -4x^2 $, $ 2x * 2 = 4x $.

Then, multiply 2 by each term: $ 2 * x^2 = 2x^2 $, $ 2 * (-2x) = -4x $, $ 2 * 2 = 4 $.

Now, add all these terms together:

- $ x^4 $
- $ -2x^3 + 2x^3 = 0 $
- $ 2x^2 - 4x^2 + 2x^2 = 0 $
- $ 4x - 4x = 0 $
- $ +4 $

So, the result is $ x^4 + 4 $. Perfect! So, that factorization is correct.

Therefore, the original polynomial factors as:

$ (x^2 + 2x + 2)(x^2 - 2x + 2)(x - 1)(x + 1)(x^2 + 1) $.

Now, I need to check if any of these factors can be factored further over the integers. Let's check each factor:

1. $ x - 1 $: linear, can't be factored further.
2. $ x + 1 $: linear, can't be factored further.
3. $ x^2 + 1 $: This is irreducible over integers because it has no real roots (discriminant is $ 0^2 - 4*1*1 = -4 < 0 $) and cannot be factored into linear terms with integer coefficients.
4. $ x^2 + 2x + 2 $: Let's check discriminant: $ (2)^2 - 4*1*2 = 4 - 8 = -4 < 0 $. So, also irreducible over integers.
5. $ x^2 - 2x + 2 $: Discriminant is $ (-2)^2 - 4*1*2 = 4 - 8 = -4 < 0 $. Also irreducible over integers.

Therefore, all factors are irreducible over integers. So, the complete factorization is into these five polynomials: $ (x - 1) $, $ (x + 1) $, $ (x^2 + 1) $, $ (x^2 + 2x + 2) $, and $ (x^2 - 2x + 2) $.

Wait, but the problem says "each non-constant polynomial $ p_i(x) $ is monic with integer coefficients, and cannot be factored further over the integers." So, all these factors are monic, have integer coefficients, and are irreducible. Therefore, the factorization is correct.

Now, the question asks for the sum $ p_1(1) + p_2(1) + \dots + p_k(1) $. So, I need to evaluate each of these polynomials at x=1 and add them up.

Let me compute each one:

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

Adding them up: 0 + 2 + 2 + 5 + 1 = 10.

Wait, let me check again:

First polynomial: 0

Second: 2

Third: 2

Fourth: 5

Fifth: 1

Total: 0 + 2 is 2, plus 2 is 4, plus 5 is 9, plus 1 is 10. So, 10?

But let me check if I have the correct factors. Let me verify the original polynomial's value at x=1 and see if the product of the factors evaluated at 1 equals the original polynomial evaluated at 1.

Original polynomial: $ x^8 + 3x^4 - 4 $. At x=1: 1 + 3 - 4 = 0. So, the product of all p_i(1) should be 0. Let me check:

Product of p_i(1): 0 * 2 * 2 * 5 * 1 = 0. Which matches. So that's correct.

But the question isn't asking for the product, it's asking for the sum. So, according to my calculation, the sum is 10. But let me check again each p_i(1):

First factor: x - 1 at 1 is 0. Correct.

Second factor: x + 1 at 1 is 2. Correct.

Third factor: x² + 1 at 1 is 2. Correct.

Fourth factor: x² + 2x + 2 at 1: 1 + 2 + 2 = 5. Correct.

Fifth factor: x² - 2x + 2 at 1: 1 - 2 + 2 = 1. Correct.

Sum: 0 + 2 + 2 + 5 + 1 = 10. So, 10. Is that the answer? Wait, but let me check again if I have all the factors. Let me check the original factorization again.

Original polynomial: x^8 + 3x^4 - 4 = (x^4 + 4)(x^4 - 1) = (x^4 + 4)(x^2 - 1)(x^2 + 1) = (x^4 + 4)(x - 1)(x + 1)(x^2 + 1). Then, x^4 + 4 factors into (x^2 + 2x + 2)(x^2 - 2x + 2). Therefore, the full factorization is (x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2). So, five factors. So, five polynomials. So, the sum is 0 + 2 + 2 + 5 + 1 = 10. So, 10. But let me check again if there's another way to factor or if I missed something.

Wait, but maybe there's another factorization? Let me think. For example, could x^4 + 4 be factored into linear terms? But over integers, no. Because x^4 + 4 has no real roots, so it can't be factored into linear terms. And as we saw, it factors into quadratics. So, the factorization is correct.

Alternatively, maybe there's a different factorization? Let me check. Suppose someone tried to factor the original polynomial differently. Let me check if there are other possible factorizations.

Alternatively, maybe the original polynomial can be factored into more factors? For example, if there are more linear factors? But the original polynomial evaluated at x=1 is 0, so x=1 is a root. Similarly, x=-1 is a root? Let me check:

Original polynomial at x=1: 1 + 3 - 4 = 0. At x=-1: (-1)^8 + 3*(-1)^4 - 4 = 1 + 3 - 4 = 0. So, both x=1 and x=-1 are roots. Therefore, (x - 1)(x + 1) are factors, which we have. Then, the rest of the factors are quadratics. So, seems correct.

Alternatively, maybe there's a way to factor the original polynomial into more polynomials? For example, if some of the quadratics can be factored further? But we checked that x^2 + 1, x^2 + 2x + 2, and x^2 - 2x + 2 are irreducible over integers. So, no. Therefore, the factorization is complete.

Therefore, the answer should be 10. Wait, but let me check again the sum:

p1(1) = 0

p2(1) = 2

p3(1) = 2

p4(1) = 5

p5(1) = 1

Total: 0 + 2 + 2 + 5 + 1 = 10. Yes. So, 10. But let me check once more if I have the correct factors. Let me compute the product of all p_i(x):

(x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2). Let me multiply them step by step.

First, (x - 1)(x + 1) = x^2 - 1.

Then, multiply by (x^2 + 1): (x^2 - 1)(x^2 + 1) = x^4 - 1.

Then, multiply by (x^2 + 2x + 2): (x^4 - 1)(x^2 + 2x + 2). Let me compute that:

Multiply x^4 by (x^2 + 2x + 2): x^6 + 2x^5 + 2x^4

Multiply -1 by (x^2 + 2x + 2): -x^2 - 2x - 2

So, adding together: x^6 + 2x^5 + 2x^4 - x^2 - 2x - 2.

Then, multiply by (x^2 - 2x + 2). Let me denote this as (A)(B), where A = x^6 + 2x^5 + 2x^4 - x^2 - 2x - 2 and B = x^2 - 2x + 2.

This seems complicated, but maybe there's a smarter way. Alternatively, since we know that the original polynomial is x^8 + 3x^4 - 4, and we have already factored it as (x^4 + 4)(x^4 - 1), which is equal to (x^4 + 4)(x^2 - 1)(x^2 + 1). Then, since x^4 + 4 factors into (x^2 + 2x + 2)(x^2 - 2x + 2), so the entire product is (x^2 + 2x + 2)(x^2 - 2x + 2)(x^2 - 1)(x^2 + 1). Which is the same as the product of all the p_i(x). Therefore, the factorization is correct.

Alternatively, maybe I can check the product of the p_i(1) equals the original polynomial evaluated at 1. Which is 0, as we saw. And the product of p_i(1) is 0 * 2 * 2 * 5 * 1 = 0. So that's correct. Therefore, the factorization is correct.

Therefore, the answer is 10. Wait, but let me check again if the problem says "non-constant polynomial p_i(x)", so all factors are non-constant, which they are. So, the answer is 10. But let me check again the sum:

0 + 2 + 2 + 5 + 1 = 10. Yes. So, I think that's correct.

But just to be thorough, let me check if there's another way to factor the original polynomial. For example, maybe into different polynomials? Let me think. Suppose someone tried to factor x^8 + 3x^4 - 4 as a product of two quartic polynomials? But since we already factored it into a product of a quartic and a quartic, but then further factored into lower degrees. But since the problem says "cannot be factored further over the integers", so the answer is as above.

Alternatively, maybe there's a different factorization? For example, suppose someone tried to factor x^8 + 3x^4 - 4 as (x^4 + ax^3 + bx^2 + cx + d)(x^4 + ...). But that would be complicated. However, since we already have a valid factorization, and the problem says "cannot be factored further", so our factorization is correct.

Therefore, I think the answer is 10. Wait, but let me check once more the individual evaluations:

For p1(x) = x - 1: 1 - 1 = 0. Correct.

p2(x) = x + 1: 1 + 1 = 2. Correct.

p3(x) = x² + 1: 1 + 1 = 2. Correct.

p4(x) = x² + 2x + 2: 1 + 2 + 2 = 5. Correct.

p5(x) = x² - 2x + 2: 1 - 2 + 2 = 1. Correct.

Sum: 0 + 2 + 2 + 5 + 1 = 10. Yes. So, I think that's correct.

But just to be 100% sure, let me check if there's a possibility that some of the factors could be combined or if there's a different factorization. For example, could x^2 + 1 be combined with another factor? But since they are all irreducible, they can't be combined further. So, the answer is 10.

Wait, but let me check if the problem says "each non-constant polynomial p_i(x) is monic with integer coefficients, and cannot be factored further over the integers." So, the factors are monic, which they are. For example, x - 1 is monic, x + 1 is monic, x² + 1 is monic, etc. So, all factors are monic. Therefore, the answer is 10.

But just to check, maybe I made a mistake in the factorization? Let me check the original polynomial with the factors:

Multiply all the factors:

First, (x - 1)(x + 1) = x² - 1.

Multiply by (x² + 1): (x² - 1)(x² + 1) = x⁴ - 1.

Multiply by (x² + 2x + 2): (x⁴ - 1)(x² + 2x + 2). Let me compute this:

Multiply x⁴ by (x² + 2x + 2): x⁶ + 2x⁵ + 2x⁴.

Multiply -1 by (x² + 2x + 2): -x² - 2x - 2.

So, total is x⁶ + 2x⁵ + 2x⁴ - x² - 2x - 2.

Now, multiply by (x² - 2x + 2):

Let me denote this as (x⁶ + 2x⁵ + 2x⁴ - x² - 2x - 2)(x² - 2x + 2). Let me compute term by term.

First, multiply x⁶ by (x² - 2x + 2): x⁸ - 2x⁷ + 2x⁶.

Then, multiply 2x⁵ by (x² - 2x + 2): 2x⁷ - 4x⁶ + 4x⁵.

Then, multiply 2x⁴ by (x² - 2x + 2): 2x⁶ - 4x⁵ + 4x⁴.

Then, multiply -x² by (x² - 2x + 2): -x⁴ + 2x³ - 2x².

Then, multiply -2x by (x² - 2x + 2): -2x³ + 4x² - 4x.

Then, multiply -2 by (x² - 2x + 2): -2x² + 4x - 4.

Now, let me add all these terms together:

Start with the highest degree term:

x⁸

For x⁷: -2x⁷ + 2x⁷ = 0.

For x⁶: 2x⁶ - 4x⁶ + 2x⁶ = 0.

For x⁵: 4x⁵ - 4x⁵ = 0.

For x⁴: 4x⁴ - x⁴ = 3x⁴.

For x³: 2x³ - 2x³ = 0.

For x²: -2x² + 4x² - 2x² = 0.

For x: -4x + 4x = 0.

Constants: -4.

So, combining all terms:

x⁸ + 3x⁴ - 4. Which matches the original polynomial. Perfect! Therefore, the factorization is correct. Therefore, the sum is indeed 10.

Therefore, the answer is 10. So, I think that's correct.

**Final Answer**
\boxed{10}

To solve the problem, we are given the polynomial:

$$
x^8 + 3x^4 - 4
$$

and asked to factor it into monic polynomials with integer coefficients that cannot be factored further over the integers. Then, we are to compute the sum of the values of these polynomials evaluated at $ x = 1 $.

---

### Step 1: Factor the Polynomial

We begin by observing that the polynomial is a biquadratic in $ x^4 $. Let’s make the substitution $ y = x^4 $, so the polynomial becomes:

$$
y^2 + 3y - 4
$$

This is a quadratic in $ y $, and we can factor it as:

$$
y^2 + 3y - 4 = (y + 4)(y - 1)
$$

Substituting back $ y = x^4 $, we get:

$$
x^8 + 3x^4 - 4 = (x^4 + 4)(x^4 - 1)
$$

Now, we factor each of these terms further:

- $ x^4 - 1 $ is a difference of squares:

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

- $ x^4 + 4 $ is a special case that can be factored using the **Sophie Germain identity**:

$$
x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)
$$

So, the full factorization of the original polynomial is:

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

---

### Step 2: Verify Irreducibility

Each of the factors is a monic polynomial with integer coefficients. We now check whether they can be factored further over the integers:

- $ x - 1 $ and $ x + 1 $ are linear and irreducible.
- $ x^2 + 1 $, $ x^2 + 2x + 2 $, and $ x^2 - 2x + 2 $ all have negative discriminants:

$$
\text{Discriminant of } x^2 + 1 = 0^2 - 4(1)(1) = -4 < 0
$$
$$
\text{Discriminant of } x^2 + 2x + 2 = 4 - 8 = -4 < 0
$$
$$
\text{Discriminant of } x^2 - 2x + 2 = 4 - 8 = -4 < 0
$$

Thus, all factors are irreducible over the integers.

---

### Step 3: Evaluate Each Factor at $ x = 1 $

We now compute the value of each factor at $ x = 1 $:

- $ (x - 1) \Rightarrow 1 - 1 = 0 $
- $ (x + 1) \Rightarrow 1 + 1 = 2 $
- $ (x^2 + 1) \Rightarrow 1^2 + 1 = 2 $
- $ (x^2 + 2x + 2) \Rightarrow 1^2 + 2(1) + 2 = 1 + 2 + 2 = 5 $
- $ (x^2 - 2x + 2) \Rightarrow 1^2 - 2(1) + 2 = 1 - 2 + 2 = 1 $

---

### Step 4: Compute the Sum

Now, summing the values:

$$
0 + 2 + 2 + 5 + 1 = \boxed{10}
$$

---

### Final Answer

$$
\boxed{10}
$$

[Assistant]
Explanation:
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, I have to compute the sum of each of these polynomials evaluated at 1. Let me start by trying to factor the given polynomial.

First, I notice that the polynomial is $ x^8 + 3x^4 - 4 $. It looks like a quadratic in terms of $ x^4 $. Maybe I can make a substitution to simplify it. Let me set $ y = x^4 $. Then the polynomial becomes $ y^2 + 3y - 4 $. Let me try to factor this quadratic.

Looking for two numbers that multiply to -4 and add up to 3. Let me think... 4 and -1? Yes, 4 * (-1) = -4 and 4 + (-1) = 3. So, the quadratic factors as $ (y + 4)(y - 1) $. Therefore, substituting back $ y = x^4 $, we get:

$ (x^4 + 4)(x^4 - 1) $.

So, the original polynomial factors into $ (x^4 + 4)(x^4 - 1) $. Now, I need to check if these factors can be factored further over the integers.

Starting with $ x^4 - 1 $. This is 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 - 1)(x + 1) $. So, putting it all together:

$ x^4 - 1 = (x - 1)(x + 1)(x^2 + 1) $.

Now, check if $ x^4 + 4 $ can be factored further. Hmm, I remember that $ x^4 + 4 $ is a special case. It's a biquadratic polynomial. Maybe it can be factored into quadratics? Let me try to factor it.

I recall that $ x^4 + 4 $ can be factored using the Sophie Germain identity. The identity is $ a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) $. Let me check if that applies here. If I set $ a = x $ and $ b = 1 $, then:

$ x^4 + 4(1)^4 = x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2) $.

Yes! So, $ x^4 + 4 $ factors into $ (x^2 + 2x + 2)(x^2 - 2x + 2) $. Let me verify this multiplication:

Multiply $ (x^2 + 2x + 2)(x^2 - 2x + 2) $:

First, multiply $ x^2 $ by each term in the second polynomial: $ x^2 * x^2 = x^4 $, $ x^2 * (-2x) = -2x^3 $, $ x^2 * 2 = 2x^2 $.

Then, multiply 2x by each term: $ 2x * x^2 = 2x^3 $, $ 2x * (-2x) = -4x^2 $, $ 2x * 2 = 4x $.

Then, multiply 2 by each term: $ 2 * x^2 = 2x^2 $, $ 2 * (-2x) = -4x $, $ 2 * 2 = 4 $.

Now, add all these terms together:

- $ x^4 $
- $ -2x^3 + 2x^3 = 0 $
- $ 2x^2 - 4x^2 + 2x^2 = 0 $
- $ 4x - 4x = 0 $
- $ +4 $

So, the result is $ x^4 + 4 $. Perfect! So, that factorization is correct.

Therefore, the original polynomial factors as:

$ (x^2 + 2x + 2)(x^2 - 2x + 2)(x - 1)(x + 1)(x^2 + 1) $.

Now, I need to check if any of these factors can be factored further over the integers. Let's check each factor:

1. $ x - 1 $: linear, can't be factored further.
2. $ x + 1 $: linear, can't be factored further.
3. $ x^2 + 1 $: This is irreducible over integers because it has no real roots (discriminant is $ 0^2 - 4*1*1 = -4 < 0 $) and cannot be factored into linear terms with integer coefficients.
4. $ x^2 + 2x + 2 $: Let's check discriminant: $ (2)^2 - 4*1*2 = 4 - 8 = -4 < 0 $. So, also irreducible over integers.
5. $ x^2 - 2x + 2 $: Discriminant is $ (-2)^2 - 4*1*2 = 4 - 8 = -4 < 0 $. Also irreducible over integers.

Therefore, all factors are irreducible over integers. So, the complete factorization is into these five polynomials: $ (x - 1) $, $ (x + 1) $, $ (x^2 + 1) $, $ (x^2 + 2x + 2) $, and $ (x^2 - 2x + 2) $.

Wait, but the problem says "each non-constant polynomial $ p_i(x) $ is monic with integer coefficients, and cannot be factored further over the integers." So, all these factors are monic, have integer coefficients, and are irreducible. Therefore, the factorization is correct.

Now, the question asks for the sum $ p_1(1) + p_2(1) + \dots + p_k(1) $. So, I need to evaluate each of these polynomials at x=1 and add them up.

Let me compute each one:

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

Adding them up: 0 + 2 + 2 + 5 + 1 = 10.

Wait, let me check again:

First polynomial: 0

Second: 2

Third: 2

Fourth: 5

Fifth: 1

Total: 0 + 2 is 2, plus 2 is 4, plus 5 is 9, plus 1 is 10. So, 10?

But let me check if I have the correct factors. Let me verify the original polynomial's value at x=1 and see if the product of the factors evaluated at 1 equals the original polynomial evaluated at 1.

Original polynomial: $ x^8 + 3x^4 - 4 $. At x=1: 1 + 3 - 4 = 0. So, the product of all p_i(1) should be 0. Let me check:

Product of p_i(1): 0 * 2 * 2 * 5 * 1 = 0. Which matches. So that's correct.

But the question isn't asking for the product, it's asking for the sum. So, according to my calculation, the sum is 10. But let me check again each p_i(1):

First factor: x - 1 at 1 is 0. Correct.

Second factor: x + 1 at 1 is 2. Correct.

Third factor: x² + 1 at 1 is 2. Correct.

Fourth factor: x² + 2x + 2 at 1: 1 + 2 + 2 = 5. Correct.

Fifth factor: x² - 2x + 2 at 1: 1 - 2 + 2 = 1. Correct.

Sum: 0 + 2 + 2 + 5 + 1 = 10. So, 10. Is that the answer? Wait, but let me check again if I have all the factors. Let me check the original factorization again.

Original polynomial: x^8 + 3x^4 - 4 = (x^4 + 4)(x^4 - 1) = (x^4 + 4)(x^2 - 1)(x^2 + 1) = (x^4 + 4)(x - 1)(x + 1)(x^2 + 1). Then, x^4 + 4 factors into (x^2 + 2x + 2)(x^2 - 2x + 2). Therefore, the full factorization is (x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2). So, five factors. So, five polynomials. So, the sum is 0 + 2 + 2 + 5 + 1 = 10. So, 10. But let me check again if there's another way to factor or if I missed something.

Wait, but maybe there's another factorization? Let me think. For example, could x^4 + 4 be factored into linear terms? But over integers, no. Because x^4 + 4 has no real roots, so it can't be factored into linear terms. And as we saw, it factors into quadratics. So, the factorization is correct.

Alternatively, maybe there's a different factorization? Let me check. Suppose someone tried to factor the original polynomial differently. Let me check if there are other possible factorizations.

Alternatively, maybe the original polynomial can be factored into more factors? For example, if there are more linear factors? But the original polynomial evaluated at x=1 is 0, so x=1 is a root. Similarly, x=-1 is a root? Let me check:

Original polynomial at x=1: 1 + 3 - 4 = 0. At x=-1: (-1)^8 + 3*(-1)^4 - 4 = 1 + 3 - 4 = 0. So, both x=1 and x=-1 are roots. Therefore, (x - 1)(x + 1) are factors, which we have. Then, the rest of the factors are quadratics. So, seems correct.

Alternatively, maybe there's a way to factor the original polynomial into more polynomials? For example, if some of the quadratics can be factored further? But we checked that x^2 + 1, x^2 + 2x + 2, and x^2 - 2x + 2 are irreducible over integers. So, no. Therefore, the factorization is complete.

Therefore, the answer should be 10. Wait, but let me check again the sum:

p1(1) = 0

p2(1) = 2

p3(1) = 2

p4(1) = 5

p5(1) = 1

Total: 0 + 2 + 2 + 5 + 1 = 10. Yes. So, 10. But let me check once more if I have the correct factors. Let me compute the product of all p_i(x):

(x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2). Let me multiply them step by step.

First, (x - 1)(x + 1) = x^2 - 1.

Then, multiply by (x^2 + 1): (x^2 - 1)(x^2 + 1) = x^4 - 1.

Then, multiply by (x^2 + 2x + 2): (x^4 - 1)(x^2 + 2x + 2). Let me compute that:

Multiply x^4 by (x^2 + 2x + 2): x^6 + 2x^5 + 2x^4

Multiply -1 by (x^2 + 2x + 2): -x^2 - 2x - 2

So, adding together: x^6 + 2x^5 + 2x^4 - x^2 - 2x - 2.

Then, multiply by (x^2 - 2x + 2). Let me denote this as (A)(B), where A = x^6 + 2x^5 + 2x^4 - x^2 - 2x - 2 and B = x^2 - 2x + 2.

This seems complicated, but maybe there's a smarter way. Alternatively, since we know that the original polynomial is x^8 + 3x^4 - 4, and we have already factored it as (x^4 + 4)(x^4 - 1), which is equal to (x^4 + 4)(x^2 - 1)(x^2 + 1). Then, since x^4 + 4 factors into (x^2 + 2x + 2)(x^2 - 2x + 2), so the entire product is (x^2 + 2x + 2)(x^2 - 2x + 2)(x^2 - 1)(x^2 + 1). Which is the same as the product of all the p_i(x). Therefore, the factorization is correct.

Alternatively, maybe I can check the product of the p_i(1) equals the original polynomial evaluated at 1. Which is 0, as we saw. And the product of p_i(1) is 0 * 2 * 2 * 5 * 1 = 0. So that's correct. Therefore, the factorization is correct.

Therefore, the answer is 10. Wait, but let me check again if the problem says "non-constant polynomial p_i(x)", so all factors are non-constant, which they are. So, the answer is 10. But let me check again the sum:

0 + 2 + 2 + 5 + 1 = 10. Yes. So, I think that's correct.

But just to be thorough, let me check if there's another way to factor the original polynomial. For example, maybe into different polynomials? Let me think. Suppose someone tried to factor x^8 + 3x^4 - 4 as a product of two quartic polynomials? But since we already factored it into a product of a quartic and a quartic, but then further factored into lower degrees. But since the problem says "cannot be factored further over the integers", so the answer is as above.

Alternatively, maybe there's a different factorization? For example, suppose someone tried to factor x^8 + 3x^4 - 4 as (x^4 + ax^3 + bx^2 + cx + d)(x^4 + ...). But that would be complicated. However, since we already have a valid factorization, and the problem says "cannot be factored further", so our factorization is correct.

Therefore, I think the answer is 10. Wait, but let me check once more the individual evaluations:

For p1(x) = x - 1: 1 - 1 = 0. Correct.

p2(x) = x + 1: 1 + 1 = 2. Correct.

p3(x) = x² + 1: 1 + 1 = 2. Correct.

p4(x) = x² + 2x + 2: 1 + 2 + 2 = 5. Correct.

p5(x) = x² - 2x + 2: 1 - 2 + 2 = 1. Correct.

Sum: 0 + 2 + 2 + 5 + 1 = 10. Yes. So, I think that's correct.

But just to be 100% sure, let me check if there's a possibility that some of the factors could be combined or if there's a different factorization. For example, could x^2 + 1 be combined with another factor? But since they are all irreducible, they can't be combined further. So, the answer is 10.

Wait, but let me check if the problem says "each non-constant polynomial p_i(x) is monic with integer coefficients, and cannot be factored further over the integers." So, the factors are monic, which they are. For example, x - 1 is monic, x + 1 is monic, x² + 1 is monic, etc. So, all factors are monic. Therefore, the answer is 10.

But just to check, maybe I made a mistake in the factorization? Let me check the original polynomial with the factors:

Multiply all the factors:

First, (x - 1)(x + 1) = x² - 1.

Multiply by (x² + 1): (x² - 1)(x² + 1) = x⁴ - 1.

Multiply by (x² + 2x + 2): (x⁴ - 1)(x² + 2x + 2). Let me compute this:

Multiply x⁴ by (x² + 2x + 2): x⁶ + 2x⁵ + 2x⁴.

Multiply -1 by (x² + 2x + 2): -x² - 2x - 2.

So, total is x⁶ + 2x⁵ + 2x⁴ - x² - 2x - 2.

Now, multiply by (x² - 2x + 2):

Let me denote this as (x⁶ + 2x⁵ + 2x⁴ - x² - 2x - 2)(x² - 2x + 2). Let me compute term by term.

First, multiply x⁶ by (x² - 2x + 2): x⁸ - 2x⁷ + 2x⁶.

Then, multiply 2x⁵ by (x² - 2x + 2): 2x⁷ - 4x⁶ + 4x⁵.

Then, multiply 2x⁴ by (x² - 2x + 2): 2x⁶ - 4x⁵ + 4x⁴.

Then, multiply -x² by (x² - 2x + 2): -x⁴ + 2x³ - 2x².

Then, multiply -2x by (x² - 2x + 2): -2x³ + 4x² - 4x.

Then, multiply -2 by (x² - 2x + 2): -2x² + 4x - 4.

Now, let me add all these terms together:

Start with the highest degree term:

x⁸

For x⁷: -2x⁷ + 2x⁷ = 0.

For x⁶: 2x⁶ - 4x⁶ + 2x⁶ = 0.

For x⁵: 4x⁵ - 4x⁵ = 0.

For x⁴: 4x⁴ - x⁴ = 3x⁴.

For x³: 2x³ - 2x³ = 0.

For x²: -2x² + 4x² - 2x² = 0.

For x: -4x + 4x = 0.

Constants: -4.

So, combining all terms:

x⁸ + 3x⁴ - 4. Which matches the original polynomial. Perfect! Therefore, the factorization is correct. Therefore, the sum is indeed 10.

Therefore, the answer is 10. So, I think that's correct.

**Final Answer**
\boxed{10}

To solve the problem, we are given the polynomial:

$$
x^8 + 3x^4 - 4
$$

and asked to factor it into monic polynomials with integer coefficients that cannot be factored further over the integers. Then, we are to compute the sum of the values of these polynomials evaluated at $ x = 1 $.

---

### Step 1: Factor the Polynomial

We begin by observing that the polynomial is a biquadratic in $ x^4 $. Let’s make the substitution $ y = x^4 $, so the polynomial becomes:

$$
y^2 + 3y - 4
$$

This is a quadratic in $ y $, and we can factor it as:

$$
y^2 + 3y - 4 = (y + 4)(y - 1)
$$

Substituting back $ y = x^4 $, we get:

$$
x^8 + 3x^4 - 4 = (x^4 + 4)(x^4 - 1)
$$

Now, we factor each of these terms further:

- $ x^4 - 1 $ is a difference of squares:

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

- $ x^4 + 4 $ is a special case that can be factored using the **Sophie Germain identity**:

$$
x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)
$$

So, the full factorization of the original polynomial is:

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

---

### Step 2: Verify Irreducibility

Each of the factors is a monic polynomial with integer coefficients. We now check whether they can be factored further over the integers:

- $ x - 1 $ and $ x + 1 $ are linear and irreducible.
- $ x^2 + 1 $, $ x^2 + 2x + 2 $, and $ x^2 - 2x + 2 $ all have negative discriminants:

$$
\text{Discriminant of } x^2 + 1 = 0^2 - 4(1)(1) = -4 < 0
$$
$$
\text{Discriminant of } x^2 + 2x + 2 = 4 - 8 = -4 < 0
$$
$$
\text{Discriminant of } x^2 - 2x + 2 = 4 - 8 = -4 < 0
$$

Thus, all factors are irreducible over the integers.

---

### Step 3: Evaluate Each Factor at $ x = 1 $

We now compute the value of each factor at $ x = 1 $:

- $ (x - 1) \Rightarrow 1 - 1 = 0 $
- $ (x + 1) \Rightarrow 1 + 1 = 2 $
- $ (x^2 + 1) \Rightarrow 1^2 + 1 = 2 $
- $ (x^2 + 2x + 2) \Rightarrow 1^2 + 2(1) + 2 = 1 + 2 + 2 = 5 $
- $ (x^2 - 2x + 2) \Rightarrow 1^2 - 2(1) + 2 = 1 - 2 + 2 = 1 $

---

### Step 4: Compute the Sum

Now, summing the values:

$$
0 + 2 + 2 + 5 + 1 = \boxed{10}
$$

---

### Final Answer

$$
\boxed{10}
$$

Solution:
First, we can factor $x^8 + 3x^4 - 4$ as $(x^4 - 1)(x^4 + 4).$  Then
\[x^4 - 1 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x - 1)(x + 1),\]and by Sophie Germain,
\[x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2).\]Thus, the full factorization is
\[x^8 + 3x^4 - 4 = (x^2 + 1)(x - 1)(x + 1)(x^2 + 2x + 2)(x^2 - 2x + 2).\]Evaluating each factor at $x = 1,$ we get $2 + 0 + 2 + 5 + 1 = \boxed{10}.$