Explanation:
Okay, so I need to find the ordered triple (p, q, r) such that the polynomial x³ - 3x² + 4x - 1 is a factor of x⁹ + p x⁶ + q x³ + r. Hmm, let me think about how to approach this.

First, if x³ - 3x² + 4x - 1 is a factor of the given polynomial, then there must be another polynomial, let's call it Q(x), such that:

x⁹ + p x⁶ + q x³ + r = (x³ - 3x² + 4x - 1) * Q(x)

Since the original polynomial is of degree 9, and the factor is degree 3, then Q(x) must be a degree 6 polynomial. So, maybe I can write Q(x) as some cubic polynomial multiplied by another cubic polynomial? Wait, no. Let me check. If the original polynomial is degree 9, and the factor is degree 3, then Q(x) must be degree 6. So, Q(x) can be written as a degree 6 polynomial. However, given that the original polynomial is x⁹ + p x⁶ + q x³ + r, which only has terms with exponents that are multiples of 3. That seems interesting. So maybe Q(x) is also a polynomial with only terms of exponents that are multiples of 3? Or maybe there is some pattern here.

Alternatively, maybe we can use polynomial division? If I divide x⁹ + p x⁶ + q x³ + r by x³ - 3x² + 4x - 1, the remainder should be zero. But dividing a 9th-degree polynomial by a 3rd-degree polynomial might be tedious. Is there a better way?

Alternatively, since x³ - 3x² + 4x - 1 divides the given polynomial, then all roots of x³ - 3x² + 4x - 1 are also roots of x⁹ + p x⁶ + q x³ + r. Let me denote the roots of the cubic polynomial as α, β, γ. Then, for each of these roots, we have:

α³ - 3α² + 4α - 1 = 0 ⇒ α³ = 3α² - 4α + 1

Similarly for β and γ. So, maybe I can use this relation to express higher powers of α in terms of lower powers, which can help me simplify the expression α⁹ + p α⁶ + q α³ + r = 0. Since if α is a root, then plugging into the original polynomial gives zero. Therefore, for each root α, we can find relations that will allow us to write α⁹ in terms of lower-degree terms, which will let me solve for p, q, r.

This seems promising. Let me try that.

First, let me note that since α³ = 3α² -4α +1, maybe I can use this to find α⁶ and α⁹ in terms of lower powers. Let me compute α³, α⁴, α⁵, etc., step by step.

Starting with α³ = 3α² - 4α + 1.

Compute α⁴:

α⁴ = α * α³ = α*(3α² - 4α +1) = 3α³ -4α² + α.

But we can substitute α³ again:

= 3*(3α² -4α +1) -4α² + α = 9α² -12α +3 -4α² + α = (9α² -4α²) + (-12α + α) + 3 = 5α² -11α +3.

Similarly, compute α⁵:

α⁵ = α * α⁴ = α*(5α² -11α +3) = 5α³ -11α² +3α.

Substitute α³ again:

=5*(3α² -4α +1) -11α² +3α = 15α² -20α +5 -11α² +3α = (15α² -11α²) + (-20α +3α) +5 = 4α² -17α +5.

Now, α⁶:

α⁶ = α * α⁵ = α*(4α² -17α +5) = 4α³ -17α² +5α.

Substitute α³:

=4*(3α² -4α +1) -17α² +5α = 12α² -16α +4 -17α² +5α = (12α² -17α²) + (-16α +5α) +4 = -5α² -11α +4.

Okay, moving on to α⁷:

α⁷ = α * α⁶ = α*(-5α² -11α +4) = -5α³ -11α² +4α.

Substitute α³:

= -5*(3α² -4α +1) -11α² +4α = -15α² +20α -5 -11α² +4α = (-15α² -11α²) + (20α +4α) -5 = -26α² +24α -5.

Now α⁸:

α⁸ = α * α⁷ = α*(-26α² +24α -5) = -26α³ +24α² -5α.

Substitute α³:

= -26*(3α² -4α +1) +24α² -5α = -78α² +104α -26 +24α² -5α = (-78α² +24α²) + (104α -5α) -26 = -54α² +99α -26.

Then α⁹:

α⁹ = α * α⁸ = α*(-54α² +99α -26) = -54α³ +99α² -26α.

Substitute α³:

= -54*(3α² -4α +1) +99α² -26α = -162α² +216α -54 +99α² -26α = (-162α² +99α²) + (216α -26α) -54 = -63α² +190α -54.

So, α⁹ = -63α² +190α -54.

Therefore, the original equation α⁹ + p α⁶ + q α³ + r = 0 becomes:

(-63α² +190α -54) + p*(-5α² -11α +4) + q*(3α² -4α +1) + r = 0.

Let me expand this:

First, expand each term:

-63α² +190α -54 + p*(-5α² -11α +4) + q*(3α² -4α +1) + r.

Let me collect like terms.

Let me compute coefficients for α², α, and constants.

For α²:

-63α² -5pα² +3qα² = (-63 -5p +3q) α².

For α:

190α -11pα -4qα = (190 -11p -4q) α.

For constants:

-54 +4p + q + r.

Therefore, the entire expression is:

[(-63 -5p +3q) α² + (190 -11p -4q) α + (-54 +4p + q + r)] = 0.

Since this must hold for each root α of the cubic polynomial x³ -3x² +4x -1, which is irreducible over the rationals (I think?), so the coefficients of α², α, and the constant term must all be zero. Therefore, we have the following system of equations:

1. Coefficient of α²: -63 -5p +3q = 0

2. Coefficient of α: 190 -11p -4q = 0

3. Constant term: -54 +4p + q + r = 0

So, we have three equations:

Equation (1): -5p +3q = 63

Equation (2): -11p -4q = -190

Equation (3): 4p + q + r = 54

So, now we need to solve equations (1) and (2) for p and q, then substitute into equation (3) to find r.

Let me write equations (1) and (2):

1. -5p + 3q = 63

2. -11p -4q = -190

We can solve this system using substitution or elimination. Let me use elimination. Let's multiply equation (1) by 4 and equation (2) by 3 to eliminate q:

Multiply equation (1) by 4: -20p + 12q = 252

Multiply equation (2) by 3: -33p -12q = -570

Now add these two equations:

(-20p +12q) + (-33p -12q) = 252 + (-570)

This gives:

-53p = -318 ⇒ p = (-318)/(-53) = 318/53. Wait, 53*6 is 318. Yes, 53*6 = 318. Therefore, p = 6.

So p is 6. Now substitute p = 6 into equation (1):

-5*(6) +3q = 63 ⇒ -30 +3q = 63 ⇒ 3q = 93 ⇒ q = 31.

So, q = 31.

Now, substitute p = 6 and q = 31 into equation (3):

4*6 +31 + r = 54 ⇒ 24 +31 + r = 54 ⇒ 55 + r = 54 ⇒ r = 54 -55 = -1.

Therefore, p = 6, q = 31, r = -1. So the ordered triple is (6, 31, -1).

Wait, let me check if these values satisfy equation (2):

Equation (2): -11p -4q = -11*6 -4*31 = -66 -124 = -190, which matches. And equation (1): -5*6 +3*31 = -30 +93 = 63, which is correct. And equation (3): 4*6 +31 + (-1) = 24 +31 -1 = 54, which is correct. So that seems consistent. Therefore, the answer is (6,31,-1).

But let me just think again if there's another way to approach this problem. For example, since the original polynomial is x⁹ + p x⁶ + q x³ + r, which is like a polynomial in x³. Let me denote y = x³. Then, the polynomial becomes y³ + p y² + q y + r. So, if x³ -3x² +4x -1 divides the original polynomial, then perhaps y³ + p y² + q y + r is divisible by (x³ -3x² +4x -1). Wait, but how does that relate?

Alternatively, if we consider that the original polynomial is a polynomial in x³, so perhaps the factor x³ -3x² +4x -1 can be related to some substitution. However, since we have both x³ and lower-degree terms, maybe not directly.

Alternatively, perhaps we can think of the original polynomial as (x³)^3 + p(x³)^2 + q(x³) + r. So if we let y = x³, then the original polynomial is y³ + p y² + q y + r. But if x³ -3x² +4x -1 divides the original polynomial, then maybe we can relate y and the cubic polynomial. But how?

Let me think. If α is a root of x³ -3x² +4x -1 = 0, then α³ = 3α² -4α +1. Then, as above, we can express higher powers of α in terms of lower powers, and so for y = α³, we can write y = 3α² -4α +1. But maybe we can find a relationship between y and α?

Alternatively, perhaps we can consider that since the original polynomial is y³ + p y² + q y + r, and since y = α³, then substituting y into the polynomial gives:

(α³)^3 + p(α³)^2 + q(α³) + r = α⁹ + p α⁶ + q α³ + r = 0, as before. Which is the same as before. Therefore, we have the same equations. So our previous approach is correct.

Alternatively, perhaps we can consider that if we have x³ - 3x² +4x -1 divides x⁹ +p x⁶ + q x³ +r, then the original polynomial can be written as (x³ - 3x² +4x -1) multiplied by some polynomial. Since the original polynomial is x⁹ + p x⁶ + q x³ + r, which is a polynomial in x³. Let me consider that maybe the other factor is also a polynomial in x³? For example, if we consider that the original polynomial is (x³ - 3x² +4x -1)(something). Let me suppose that something is of the form x⁶ + a x³ + b. Because multiplying (x³ - 3x² +4x -1) by (x⁶ + a x³ + b) would give:

(x³)(x⁶ + a x³ + b) - 3x²(x⁶ + a x³ + b) +4x(x⁶ + a x³ + b) -1*(x⁶ + a x³ + b)

Let me compute each term:

First term: x³ * x⁶ + x³ * a x³ + x³ * b = x⁹ + a x⁶ + b x³

Second term: -3x² * x⁶ -3x² * a x³ -3x² * b = -3x⁸ -3a x⁵ -3b x²

Third term: 4x * x⁶ + 4x * a x³ + 4x * b = 4x⁷ + 4a x⁴ +4b x

Fourth term: -1 * x⁶ -1 * a x³ -1 * b = -x⁶ -a x³ -b

Now, adding all these together:

x⁹ + a x⁶ + b x³ -3x⁸ -3a x⁵ -3b x² +4x⁷ +4a x⁴ +4b x -x⁶ -a x³ -b.

Now, combine like terms:

x⁹

-3x⁸

+4x⁷

-3a x⁵

+4a x⁴

+a x⁶ -x⁶ = (a -1)x⁶

b x³ -a x³ = (b -a)x³

-3b x²

+4b x

- b

So overall, the product is:

x⁹ - 3x⁸ +4x⁷ + (a -1)x⁶ -3a x⁵ +4a x⁴ + (b -a)x³ -3b x² +4b x -b.

But the original polynomial is x⁹ + p x⁶ + q x³ + r. Comparing coefficients, we need all coefficients of x⁸, x⁷, x⁵, x⁴, x², x to be zero, and the coefficients of x⁹, x⁶, x³, and constant term to match. Therefore:

Coefficient of x⁸: -3 = 0? Wait, but the original polynomial has no x⁸ term. Therefore, -3 must be zero? That can't be. So clearly, this approach is flawed. Because if we assume that the other factor is x⁶ + a x³ + b, then the product has terms with x⁸, x⁷, etc., which are not present in the original polynomial. Therefore, this suggests that my assumption about the form of Q(x) is incorrect.

Therefore, perhaps Q(x) is not a simple cubic in x³, but something else. Alternatively, maybe there is some symmetry or pattern here. Wait, but the original polynomial is x⁹ + p x⁶ + q x³ + r. If we consider that perhaps the original polynomial can be written as (x³)^3 + p (x³)^2 + q (x³) + r, which is similar to a cubic in y = x³. Therefore, if the original polynomial is divisible by x³ -3x² +4x -1, then maybe substituting y = x³, and considering that y = 3x² -4x +1, as per the equation x³ = 3x² -4x +1. So if we can express y in terms of x, then perhaps we can find a relation for the polynomial in y?

Let me try that. Let me set y = x³. Then, from the cubic equation, we have:

y = 3x² -4x +1. So, perhaps we can express x² in terms of y and x? Let me rearrange:

From y = 3x² -4x +1 ⇒ 3x² = y +4x -1 ⇒ x² = (y +4x -1)/3.

But then, if we have expressions involving higher powers of x, but since our original polynomial is in terms of y, maybe we can find a relationship between y and the polynomial?

Alternatively, since the original polynomial is y³ + p y² + q y + r, and since y = 3x² -4x +1, substituting into the polynomial would give:

(3x² -4x +1)^3 + p(3x² -4x +1)^2 + q(3x² -4x +1) + r.

But this should be equal to zero for all x that satisfy the original cubic equation. But this seems more complicated. However, since we already solved the problem earlier by using the root substitution, which gave us p=6, q=31, r=-1, maybe we can check whether this works here.

Alternatively, since the original polynomial is divisible by the cubic, then the polynomial y³ + p y² + q y + r must be divisible by the cubic equation expressed in terms of y. Wait, but how?

Alternatively, since y = x³, and we have x³ -3x² +4x -1 = 0 ⇒ y -3x² +4x -1 =0 ⇒ y = 3x² -4x +1. Then, if we can eliminate x from these equations? Maybe?

But if I consider y = 3x² -4x +1, and also have to relate to the polynomial y³ + p y² + q y + r = 0. So, substituting y into that equation:

(3x² -4x +1)^3 + p(3x² -4x +1)^2 + q(3x² -4x +1) + r = 0.

But since this must hold for all x that satisfy the original cubic equation. However, since the original cubic equation is x³ -3x² +4x -1 = 0, which is equivalent to x³ = 3x² -4x +1. Therefore, maybe we can express higher powers of x in terms of lower powers. But if we have this equation involving y, which is already expressed in terms of x² and x, perhaps we can eliminate x? This might be complicated. However, maybe we can consider that the equation above must be identically zero for all x, but given that x is constrained by the original cubic. However, since the original cubic has three roots, and we already used the root method which gave us a solution, maybe there is no need to go further here.

But since my previous method gave me p=6, q=31, r=-1, which satisfies the equations, and the check with equations (1), (2), (3) worked, then maybe that's correct. However, perhaps I should verify with a specific value.

Alternatively, let me check if with p=6, q=31, r=-1, then the original polynomial is divisible by x³ -3x² +4x -1.

Let me perform polynomial division or check if the remainder is zero.

But with such high degrees, it might be tedious, but perhaps I can do it. Alternatively, let me take the original polynomial x⁹ +6x⁶ +31x³ -1 and divide by x³ -3x² +4x -1. But doing polynomial division for such high degrees is going to be lengthy. Alternatively, perhaps I can use synthetic division or some other trick. Wait, but since we already verified the equations through root substitution, which is a solid method, and solved the system correctly, then this should be correct.

Alternatively, let me check with the minimal polynomial. Suppose that (x³ -3x² +4x -1) divides the original polynomial. Let me compute the original polynomial evaluated at the roots of the cubic. But since we already did that, and the equations were satisfied, then it's okay.

Alternatively, let me check with a specific root. Suppose α is a root of x³ -3x² +4x -1 = 0. Then, as we found, α³ = 3α² -4α +1. Then, substituting into the original polynomial:

α⁹ +6α⁶ +31α³ -1 = ?

Earlier, we found that α⁹ = -63α² +190α -54

And α⁶ = -5α² -11α +4

And α³ = 3α² -4α +1.

So substituting:

(-63α² +190α -54) +6*(-5α² -11α +4) +31*(3α² -4α +1) -1.

Let me compute each term:

First term: -63α² +190α -54

Second term: 6*(-5α² -11α +4) = -30α² -66α +24

Third term: 31*(3α² -4α +1) = 93α² -124α +31

Fourth term: -1

Adding them all together:

Let me collect like terms:

For α²:

-63α² -30α² +93α² = (-63 -30 +93)α² = 0α²

For α:

190α -66α -124α = (190 -66 -124)α = 0α

Constants:

-54 +24 +31 -1 = (-54 +24) + (31 -1) = -30 +30 = 0

So indeed, the total is 0. So this confirms that with p=6, q=31, r=-1, the original polynomial is zero for each root α of the cubic, which means that the cubic divides the original polynomial. Therefore, the answer is (6, 31, -1).

Therefore, I think this is correct. I can't find any mistakes in the reasoning. So the ordered triple is (6, 31, -1).

**Final Answer**
The ordered triple is \boxed{(6, 31, -1)}.

To determine the ordered triple $(p, q, r)$ such that the polynomial $x^9 + p x^6 + q x^3 + r$ is divisible by $x^3 - 3x^2 + 4x - 1$, we proceed by analyzing their relationship algebraically.

---

### Step 1: Factor Structure

Let’s denote:

$$
P(x) = x^9 + p x^6 + q x^3 + r
$$
$$
F(x) = x^3 - 3x^2 + 4x - 1
$$

If $F(x)$ divides $P(x)$, then all roots of $F(x)$ are also roots of $P(x)$. Let $\alpha$ be a root of $F(x)$, hence $\alpha^3 = 3\alpha^2 - 4\alpha + 1$, which we'll use to reduce powers of $\alpha$ in $P(\alpha)$.

---

### Step 2: Express $\alpha^3$ and Higher Powers

Using $\alpha^3 = 3\alpha^2 - 4\alpha + 1$, we compute higher powers of $\alpha$:

- $\alpha^4 = \alpha \cdot \alpha^3 = \alpha(3\alpha^2 - 4\alpha + 1) = 3\alpha^3 - 4\alpha^2 + \alpha$
  - Substitute $\alpha^3 = 3\alpha^2 - 4\alpha + 1$
  - $\alpha^4 = 3(3\alpha^2 - 4\alpha + 1) - 4\alpha^2 + \alpha = 5\alpha^2 - 11\alpha + 3$

- $\alpha^5 = \alpha \cdot \alpha^4 = \alpha(5\alpha^2 - 11\alpha + 3) = 5\alpha^3 - 11\alpha^2 + 3\alpha$
  - Substitute $\alpha^3$
  - $\alpha^5 = 5(3\alpha^2 - 4\alpha + 1) - 11\alpha^2 + 3\alpha = 4\alpha^2 - 17\alpha + 5$

- $\alpha^6 = \alpha \cdot \alpha^5 = \alpha(4\alpha^2 - 17\alpha + 5) = 4\alpha^3 - 17\alpha^2 + 5\alpha$
  - Substitute $\alpha^3$
  - $\alpha^6 = 4(3\alpha^2 - 4\alpha + 1) - 17\alpha^2 + 5\alpha = -5\alpha^2 - 11\alpha + 4$

- $\alpha^9 = \alpha \cdot \alpha^8$ and can be derived as $-63\alpha^2 + 190\alpha - 54$ (as shown in the detailed process).

---

### Step 3: Apply to the Polynomial

We now substitute the expressions for $\alpha^9$, $\alpha^6$, and $\alpha^3$ into:

$$
P(\alpha) = \alpha^9 + p\alpha^6 + q\alpha^3 + r
$$

Substitute known values:

$$
P(\alpha) = (-63\alpha^2 + 190\alpha - 54) + p(-5\alpha^2 - 11\alpha + 4) + q(3\alpha^2 - 4\alpha + 1) + r
$$

Group coefficients:

- Coefficient of $\alpha^2$: $-63 - 5p + 3q$
- Coefficient of $\alpha$: $190 - 11p - 4q$
- Constant term: $-54 + 4p + q + r$

Set each to zero:

1. $-63 - 5p + 3q = 0$
2. $190 - 11p - 4q = 0$
3. $-54 + 4p + q + r = 0$

---

### Step 4: Solve the System of Equations

From equations (1) and (2):

- Equation (1): $-5p + 3q = 63$
- Equation (2): $-11p - 4q = -190$

Multiply equation (1) by 4 and equation (2) by 3:

- $-20p + 12q = 252$
- $-33p - 12q = -570$

Add:

$$
-53p = -318 \Rightarrow p = 6
$$

Substitute $p = 6$ into equation (1):

$$
-5(6) + 3q = 63 \Rightarrow -30 + 3q = 63 \Rightarrow 3q = 93 \Rightarrow q = 31
$$

Now substitute $p = 6$, $q = 31$ into equation (3):

$$
-54 + 4(6) + 31 + r = 0 \Rightarrow -54 + 24 + 31 + r = 0 \Rightarrow 55 + r = 0 \Rightarrow r = -1
$$

---

### Final Result

Thus, the ordered triple $(p, q, r)$ is:

$$
\boxed{(6, 31, -1)}
$$

[Assistant]
Explanation:
Okay, so I need to find the ordered triple (p, q, r) such that the polynomial x³ - 3x² + 4x - 1 is a factor of x⁹ + p x⁶ + q x³ + r. Hmm, let me think about how to approach this.

First, if x³ - 3x² + 4x - 1 is a factor of the given polynomial, then there must be another polynomial, let's call it Q(x), such that:

x⁹ + p x⁶ + q x³ + r = (x³ - 3x² + 4x - 1) * Q(x)

Since the original polynomial is of degree 9, and the factor is degree 3, then Q(x) must be a degree 6 polynomial. So, maybe I can write Q(x) as some cubic polynomial multiplied by another cubic polynomial? Wait, no. Let me check. If the original polynomial is degree 9, and the factor is degree 3, then Q(x) must be degree 6. So, Q(x) can be written as a degree 6 polynomial. However, given that the original polynomial is x⁹ + p x⁶ + q x³ + r, which only has terms with exponents that are multiples of 3. That seems interesting. So maybe Q(x) is also a polynomial with only terms of exponents that are multiples of 3? Or maybe there is some pattern here.

Alternatively, maybe we can use polynomial division? If I divide x⁹ + p x⁶ + q x³ + r by x³ - 3x² + 4x - 1, the remainder should be zero. But dividing a 9th-degree polynomial by a 3rd-degree polynomial might be tedious. Is there a better way?

Alternatively, since x³ - 3x² + 4x - 1 divides the given polynomial, then all roots of x³ - 3x² + 4x - 1 are also roots of x⁹ + p x⁶ + q x³ + r. Let me denote the roots of the cubic polynomial as α, β, γ. Then, for each of these roots, we have:

α³ - 3α² + 4α - 1 = 0 ⇒ α³ = 3α² - 4α + 1

Similarly for β and γ. So, maybe I can use this relation to express higher powers of α in terms of lower powers, which can help me simplify the expression α⁹ + p α⁶ + q α³ + r = 0. Since if α is a root, then plugging into the original polynomial gives zero. Therefore, for each root α, we can find relations that will allow us to write α⁹ in terms of lower-degree terms, which will let me solve for p, q, r.

This seems promising. Let me try that.

First, let me note that since α³ = 3α² -4α +1, maybe I can use this to find α⁶ and α⁹ in terms of lower powers. Let me compute α³, α⁴, α⁵, etc., step by step.

Starting with α³ = 3α² - 4α + 1.

Compute α⁴:

α⁴ = α * α³ = α*(3α² - 4α +1) = 3α³ -4α² + α.

But we can substitute α³ again:

= 3*(3α² -4α +1) -4α² + α = 9α² -12α +3 -4α² + α = (9α² -4α²) + (-12α + α) + 3 = 5α² -11α +3.

Similarly, compute α⁵:

α⁵ = α * α⁴ = α*(5α² -11α +3) = 5α³ -11α² +3α.

Substitute α³ again:

=5*(3α² -4α +1) -11α² +3α = 15α² -20α +5 -11α² +3α = (15α² -11α²) + (-20α +3α) +5 = 4α² -17α +5.

Now, α⁶:

α⁶ = α * α⁵ = α*(4α² -17α +5) = 4α³ -17α² +5α.

Substitute α³:

=4*(3α² -4α +1) -17α² +5α = 12α² -16α +4 -17α² +5α = (12α² -17α²) + (-16α +5α) +4 = -5α² -11α +4.

Okay, moving on to α⁷:

α⁷ = α * α⁶ = α*(-5α² -11α +4) = -5α³ -11α² +4α.

Substitute α³:

= -5*(3α² -4α +1) -11α² +4α = -15α² +20α -5 -11α² +4α = (-15α² -11α²) + (20α +4α) -5 = -26α² +24α -5.

Now α⁸:

α⁸ = α * α⁷ = α*(-26α² +24α -5) = -26α³ +24α² -5α.

Substitute α³:

= -26*(3α² -4α +1) +24α² -5α = -78α² +104α -26 +24α² -5α = (-78α² +24α²) + (104α -5α) -26 = -54α² +99α -26.

Then α⁹:

α⁹ = α * α⁸ = α*(-54α² +99α -26) = -54α³ +99α² -26α.

Substitute α³:

= -54*(3α² -4α +1) +99α² -26α = -162α² +216α -54 +99α² -26α = (-162α² +99α²) + (216α -26α) -54 = -63α² +190α -54.

So, α⁹ = -63α² +190α -54.

Therefore, the original equation α⁹ + p α⁶ + q α³ + r = 0 becomes:

(-63α² +190α -54) + p*(-5α² -11α +4) + q*(3α² -4α +1) + r = 0.

Let me expand this:

First, expand each term:

-63α² +190α -54 + p*(-5α² -11α +4) + q*(3α² -4α +1) + r.

Let me collect like terms.

Let me compute coefficients for α², α, and constants.

For α²:

-63α² -5pα² +3qα² = (-63 -5p +3q) α².

For α:

190α -11pα -4qα = (190 -11p -4q) α.

For constants:

-54 +4p + q + r.

Therefore, the entire expression is:

[(-63 -5p +3q) α² + (190 -11p -4q) α + (-54 +4p + q + r)] = 0.

Since this must hold for each root α of the cubic polynomial x³ -3x² +4x -1, which is irreducible over the rationals (I think?), so the coefficients of α², α, and the constant term must all be zero. Therefore, we have the following system of equations:

1. Coefficient of α²: -63 -5p +3q = 0

2. Coefficient of α: 190 -11p -4q = 0

3. Constant term: -54 +4p + q + r = 0

So, we have three equations:

Equation (1): -5p +3q = 63

Equation (2): -11p -4q = -190

Equation (3): 4p + q + r = 54

So, now we need to solve equations (1) and (2) for p and q, then substitute into equation (3) to find r.

Let me write equations (1) and (2):

1. -5p + 3q = 63

2. -11p -4q = -190

We can solve this system using substitution or elimination. Let me use elimination. Let's multiply equation (1) by 4 and equation (2) by 3 to eliminate q:

Multiply equation (1) by 4: -20p + 12q = 252

Multiply equation (2) by 3: -33p -12q = -570

Now add these two equations:

(-20p +12q) + (-33p -12q) = 252 + (-570)

This gives:

-53p = -318 ⇒ p = (-318)/(-53) = 318/53. Wait, 53*6 is 318. Yes, 53*6 = 318. Therefore, p = 6.

So p is 6. Now substitute p = 6 into equation (1):

-5*(6) +3q = 63 ⇒ -30 +3q = 63 ⇒ 3q = 93 ⇒ q = 31.

So, q = 31.

Now, substitute p = 6 and q = 31 into equation (3):

4*6 +31 + r = 54 ⇒ 24 +31 + r = 54 ⇒ 55 + r = 54 ⇒ r = 54 -55 = -1.

Therefore, p = 6, q = 31, r = -1. So the ordered triple is (6, 31, -1).

Wait, let me check if these values satisfy equation (2):

Equation (2): -11p -4q = -11*6 -4*31 = -66 -124 = -190, which matches. And equation (1): -5*6 +3*31 = -30 +93 = 63, which is correct. And equation (3): 4*6 +31 + (-1) = 24 +31 -1 = 54, which is correct. So that seems consistent. Therefore, the answer is (6,31,-1).

But let me just think again if there's another way to approach this problem. For example, since the original polynomial is x⁹ + p x⁶ + q x³ + r, which is like a polynomial in x³. Let me denote y = x³. Then, the polynomial becomes y³ + p y² + q y + r. So, if x³ -3x² +4x -1 divides the original polynomial, then perhaps y³ + p y² + q y + r is divisible by (x³ -3x² +4x -1). Wait, but how does that relate?

Alternatively, if we consider that the original polynomial is a polynomial in x³, so perhaps the factor x³ -3x² +4x -1 can be related to some substitution. However, since we have both x³ and lower-degree terms, maybe not directly.

Alternatively, perhaps we can think of the original polynomial as (x³)^3 + p(x³)^2 + q(x³) + r. So if we let y = x³, then the original polynomial is y³ + p y² + q y + r. But if x³ -3x² +4x -1 divides the original polynomial, then maybe we can relate y and the cubic polynomial. But how?

Let me think. If α is a root of x³ -3x² +4x -1 = 0, then α³ = 3α² -4α +1. Then, as above, we can express higher powers of α in terms of lower powers, and so for y = α³, we can write y = 3α² -4α +1. But maybe we can find a relationship between y and α?

Alternatively, perhaps we can consider that since the original polynomial is y³ + p y² + q y + r, and since y = α³, then substituting y into the polynomial gives:

(α³)^3 + p(α³)^2 + q(α³) + r = α⁹ + p α⁶ + q α³ + r = 0, as before. Which is the same as before. Therefore, we have the same equations. So our previous approach is correct.

Alternatively, perhaps we can consider that if we have x³ - 3x² +4x -1 divides x⁹ +p x⁶ + q x³ +r, then the original polynomial can be written as (x³ - 3x² +4x -1) multiplied by some polynomial. Since the original polynomial is x⁹ + p x⁶ + q x³ + r, which is a polynomial in x³. Let me consider that maybe the other factor is also a polynomial in x³? For example, if we consider that the original polynomial is (x³ - 3x² +4x -1)(something). Let me suppose that something is of the form x⁶ + a x³ + b. Because multiplying (x³ - 3x² +4x -1) by (x⁶ + a x³ + b) would give:

(x³)(x⁶ + a x³ + b) - 3x²(x⁶ + a x³ + b) +4x(x⁶ + a x³ + b) -1*(x⁶ + a x³ + b)

Let me compute each term:

First term: x³ * x⁶ + x³ * a x³ + x³ * b = x⁹ + a x⁶ + b x³

Second term: -3x² * x⁶ -3x² * a x³ -3x² * b = -3x⁸ -3a x⁵ -3b x²

Third term: 4x * x⁶ + 4x * a x³ + 4x * b = 4x⁷ + 4a x⁴ +4b x

Fourth term: -1 * x⁶ -1 * a x³ -1 * b = -x⁶ -a x³ -b

Now, adding all these together:

x⁹ + a x⁶ + b x³ -3x⁸ -3a x⁵ -3b x² +4x⁷ +4a x⁴ +4b x -x⁶ -a x³ -b.

Now, combine like terms:

x⁹

-3x⁸

+4x⁷

-3a x⁵

+4a x⁴

+a x⁶ -x⁶ = (a -1)x⁶

b x³ -a x³ = (b -a)x³

-3b x²

+4b x

- b

So overall, the product is:

x⁹ - 3x⁸ +4x⁷ + (a -1)x⁶ -3a x⁵ +4a x⁴ + (b -a)x³ -3b x² +4b x -b.

But the original polynomial is x⁹ + p x⁶ + q x³ + r. Comparing coefficients, we need all coefficients of x⁸, x⁷, x⁵, x⁴, x², x to be zero, and the coefficients of x⁹, x⁶, x³, and constant term to match. Therefore:

Coefficient of x⁸: -3 = 0? Wait, but the original polynomial has no x⁸ term. Therefore, -3 must be zero? That can't be. So clearly, this approach is flawed. Because if we assume that the other factor is x⁶ + a x³ + b, then the product has terms with x⁸, x⁷, etc., which are not present in the original polynomial. Therefore, this suggests that my assumption about the form of Q(x) is incorrect.

Therefore, perhaps Q(x) is not a simple cubic in x³, but something else. Alternatively, maybe there is some symmetry or pattern here. Wait, but the original polynomial is x⁹ + p x⁶ + q x³ + r. If we consider that perhaps the original polynomial can be written as (x³)^3 + p (x³)^2 + q (x³) + r, which is similar to a cubic in y = x³. Therefore, if the original polynomial is divisible by x³ -3x² +4x -1, then maybe substituting y = x³, and considering that y = 3x² -4x +1, as per the equation x³ = 3x² -4x +1. So if we can express y in terms of x, then perhaps we can find a relation for the polynomial in y?

Let me try that. Let me set y = x³. Then, from the cubic equation, we have:

y = 3x² -4x +1. So, perhaps we can express x² in terms of y and x? Let me rearrange:

From y = 3x² -4x +1 ⇒ 3x² = y +4x -1 ⇒ x² = (y +4x -1)/3.

But then, if we have expressions involving higher powers of x, but since our original polynomial is in terms of y, maybe we can find a relationship between y and the polynomial?

Alternatively, since the original polynomial is y³ + p y² + q y + r, and since y = 3x² -4x +1, substituting into the polynomial would give:

(3x² -4x +1)^3 + p(3x² -4x +1)^2 + q(3x² -4x +1) + r.

But this should be equal to zero for all x that satisfy the original cubic equation. But this seems more complicated. However, since we already solved the problem earlier by using the root substitution, which gave us p=6, q=31, r=-1, maybe we can check whether this works here.

Alternatively, since the original polynomial is divisible by the cubic, then the polynomial y³ + p y² + q y + r must be divisible by the cubic equation expressed in terms of y. Wait, but how?

Alternatively, since y = x³, and we have x³ -3x² +4x -1 = 0 ⇒ y -3x² +4x -1 =0 ⇒ y = 3x² -4x +1. Then, if we can eliminate x from these equations? Maybe?

But if I consider y = 3x² -4x +1, and also have to relate to the polynomial y³ + p y² + q y + r = 0. So, substituting y into that equation:

(3x² -4x +1)^3 + p(3x² -4x +1)^2 + q(3x² -4x +1) + r = 0.

But since this must hold for all x that satisfy the original cubic equation. However, since the original cubic equation is x³ -3x² +4x -1 = 0, which is equivalent to x³ = 3x² -4x +1. Therefore, maybe we can express higher powers of x in terms of lower powers. But if we have this equation involving y, which is already expressed in terms of x² and x, perhaps we can eliminate x? This might be complicated. However, maybe we can consider that the equation above must be identically zero for all x, but given that x is constrained by the original cubic. However, since the original cubic has three roots, and we already used the root method which gave us a solution, maybe there is no need to go further here.

But since my previous method gave me p=6, q=31, r=-1, which satisfies the equations, and the check with equations (1), (2), (3) worked, then maybe that's correct. However, perhaps I should verify with a specific value.

Alternatively, let me check if with p=6, q=31, r=-1, then the original polynomial is divisible by x³ -3x² +4x -1.

Let me perform polynomial division or check if the remainder is zero.

But with such high degrees, it might be tedious, but perhaps I can do it. Alternatively, let me take the original polynomial x⁹ +6x⁶ +31x³ -1 and divide by x³ -3x² +4x -1. But doing polynomial division for such high degrees is going to be lengthy. Alternatively, perhaps I can use synthetic division or some other trick. Wait, but since we already verified the equations through root substitution, which is a solid method, and solved the system correctly, then this should be correct.

Alternatively, let me check with the minimal polynomial. Suppose that (x³ -3x² +4x -1) divides the original polynomial. Let me compute the original polynomial evaluated at the roots of the cubic. But since we already did that, and the equations were satisfied, then it's okay.

Alternatively, let me check with a specific root. Suppose α is a root of x³ -3x² +4x -1 = 0. Then, as we found, α³ = 3α² -4α +1. Then, substituting into the original polynomial:

α⁹ +6α⁶ +31α³ -1 = ?

Earlier, we found that α⁹ = -63α² +190α -54

And α⁶ = -5α² -11α +4

And α³ = 3α² -4α +1.

So substituting:

(-63α² +190α -54) +6*(-5α² -11α +4) +31*(3α² -4α +1) -1.

Let me compute each term:

First term: -63α² +190α -54

Second term: 6*(-5α² -11α +4) = -30α² -66α +24

Third term: 31*(3α² -4α +1) = 93α² -124α +31

Fourth term: -1

Adding them all together:

Let me collect like terms:

For α²:

-63α² -30α² +93α² = (-63 -30 +93)α² = 0α²

For α:

190α -66α -124α = (190 -66 -124)α = 0α

Constants:

-54 +24 +31 -1 = (-54 +24) + (31 -1) = -30 +30 = 0

So indeed, the total is 0. So this confirms that with p=6, q=31, r=-1, the original polynomial is zero for each root α of the cubic, which means that the cubic divides the original polynomial. Therefore, the answer is (6, 31, -1).

Therefore, I think this is correct. I can't find any mistakes in the reasoning. So the ordered triple is (6, 31, -1).

**Final Answer**
The ordered triple is \boxed{(6, 31, -1)}.

To determine the ordered triple $(p, q, r)$ such that the polynomial $x^9 + p x^6 + q x^3 + r$ is divisible by $x^3 - 3x^2 + 4x - 1$, we proceed by analyzing their relationship algebraically.

---

### Step 1: Factor Structure

Let’s denote:

$$
P(x) = x^9 + p x^6 + q x^3 + r
$$
$$
F(x) = x^3 - 3x^2 + 4x - 1
$$

If $F(x)$ divides $P(x)$, then all roots of $F(x)$ are also roots of $P(x)$. Let $\alpha$ be a root of $F(x)$, hence $\alpha^3 = 3\alpha^2 - 4\alpha + 1$, which we'll use to reduce powers of $\alpha$ in $P(\alpha)$.

---

### Step 2: Express $\alpha^3$ and Higher Powers

Using $\alpha^3 = 3\alpha^2 - 4\alpha + 1$, we compute higher powers of $\alpha$:

- $\alpha^4 = \alpha \cdot \alpha^3 = \alpha(3\alpha^2 - 4\alpha + 1) = 3\alpha^3 - 4\alpha^2 + \alpha$
  - Substitute $\alpha^3 = 3\alpha^2 - 4\alpha + 1$
  - $\alpha^4 = 3(3\alpha^2 - 4\alpha + 1) - 4\alpha^2 + \alpha = 5\alpha^2 - 11\alpha + 3$

- $\alpha^5 = \alpha \cdot \alpha^4 = \alpha(5\alpha^2 - 11\alpha + 3) = 5\alpha^3 - 11\alpha^2 + 3\alpha$
  - Substitute $\alpha^3$
  - $\alpha^5 = 5(3\alpha^2 - 4\alpha + 1) - 11\alpha^2 + 3\alpha = 4\alpha^2 - 17\alpha + 5$

- $\alpha^6 = \alpha \cdot \alpha^5 = \alpha(4\alpha^2 - 17\alpha + 5) = 4\alpha^3 - 17\alpha^2 + 5\alpha$
  - Substitute $\alpha^3$
  - $\alpha^6 = 4(3\alpha^2 - 4\alpha + 1) - 17\alpha^2 + 5\alpha = -5\alpha^2 - 11\alpha + 4$

- $\alpha^9 = \alpha \cdot \alpha^8$ and can be derived as $-63\alpha^2 + 190\alpha - 54$ (as shown in the detailed process).

---

### Step 3: Apply to the Polynomial

We now substitute the expressions for $\alpha^9$, $\alpha^6$, and $\alpha^3$ into:

$$
P(\alpha) = \alpha^9 + p\alpha^6 + q\alpha^3 + r
$$

Substitute known values:

$$
P(\alpha) = (-63\alpha^2 + 190\alpha - 54) + p(-5\alpha^2 - 11\alpha + 4) + q(3\alpha^2 - 4\alpha + 1) + r
$$

Group coefficients:

- Coefficient of $\alpha^2$: $-63 - 5p + 3q$
- Coefficient of $\alpha$: $190 - 11p - 4q$
- Constant term: $-54 + 4p + q + r$

Set each to zero:

1. $-63 - 5p + 3q = 0$
2. $190 - 11p - 4q = 0$
3. $-54 + 4p + q + r = 0$

---

### Step 4: Solve the System of Equations

From equations (1) and (2):

- Equation (1): $-5p + 3q = 63$
- Equation (2): $-11p - 4q = -190$

Multiply equation (1) by 4 and equation (2) by 3:

- $-20p + 12q = 252$
- $-33p - 12q = -570$

Add:

$$
-53p = -318 \Rightarrow p = 6
$$

Substitute $p = 6$ into equation (1):

$$
-5(6) + 3q = 63 \Rightarrow -30 + 3q = 63 \Rightarrow 3q = 93 \Rightarrow q = 31
$$

Now substitute $p = 6$, $q = 31$ into equation (3):

$$
-54 + 4(6) + 31 + r = 0 \Rightarrow -54 + 24 + 31 + r = 0 \Rightarrow 55 + r = 0 \Rightarrow r = -1
$$

---

### Final Result

Thus, the ordered triple $(p, q, r)$ is:

$$
\boxed{(6, 31, -1)}
$$

Solution:
Let $\alpha$ be a root of $x^3 - 3x^2 + 4x - 1 = 0,$ so $\alpha^3 = 3 \alpha^2 - 4 \alpha + 1.$  Then
\[\alpha^4 = 3 \alpha^3 - 4 \alpha^2 + \alpha = 3 (3 \alpha^2 - 4 \alpha + 1) - 4 \alpha^2 + \alpha = 5 \alpha^2 - 11 \alpha + 3.\]Hence,
\begin{align*}
\alpha^6 &= (3 \alpha^2 - 4 \alpha + 1)^2 \\
&= 9 \alpha^4 - 24 \alpha^3 + 22 \alpha^2 - 8 \alpha + 1 \\
&= 9 (5 \alpha^2 - 11 \alpha + 3) - 24 (3 \alpha^2 - 4 \alpha + 1) + 22 \alpha^2 - 8 \alpha + 1 \\
&= -5 \alpha^2 - 11 \alpha + 4,
\end{align*}and
\begin{align*}
\alpha^9 &= \alpha^3 \cdot \alpha^6 \\
&= (3 \alpha^2 - 4 \alpha + 1)(-5 \alpha^2 - 11 \alpha + 4) \\
&= -15 \alpha^4 - 13 \alpha^3 + 51 \alpha^2 - 27 \alpha + 4 \\
&= -15 (5 \alpha^2 - 11 \alpha + 3) - 13 (3 \alpha^2 - 4 \alpha + 1) + 51 \alpha^2 - 27 \alpha + 4 \\
&= -63 \alpha^2 + 190 \alpha - 54.
\end{align*}Then
\begin{align*}
\alpha^9 + p \alpha^6 + q \alpha^3 + r &= (-63 \alpha^2 + 190 \alpha - 54) + p (-5 \alpha^2 - 11 \alpha + 4) + q (3 \alpha^2 - 4 \alpha + 1) + r \\
&= (-5p + 3q - 63) \alpha^2 + (-11p - 4q + 190) \alpha + (4p + q + r - 54).
\end{align*}We want this to reduce to 0, so we set
\begin{align*}
-5p + 3q &= 63, \\
11p + 4q &= 190, \\
4p + q + r &= 54.
\end{align*}Solving, we find $(p,q,r) = \boxed{(6,31,-1)}.$   For these values, $\alpha^9 + p \alpha^6 + q \alpha^3 + r$ reduces to 0 for any root $\alpha$ of $x^3 - 3x^2 + 4x - 1,$ so $x^9 + px^6 + qx^3 + r$ will be divisible by $x^3 - 3x^2 + 4x - 1.$


Explanation:
Okay, so I need to find the smallest positive real number C such that the norm of the matrix times any vector v is less than or equal to C times the norm of v. The matrix given is [[2, 3], [0, -2]]. Hmm, okay. Let me recall what this means.

First, the problem is about matrix norms. Specifically, it's asking for the operator norm of the matrix, right? The operator norm is defined as the smallest C such that ||A v|| ≤ C ||v|| for all vectors v. So, I need to compute the operator norm of this matrix.

I remember that for a matrix A, the operator norm induced by the Euclidean norm (which is what they're using here, since they mentioned the magnitude of a vector) is equal to the largest singular value of A. Alternatively, another way to think about it is that the operator norm is the maximum value of ||A v|| / ||v|| over all non-zero vectors v. Therefore, the smallest such C is the maximum singular value of A.

But maybe I can approach this without knowing singular values. Let me think. If I consider the matrix A = [[2, 3], [0, -2]], then for any vector v = [x; y], Av = [2x + 3y; -2y]. Then the norm of Av is sqrt[(2x + 3y)^2 + (-2y)^2]. So, the norm squared is (2x + 3y)^2 + ( -2y)^2. Let me compute that:

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

Therefore, the norm squared of Av is 4x² + 12xy + 13y². So, the norm of Av is sqrt(4x² + 12xy + 13y²).

We need this to be ≤ C sqrt(x² + y²) for all x, y. Therefore, squaring both sides, we get 4x² + 12xy + 13y² ≤ C²(x² + y²).

So, rearranged, this inequality becomes (4 - C²)x² + 12xy + (13 - C²)y² ≤ 0 for all x, y.

But this inequality must hold for all real numbers x and y. Wait, but if we want this to be true for all vectors v, then the quadratic form (4 - C²)x² + 12xy + (13 - C²)y² must be ≤ 0 for all x, y.

But a quadratic form is non-positive definite for all x, y only if the matrix of the quadratic form is negative semi-definite. However, since the original expression is 4x² + 12xy + 13y², which is a quadratic form, and we want it to be less than or equal to C²(x² + y²). Therefore, the difference between the original quadratic form and C²(x² + y²) must be negative semi-definite.

Alternatively, perhaps I can think of the maximum value of ||Av|| / ||v||, which is the operator norm.

Alternatively, maybe I should parametrize the vector v. Let me think. Let me write v as [x; y], then the ratio ||Av|| / ||v|| is sqrt(4x² + 12xy + 13y²)/sqrt(x² + y²). So, to find the maximum of this ratio over all non-zero vectors [x; y].

To find the maximum value of this ratio, it's equivalent to finding the maximum eigenvalue of A^T A, since for any matrix A, the operator norm squared is the largest eigenvalue of A^T A.

Let me compute A^T A.

First, A is [[2, 3], [0, -2]]. Then A^T is [[2, 0], [3, -2]]. Then A^T A is:

First row first column: 2*2 + 0*3 = 4 + 0 = 4

First row second column: 2*0 + 0*(-2) = 0 + 0 = 0? Wait, no. Wait, A^T A is computed as:

Let me do the multiplication properly. A^T is [[2, 0], [3, -2]]. Multiplying A^T by A:

First row of A^T is [2, 0], multiplied by first column of A, which is [2, 0]^T: 2*2 + 0*0 = 4

First row of A^T multiplied by second column of A, which is [3, -2]^T: 2*3 + 0*(-2) = 6 + 0 = 6

Second row of A^T is [3, -2], multiplied by first column of A: 3*2 + (-2)*0 = 6 + 0 = 6

Second row of A^T multiplied by second column of A: 3*3 + (-2)*(-2) = 9 + 4 = 13

Therefore, A^T A is [[4, 6], [6, 13]].

So, the operator norm squared is the largest eigenvalue of this matrix. Therefore, we need to compute the eigenvalues of [[4, 6], [6, 13]].

To find the eigenvalues, we need to solve the characteristic equation det(A^T A - λ I) = 0.

So the matrix is:

[4 - λ, 6]

[6, 13 - λ]

The determinant is (4 - λ)(13 - λ) - 36. Let me compute that:

(4 - λ)(13 - λ) = 4*13 - 4λ -13λ + λ² = 52 - 17λ + λ²

Subtract 36: 52 - 17λ + λ² - 36 = λ² - 17λ + 16

Set this equal to zero: λ² -17λ + 16 = 0

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

So, the solutions are (17 +15)/2 = 32/2 = 16 and (17 -15)/2 = 2/2 = 1.

Therefore, the eigenvalues of A^T A are 16 and 1. Therefore, the operator norm squared is 16, so the operator norm is sqrt(16) = 4. Therefore, the minimal C is 4? Wait, wait. Let me confirm.

Wait, the operator norm of A is the square root of the largest eigenvalue of A^T A. So yes, the largest eigenvalue of A^T A is 16, so operator norm is sqrt(16) = 4. Therefore, the minimal C is 4.

But let me check if that makes sense. Let me see. Alternatively, maybe I can check with specific vectors. Let me see. Let me check if for some vector v, ||Av|| / ||v|| is 4.

Suppose that the maximum ratio is 4, then there exists some vector v where this ratio is 4. Let me find such a vector.

The maximum is achieved when v is an eigenvector of A^T A corresponding to the largest eigenvalue, which is 16. So, let's find the eigenvector for λ = 16.

So, solving (A^T A - 16 I) v = 0.

A^T A is [[4, 6], [6, 13]], so subtracting 16 I gives [[4 -16, 6], [6, 13 -16]] = [[-12, 6], [6, -3]]

So, the equations are -12x + 6y = 0 and 6x - 3y = 0. Both equations reduce to -12x + 6y = 0 => -2x + y = 0 => y = 2x.

Therefore, the eigenvectors are scalar multiples of [1; 2]. Let me check with v = [1; 2]. Then ||v|| = sqrt(1 + 4) = sqrt(5). Then Av = [[2, 3], [0, -2]] [1; 2] = [2*1 + 3*2, 0*1 + (-2)*2] = [2 + 6, 0 -4] = [8; -4]. Then ||Av|| is sqrt(8² + (-4)^2) = sqrt(64 + 16) = sqrt(80) = 4*sqrt(5). Therefore, ||Av|| / ||v|| is (4 sqrt(5)) / sqrt(5)) = 4. Perfect! So that checks out. Therefore, the operator norm is indeed 4.

But just to make sure there isn't a larger value. Suppose we take another vector. Let me try v = [1; 0]. Then Av = [2; 0], ||Av|| is 2, ||v|| is 1, ratio is 2.

Another vector, say v = [0; 1]. Then Av = [3; -2], ||Av|| is sqrt(9 + 4) = sqrt(13), ratio is sqrt(13) ≈ 3.605, which is less than 4.

Another vector, let's take v = [1; 1]. Then Av = [2*1 + 3*1, 0*1 + (-2)*1] = [5; -2], norm is sqrt(25 + 4) = sqrt(29) ≈ 5.385, ||v|| is sqrt(2), so ratio ≈ 5.385 / 1.414 ≈ 3.807, which is still less than 4.

Another test vector, maybe [1; 2], which we already did, gives 4.

What about if I take v = [2; 4], which is just a scalar multiple of [1; 2]. The ratio remains the same, 4.

So seems like 4 is indeed the maximum.

Alternatively, maybe I can think of the original matrix. The matrix A is upper triangular, so its eigenvalues are the diagonal entries. Wait, the eigenvalues of A are 2 and -2. Wait, but the operator norm is not directly the maximum of the eigenvalues in absolute value. Wait, that's for the spectral norm? Wait, actually, if the matrix is normal, then the operator norm is the maximum absolute value of the eigenvalues. But A is not normal. Wait, A is upper triangular, but is it normal? Let me check.

A = [[2, 3], [0, -2]]. A^T A is [[4, 6], [6, 13]], as before. AA^T is [[4, -6], [-6, 4]]? Wait, no. Let me compute AA^T.

Wait, A is [[2, 3], [0, -2]]. A^T is [[2, 0], [3, -2]]. Then AA^T is:

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

Therefore, AA^T is [[13, -6], [-6, 4]]. Which is different from A^T A. So, A is not normal. Therefore, the operator norm is not equal to the maximum absolute value of the eigenvalues. The eigenvalues of A are 2 and -2, so maximum absolute value is 2, but we found that the operator norm is 4. Therefore, that approach was wrong.

But that makes sense because operator norm is about the maximum stretching factor, which can be different from the eigenvalues. For example, in this case, the matrix stretches vectors in some direction more than the eigenvalues suggest.

Another way to think about the operator norm is that for any vector v, ||Av|| ≤ ||A|| ||v||, with equality when v is in the direction of the singular vector corresponding to the largest singular value.

Since we computed the singular values via A^T A, which gave us singular values sqrt(16) = 4 and sqrt(1) = 1, so the largest singular value is 4. Therefore, the minimal C is 4.

Alternatively, if I wanted to do this without eigenvalues, maybe I can parametrize the vector v as a unit vector. Let me think. Suppose I set ||v|| = 1, then I need to maximize ||Av||.

So, for unit vectors [x, y], with x² + y² = 1, find the maximum of sqrt(4x² + 12xy + 13y²). Which is equivalent to maximizing the expression 4x² + 12xy + 13y².

To maximize this quadratic form on the unit circle.

Alternatively, we can use Lagrange multipliers. Let me try that. Let me consider the function f(x, y) = 4x² + 12xy + 13y² with the constraint g(x, y) = x² + y² - 1 = 0.

The Lagrangian is 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:

(1) 8x + 12y = 2λx

(2) 12x + 26y = 2λy

(3) x² + y² = 1

Let me rearrange equations (1) and (2):

From (1): 8x + 12y = 2λx => 8x - 2λx + 12y = 0 => x(8 - 2λ) + 12y = 0

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

So, we have a system:

x(8 - 2λ) + 12y = 0

12x + y(26 - 2λ) = 0

We can write this as a matrix equation:

[8 - 2λ    12        ] [x]   = [0]

[12      26 - 2λ  ] [y]     [0]

For non-trivial solutions, determinant must be zero:

(8 - 2λ)(26 - 2λ) - (12)(12) = 0

Compute this determinant:

First, expand (8 - 2λ)(26 - 2λ):

= 8*26 - 8*2λ -2λ*26 + (2λ)^2

= 208 - 16λ -52λ + 4λ²

= 208 - 68λ + 4λ²

Then subtract 12*12 = 144:

So determinant is 208 -68λ + 4λ² -144 = 64 -68λ +4λ²

Set equal to zero:

4λ² -68λ +64 = 0

Divide both sides by 4:

λ² - 17λ +16 = 0

Which is the same equation as before! So, λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± 15]/2, so 16 and 1. Therefore, the maximum value of the quadratic form is 16, so the maximum of ||Av|| is sqrt(16) = 4 when ||v||=1. Therefore, C=4.

Therefore, regardless of the approach, the minimal C is 4.

Alternatively, perhaps I can think of the matrix as a linear transformation and analyze its effect on vectors. The matrix is upper triangular, so it's a combination of scaling and shearing. The first column is [2, 0], which scales the x-axis by 2, and the second column is [3, -2], which is a combination of scaling and shearing.

But maybe not necessary. Since we've already verified multiple ways that the operator norm is 4, and the test vectors also confirm that, I think C=4 is the correct answer.

Just to make sure there isn't a lower value. Suppose someone says maybe C is sqrt( (max eigenvalue of A^T A) ), which we have as sqrt(16) = 4. Since the operator norm is defined as the smallest C such that ||Av|| ≤ C ||v|| for all v, and since we have a vector where ||Av|| / ||v|| = 4, then C cannot be smaller than 4. Hence, 4 is indeed the minimal value.

Another thought: sometimes people confuse the operator norm with the maximum absolute value of the entries, but in this case, the maximum entry is 3 in the off-diagonal, but that doesn't directly relate.

Alternatively, if we consider the maximum of the absolute values of the entries of A, which is 3, but since we have a vector that gives a higher ratio, like 4, that's not the case.

Alternatively, maybe someone might think that since the eigenvalues are 2 and -2, then the maximum norm is 2, but that's incorrect because the operator norm isn't directly the maximum of the eigenvalues.

Thus, after all these considerations, I think the answer is 4.

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

To find the smallest positive real number $ C $ such that
$$
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| \le C \|\bold{v}\|
$$
for all two-dimensional vectors $ \bold{v} $, we are essentially determining the **operator norm** of the matrix
$$
A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}.
$$

---

### Step 1: Understanding the Problem

The inequality
$$
\|A \bold{v}\| \le C \|\bold{v}\|
$$
holds for all vectors $ \bold{v} $ if and only if the matrix $ A $ is **bounded**, and the **smallest** such $ C $ is the **operator norm** of $ A $, denoted $ \|A\| $. This is equivalent to computing:

$$
\|A\| = \sup_{\bold{v} \ne 0} \frac{\|A \bold{v}\|}{\|\bold{v}\|}.
$$

This quantity is also known as the **maximum singular value** of $ A $, and it can be computed as the largest eigenvalue of $ A^T A $, which is a symmetric matrix.

---

### Step 2: Compute $ A^T A $

We first compute the transpose of $ A $:

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

Now we compute $ A^T A $:

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

---

### Step 3: Find the Eigenvalues of $ A^T A $

To find the operator norm, compute the **eigenvalues** of $ A^T A $. The characteristic equation is:
$$
\det(A^T A - \lambda I) = 0
\Rightarrow \det\begin{pmatrix} 4 - \lambda & 6 \\ 6 & 13 - \lambda \end{pmatrix} = 0
$$
$$
(4 - \lambda)(13 - \lambda) - 36 = 0
\Rightarrow \lambda^2 - 17\lambda + 16 = 0
$$

Solve the quadratic equation:
$$
\lambda = \frac{17 \pm \sqrt{289 - 64}}{2} = \frac{17 \pm \sqrt{225}}{2} = \frac{17 \pm 15}{2}
$$
$$
\Rightarrow \lambda = 16 \quad \text{or} \quad 1
$$

So the **largest eigenvalue** of $ A^T A $ is $ 16 $, and therefore the **operator norm** of $ A $ is:

$$
\|A\| = \sqrt{16} = 4
$$

---

### Step 4: Justifying the Final Answer

This result means that:
$$
\|A \bold{v}\| \le 4 \|\bold{v}\| \quad \text{for all } \bold{v} \in \mathbb{R}^2.
$$
Moreover, since the operator norm is the smallest such constant $ C $, the minimal possible $ C $ is **4**.

---

### ✅ Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest positive real number C such that the norm of the matrix times any vector v is less than or equal to C times the norm of v. The matrix given is [[2, 3], [0, -2]]. Hmm, okay. Let me recall what this means.

First, the problem is about matrix norms. Specifically, it's asking for the operator norm of the matrix, right? The operator norm is defined as the smallest C such that ||A v|| ≤ C ||v|| for all vectors v. So, I need to compute the operator norm of this matrix.

I remember that for a matrix A, the operator norm induced by the Euclidean norm (which is what they're using here, since they mentioned the magnitude of a vector) is equal to the largest singular value of A. Alternatively, another way to think about it is that the operator norm is the maximum value of ||A v|| / ||v|| over all non-zero vectors v. Therefore, the smallest such C is the maximum singular value of A.

But maybe I can approach this without knowing singular values. Let me think. If I consider the matrix A = [[2, 3], [0, -2]], then for any vector v = [x; y], Av = [2x + 3y; -2y]. Then the norm of Av is sqrt[(2x + 3y)^2 + (-2y)^2]. So, the norm squared is (2x + 3y)^2 + ( -2y)^2. Let me compute that:

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

Therefore, the norm squared of Av is 4x² + 12xy + 13y². So, the norm of Av is sqrt(4x² + 12xy + 13y²).

We need this to be ≤ C sqrt(x² + y²) for all x, y. Therefore, squaring both sides, we get 4x² + 12xy + 13y² ≤ C²(x² + y²).

So, rearranged, this inequality becomes (4 - C²)x² + 12xy + (13 - C²)y² ≤ 0 for all x, y.

But this inequality must hold for all real numbers x and y. Wait, but if we want this to be true for all vectors v, then the quadratic form (4 - C²)x² + 12xy + (13 - C²)y² must be ≤ 0 for all x, y.

But a quadratic form is non-positive definite for all x, y only if the matrix of the quadratic form is negative semi-definite. However, since the original expression is 4x² + 12xy + 13y², which is a quadratic form, and we want it to be less than or equal to C²(x² + y²). Therefore, the difference between the original quadratic form and C²(x² + y²) must be negative semi-definite.

Alternatively, perhaps I can think of the maximum value of ||Av|| / ||v||, which is the operator norm.

Alternatively, maybe I should parametrize the vector v. Let me think. Let me write v as [x; y], then the ratio ||Av|| / ||v|| is sqrt(4x² + 12xy + 13y²)/sqrt(x² + y²). So, to find the maximum of this ratio over all non-zero vectors [x; y].

To find the maximum value of this ratio, it's equivalent to finding the maximum eigenvalue of A^T A, since for any matrix A, the operator norm squared is the largest eigenvalue of A^T A.

Let me compute A^T A.

First, A is [[2, 3], [0, -2]]. Then A^T is [[2, 0], [3, -2]]. Then A^T A is:

First row first column: 2*2 + 0*3 = 4 + 0 = 4

First row second column: 2*0 + 0*(-2) = 0 + 0 = 0? Wait, no. Wait, A^T A is computed as:

Let me do the multiplication properly. A^T is [[2, 0], [3, -2]]. Multiplying A^T by A:

First row of A^T is [2, 0], multiplied by first column of A, which is [2, 0]^T: 2*2 + 0*0 = 4

First row of A^T multiplied by second column of A, which is [3, -2]^T: 2*3 + 0*(-2) = 6 + 0 = 6

Second row of A^T is [3, -2], multiplied by first column of A: 3*2 + (-2)*0 = 6 + 0 = 6

Second row of A^T multiplied by second column of A: 3*3 + (-2)*(-2) = 9 + 4 = 13

Therefore, A^T A is [[4, 6], [6, 13]].

So, the operator norm squared is the largest eigenvalue of this matrix. Therefore, we need to compute the eigenvalues of [[4, 6], [6, 13]].

To find the eigenvalues, we need to solve the characteristic equation det(A^T A - λ I) = 0.

So the matrix is:

[4 - λ, 6]

[6, 13 - λ]

The determinant is (4 - λ)(13 - λ) - 36. Let me compute that:

(4 - λ)(13 - λ) = 4*13 - 4λ -13λ + λ² = 52 - 17λ + λ²

Subtract 36: 52 - 17λ + λ² - 36 = λ² - 17λ + 16

Set this equal to zero: λ² -17λ + 16 = 0

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

So, the solutions are (17 +15)/2 = 32/2 = 16 and (17 -15)/2 = 2/2 = 1.

Therefore, the eigenvalues of A^T A are 16 and 1. Therefore, the operator norm squared is 16, so the operator norm is sqrt(16) = 4. Therefore, the minimal C is 4? Wait, wait. Let me confirm.

Wait, the operator norm of A is the square root of the largest eigenvalue of A^T A. So yes, the largest eigenvalue of A^T A is 16, so operator norm is sqrt(16) = 4. Therefore, the minimal C is 4.

But let me check if that makes sense. Let me see. Alternatively, maybe I can check with specific vectors. Let me see. Let me check if for some vector v, ||Av|| / ||v|| is 4.

Suppose that the maximum ratio is 4, then there exists some vector v where this ratio is 4. Let me find such a vector.

The maximum is achieved when v is an eigenvector of A^T A corresponding to the largest eigenvalue, which is 16. So, let's find the eigenvector for λ = 16.

So, solving (A^T A - 16 I) v = 0.

A^T A is [[4, 6], [6, 13]], so subtracting 16 I gives [[4 -16, 6], [6, 13 -16]] = [[-12, 6], [6, -3]]

So, the equations are -12x + 6y = 0 and 6x - 3y = 0. Both equations reduce to -12x + 6y = 0 => -2x + y = 0 => y = 2x.

Therefore, the eigenvectors are scalar multiples of [1; 2]. Let me check with v = [1; 2]. Then ||v|| = sqrt(1 + 4) = sqrt(5). Then Av = [[2, 3], [0, -2]] [1; 2] = [2*1 + 3*2, 0*1 + (-2)*2] = [2 + 6, 0 -4] = [8; -4]. Then ||Av|| is sqrt(8² + (-4)^2) = sqrt(64 + 16) = sqrt(80) = 4*sqrt(5). Therefore, ||Av|| / ||v|| is (4 sqrt(5)) / sqrt(5)) = 4. Perfect! So that checks out. Therefore, the operator norm is indeed 4.

But just to make sure there isn't a larger value. Suppose we take another vector. Let me try v = [1; 0]. Then Av = [2; 0], ||Av|| is 2, ||v|| is 1, ratio is 2.

Another vector, say v = [0; 1]. Then Av = [3; -2], ||Av|| is sqrt(9 + 4) = sqrt(13), ratio is sqrt(13) ≈ 3.605, which is less than 4.

Another vector, let's take v = [1; 1]. Then Av = [2*1 + 3*1, 0*1 + (-2)*1] = [5; -2], norm is sqrt(25 + 4) = sqrt(29) ≈ 5.385, ||v|| is sqrt(2), so ratio ≈ 5.385 / 1.414 ≈ 3.807, which is still less than 4.

Another test vector, maybe [1; 2], which we already did, gives 4.

What about if I take v = [2; 4], which is just a scalar multiple of [1; 2]. The ratio remains the same, 4.

So seems like 4 is indeed the maximum.

Alternatively, maybe I can think of the original matrix. The matrix A is upper triangular, so its eigenvalues are the diagonal entries. Wait, the eigenvalues of A are 2 and -2. Wait, but the operator norm is not directly the maximum of the eigenvalues in absolute value. Wait, that's for the spectral norm? Wait, actually, if the matrix is normal, then the operator norm is the maximum absolute value of the eigenvalues. But A is not normal. Wait, A is upper triangular, but is it normal? Let me check.

A = [[2, 3], [0, -2]]. A^T A is [[4, 6], [6, 13]], as before. AA^T is [[4, -6], [-6, 4]]? Wait, no. Let me compute AA^T.

Wait, A is [[2, 3], [0, -2]]. A^T is [[2, 0], [3, -2]]. Then AA^T is:

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

Therefore, AA^T is [[13, -6], [-6, 4]]. Which is different from A^T A. So, A is not normal. Therefore, the operator norm is not equal to the maximum absolute value of the eigenvalues. The eigenvalues of A are 2 and -2, so maximum absolute value is 2, but we found that the operator norm is 4. Therefore, that approach was wrong.

But that makes sense because operator norm is about the maximum stretching factor, which can be different from the eigenvalues. For example, in this case, the matrix stretches vectors in some direction more than the eigenvalues suggest.

Another way to think about the operator norm is that for any vector v, ||Av|| ≤ ||A|| ||v||, with equality when v is in the direction of the singular vector corresponding to the largest singular value.

Since we computed the singular values via A^T A, which gave us singular values sqrt(16) = 4 and sqrt(1) = 1, so the largest singular value is 4. Therefore, the minimal C is 4.

Alternatively, if I wanted to do this without eigenvalues, maybe I can parametrize the vector v as a unit vector. Let me think. Suppose I set ||v|| = 1, then I need to maximize ||Av||.

So, for unit vectors [x, y], with x² + y² = 1, find the maximum of sqrt(4x² + 12xy + 13y²). Which is equivalent to maximizing the expression 4x² + 12xy + 13y².

To maximize this quadratic form on the unit circle.

Alternatively, we can use Lagrange multipliers. Let me try that. Let me consider the function f(x, y) = 4x² + 12xy + 13y² with the constraint g(x, y) = x² + y² - 1 = 0.

The Lagrangian is 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:

(1) 8x + 12y = 2λx

(2) 12x + 26y = 2λy

(3) x² + y² = 1

Let me rearrange equations (1) and (2):

From (1): 8x + 12y = 2λx => 8x - 2λx + 12y = 0 => x(8 - 2λ) + 12y = 0

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

So, we have a system:

x(8 - 2λ) + 12y = 0

12x + y(26 - 2λ) = 0

We can write this as a matrix equation:

[8 - 2λ    12        ] [x]   = [0]

[12      26 - 2λ  ] [y]     [0]

For non-trivial solutions, determinant must be zero:

(8 - 2λ)(26 - 2λ) - (12)(12) = 0

Compute this determinant:

First, expand (8 - 2λ)(26 - 2λ):

= 8*26 - 8*2λ -2λ*26 + (2λ)^2

= 208 - 16λ -52λ + 4λ²

= 208 - 68λ + 4λ²

Then subtract 12*12 = 144:

So determinant is 208 -68λ + 4λ² -144 = 64 -68λ +4λ²

Set equal to zero:

4λ² -68λ +64 = 0

Divide both sides by 4:

λ² - 17λ +16 = 0

Which is the same equation as before! So, λ = [17 ± sqrt(289 - 64)] / 2 = [17 ± 15]/2, so 16 and 1. Therefore, the maximum value of the quadratic form is 16, so the maximum of ||Av|| is sqrt(16) = 4 when ||v||=1. Therefore, C=4.

Therefore, regardless of the approach, the minimal C is 4.

Alternatively, perhaps I can think of the matrix as a linear transformation and analyze its effect on vectors. The matrix is upper triangular, so it's a combination of scaling and shearing. The first column is [2, 0], which scales the x-axis by 2, and the second column is [3, -2], which is a combination of scaling and shearing.

But maybe not necessary. Since we've already verified multiple ways that the operator norm is 4, and the test vectors also confirm that, I think C=4 is the correct answer.

Just to make sure there isn't a lower value. Suppose someone says maybe C is sqrt( (max eigenvalue of A^T A) ), which we have as sqrt(16) = 4. Since the operator norm is defined as the smallest C such that ||Av|| ≤ C ||v|| for all v, and since we have a vector where ||Av|| / ||v|| = 4, then C cannot be smaller than 4. Hence, 4 is indeed the minimal value.

Another thought: sometimes people confuse the operator norm with the maximum absolute value of the entries, but in this case, the maximum entry is 3 in the off-diagonal, but that doesn't directly relate.

Alternatively, if we consider the maximum of the absolute values of the entries of A, which is 3, but since we have a vector that gives a higher ratio, like 4, that's not the case.

Alternatively, maybe someone might think that since the eigenvalues are 2 and -2, then the maximum norm is 2, but that's incorrect because the operator norm isn't directly the maximum of the eigenvalues.

Thus, after all these considerations, I think the answer is 4.

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

To find the smallest positive real number $ C $ such that
$$
\left\| \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix} \bold{v} \right\| \le C \|\bold{v}\|
$$
for all two-dimensional vectors $ \bold{v} $, we are essentially determining the **operator norm** of the matrix
$$
A = \begin{pmatrix} 2 & 3 \\ 0 & -2 \end{pmatrix}.
$$

---

### Step 1: Understanding the Problem

The inequality
$$
\|A \bold{v}\| \le C \|\bold{v}\|
$$
holds for all vectors $ \bold{v} $ if and only if the matrix $ A $ is **bounded**, and the **smallest** such $ C $ is the **operator norm** of $ A $, denoted $ \|A\| $. This is equivalent to computing:

$$
\|A\| = \sup_{\bold{v} \ne 0} \frac{\|A \bold{v}\|}{\|\bold{v}\|}.
$$

This quantity is also known as the **maximum singular value** of $ A $, and it can be computed as the largest eigenvalue of $ A^T A $, which is a symmetric matrix.

---

### Step 2: Compute $ A^T A $

We first compute the transpose of $ A $:

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

Now we compute $ A^T A $:

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

---

### Step 3: Find the Eigenvalues of $ A^T A $

To find the operator norm, compute the **eigenvalues** of $ A^T A $. The characteristic equation is:
$$
\det(A^T A - \lambda I) = 0
\Rightarrow \det\begin{pmatrix} 4 - \lambda & 6 \\ 6 & 13 - \lambda \end{pmatrix} = 0
$$
$$
(4 - \lambda)(13 - \lambda) - 36 = 0
\Rightarrow \lambda^2 - 17\lambda + 16 = 0
$$

Solve the quadratic equation:
$$
\lambda = \frac{17 \pm \sqrt{289 - 64}}{2} = \frac{17 \pm \sqrt{225}}{2} = \frac{17 \pm 15}{2}
$$
$$
\Rightarrow \lambda = 16 \quad \text{or} \quad 1
$$

So the **largest eigenvalue** of $ A^T A $ is $ 16 $, and therefore the **operator norm** of $ A $ is:

$$
\|A\| = \sqrt{16} = 4
$$

---

### Step 4: Justifying the Final Answer

This result means that:
$$
\|A \bold{v}\| \le 4 \|\bold{v}\| \quad \text{for all } \bold{v} \in \mathbb{R}^2.
$$
Moreover, since the operator norm is the smallest such constant $ C $, the minimal possible $ C $ is **4**.

---

### ✅ Final Answer

$$
\boxed{4}
$$

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 polynomial evaluated at 1. Let me start by trying to factor this polynomial step by step.

First, I notice that the polynomial is of degree 8, and it has only even powers of x: $ x^8 $, $ x^4 $, and the constant term. That makes me think of substitution. Maybe I can let $ y = x^4 $, so the polynomial becomes $ y^2 + 3y - 4 $. Let me try that.

So substituting $ y = x^4 $, we have:

$ y^2 + 3y - 4 $.

Now, let me factor this quadratic in y. To factor $ y^2 + 3y - 4 $, I need two numbers that multiply to -4 and add to 3. Let's see: 4 and -1. Yes, 4 * (-1) = -4 and 4 + (-1) = 3. Therefore, the quadratic factors as:

$ (y + 4)(y - 1) $.

So 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^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) $. Then, $ x^2 - 1 $ is also a difference of squares: $ (x - 1)(x + 1) $. Therefore, $ x^4 - 1 $ factors into $ (x - 1)(x + 1)(x^2 + 1) $.

Now, the other factor is $ x^4 + 4 $. Hmm, this might be more challenging. Let me recall that $ x^4 + 4 $ can be factored as a sum of squares, but over the integers? Wait, sum of squares can't be factored over the real numbers unless we use complex numbers. However, there might be a way to factor it into quadratics with integer coefficients.

Let me try to factor $ x^4 + 4 $. I remember something about factoring expressions of the form $ x^4 + a^4 $. There's a formula for that, isn't there? Let me recall. It's called the Sophie Germain identity:

$ x^4 + 4y^4 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2) $.

Yes, that's right. So if I apply that here, where $ y = 1 $, since $ x^4 + 4 = x^4 + 4(1)^4 $. Therefore, using the identity:

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

Let me check that. Multiplying those two quadratics:

First, $ (x^2 + 2x + 2)(x^2 - 2x + 2) $. Multiply them out:

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

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

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

Now, combine all these terms:

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

So indeed, the product is $ x^4 + 4 $. Perfect! So that works. Therefore, $ x^4 + 4 $ factors into $ (x^2 + 2x + 2)(x^2 - 2x + 2) $.

Therefore, putting it all together, the original polynomial factors as:

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

Now, I need to check if each of these factors can be factored further over the integers. Let me check each one:

1. $ x - 1 $: linear, so 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 $ 0 - 4 = -4 $), so it can't be factored into linear terms with integer coefficients.
4. $ x^2 + 2x + 2 $: discriminant is $ 4 - 8 = -4 $, so also irreducible over integers.
5. $ x^2 - 2x + 2 $: discriminant is $ 4 - 8 = -4 $, also irreducible over integers.

Therefore, all factors are irreducible over integers. Therefore, the complete factorization is:

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

So the polynomials $ p_1, p_2, ..., p_k $ are:

1. $ x - 1 $
2. $ x + 1 $
3. $ x^2 + 1 $
4. $ x^2 + 2x + 2 $
5. $ x^2 - 2x + 2 $

Now, the problem says each non-constant polynomial is monic with integer coefficients, which they are. So now, I need to compute the sum $ p_1(1) + p_2(1) + \dots + p_k(1) $.

Let me compute each polynomial evaluated at 1:

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 $

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

Wait, but let me check again. Let me recompute each value step by step to make sure I didn't make any mistakes.

First polynomial: $ x - 1 $. At x=1: 1 -1 = 0. Correct.

Second polynomial: x + 1. At x=1: 1 + 1 = 2. Correct.

Third polynomial: x² + 1. At x=1: 1 + 1 = 2. Correct.

Fourth polynomial: x² + 2x + 2. At x=1: 1 + 2 + 2 = 5. Correct.

Fifth polynomial: x² - 2x + 2. At x=1: 1 - 2 + 2 = 1. Correct.

Adding them up: 0 + 2 = 2; 2 + 2 = 4; 4 + 5 = 9; 9 + 1 = 10. So total is 10.

But let me think again. Is there a possibility that the problem considers the factors in a different order, or maybe I missed some factors?

Wait, let me check the original factorization again. The original polynomial was factored as:

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

Yes, that's five factors. So there are five polynomials. So the sum is 0 + 2 + 2 + 5 + 1 = 10. Therefore, the answer is 10?

But wait, let me check if I need to consider something else. Wait, 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 factorization is correct, since all factors are irreducible. Therefore, the answer should be 10.

Wait, but let me check if there is another possible factorization. For example, maybe there's a different way to factor the original polynomial into more factors? But since we have already factored it completely, into linear and irreducible quadratic factors, so I think that's the unique factorization.

Alternatively, let me check the original polynomial. Let me compute the value of the original polynomial at x=1 to see if the sum of the p_i(1) equals something else? Wait, the original polynomial is x^8 + 3x^4 -4. At x=1, that's 1 + 3 -4 = 0. However, the product of all p_i(1) should equal 0, since the original polynomial evaluated at 1 is 0. Let me check that.

Product of p_i(1) is 0 * 2 * 2 * 5 * 1 = 0, which matches the original polynomial evaluated at 1. So that seems correct.

But the question is not asking for the product, but the sum. So sum is 10. Let me check again, maybe my arithmetic is wrong. Let me write down each term:

p1(1) = 0

p2(1) = 2

p3(1) = 2

p4(1) = 5

p5(1) = 1

Total: 0 + 2 + 2 + 5 + 1 = 10. Yes, that's correct.

Wait, but maybe there's a mistake in the factorization? Let me check if the original polynomial equals the product of the factors. Let me compute (x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2). Let me compute 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 this:

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 altogether: x^6 + 2x^5 + 2x^4 - x^2 - 2x - 2.

Then, multiply by (x^2 - 2x + 2). Let me denote that as:

Let me call the previous result as A = x^6 + 2x^5 + 2x^4 - x^2 - 2x - 2.

Multiply A by (x^2 - 2x + 2). This seems tedious, but maybe there's a smarter way. Alternatively, note that earlier steps showed that x^4 + 4 factors into (x^2 + 2x + 2)(x^2 - 2x + 2), and x^4 - 1 factors into (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1). Therefore, multiplying them together gives the original polynomial. Therefore, the product is indeed x^8 + 3x^4 -4. Therefore, the factorization is correct.

Thus, the sum of p_i(1) is 10. Therefore, the answer should be 10. But let me check again with another approach.

Alternatively, if I have the factorization into p1, p2,..., pk, then p1(1) + ... + pk(1) is equal to the sum of each factor evaluated at 1. Alternatively, since the original polynomial is the product of the p_i(x), then the product of p_i(1) is the original polynomial evaluated at 1, which is 0, as we saw.

But the question is about the sum. Is there a way to relate the sum of p_i(1) with something else? Not directly that I can think of. So I think 10 is the correct answer. Wait, but maybe I made a mistake in the factorization? Let me check with specific values.

Take x = 1: original polynomial is 1 + 3 -4 = 0. The product of p_i(1) is 0 * 2 * 2 * 5 * 1 = 0. Correct.

Take x = 0: original polynomial is 0 + 0 -4 = -4. Let me compute the product of p_i(0):

p1(0) = -1, p2(0) = 1, p3(0) = 0 + 1 = 1, p4(0) = 0 + 0 + 2 = 2, p5(0) = 0 - 0 + 2 = 2. Then product is (-1)(1)(1)(2)(2) = -4, which matches the original polynomial evaluated at 0. So that's correct.

Another check: x = 2. Original polynomial: 2^8 + 3*2^4 -4 = 256 + 3*16 -4 = 256 + 48 -4 = 256 + 44 = 300.

Now compute product of p_i(2):

p1(2) = 2 - 1 = 1

p2(2) = 2 + 1 = 3

p3(2) = 4 + 1 = 5

p4(2) = 4 + 4 + 2 = 10

p5(2) = 4 - 4 + 2 = 2

Multiply them: 1 * 3 = 3; 3 * 5 = 15; 15 * 10 = 150; 150 * 2 = 300. Correct, which matches the original polynomial evaluated at 2. Therefore, the factorization is correct. Hence, my previous calculation of the sum of p_i(1) as 10 is correct.

But wait, just to be thorough, let me check if there's another possible factorization. For example, could x^4 + 4 be factored differently?

Well, we used the Sophie Germain identity, which is standard for such cases. Also, trying to factor x^4 + 4 into two quadratics with integer coefficients. Suppose we attempt to factor it as (x^2 + ax + b)(x^2 + cx + d). Then expanding:

x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd.

This should equal x^4 + 0x^3 + 0x^2 + 0x + 4. Therefore, we have the equations:

1. a + c = 0
2. ac + b + d = 0
3. ad + bc = 0
4. bd = 4

From equation 1: c = -a.

Substitute into equation 3: ad + b(-a) = 0 => a(d - b) = 0.

So either a = 0 or d = b.

Case 1: a = 0. Then c = 0. Then equation 2: 0 + b + d = 0 => b + d = 0. Equation 4: bd = 4. So solving b + d = 0 and bd = 4. Then, d = -b, so b(-b) = -b² = 4 => b² = -4, which has no real solutions, let alone integer ones. Therefore, Case 1 invalid.

Case 2: d = b. Then equation 4: b*d = b² = 4. Therefore, b = ±2. Then, equation 2: ac + b + d = (-a²) + b + b = -a² + 2b = 0. Therefore, -a² + 2b = 0 => a² = 2b.

Since b = ±2, then:

If b = 2: a² = 4 => a = ±2.

If b = -2: a² = -4, which is impossible.

Therefore, b = 2, and a = ±2. Therefore, possible solutions:

a = 2, then c = -2, and d = b = 2.

Therefore, the factors would be (x² + 2x + 2)(x² - 2x + 2), which is exactly what we found earlier. Similarly, if a = -2, then c = 2, and the factors would be (x² - 2x + 2)(x² + 2x + 2), same thing. Therefore, no other factorizations are possible. Therefore, our factorization of x^4 + 4 is correct and unique over integers.

Therefore, the factorization is correct, and the sum is indeed 10. So, I think the answer is 10.

But hold on, let me check once more if I have the correct number of factors. The original polynomial factors into (x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2). So that's five factors. So when we evaluate each at 1, we get 0, 2, 2, 5, 1. Summing to 10. Therefore, unless there's a miscalculation here, which I don't see, the answer is 10.

Alternatively, maybe the problem expects more factors? For example, if there is a different factorization? Let me think again. Suppose someone tried to factor x^4 -1 further, but we already did that. x^4 -1 factors into (x^2 -1)(x^2 + 1), and x^2 -1 factors into (x -1)(x +1). So that's all.

Alternatively, is there a way to factor the original polynomial into more factors? For example, if we factor x^2 + 1 further? But it's irreducible over integers. Similarly for the quadratics.

Therefore, I think the factorization is complete. Therefore, the answer is 10.

But just to make sure, let me check if the problem says "non-constant polynomial p_i(x)", so we have five factors, all non-constant. Therefore, the sum is 10. So I think that's the answer. Therefore, the answer is \boxed{10}.

**Final Answer**
\boxed{10}

To solve the problem, we start with the given polynomial:

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

---

### Step 1: Factor the Polynomial

We notice the polynomial only includes even powers of $x$, suggesting we can use substitution. Let:

$$
y = x^4
$$

Then the expression becomes:

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

This is a quadratic in $y$. Factoring, we find:

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

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

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

---

### Step 2: Factor Each Component

Now, we factor each term further:

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

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

- $x^2 - 1$ is again a difference of squares:

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

So:

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

Now, we handle $x^4 + 4$. This appears irreducible over the integers, but we can factor it using the **Sophie Germain identity**:

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

---

### Step 3: Final Factorization

Putting everything together, we have:

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

All of these polynomials are:

- Monic
- Have integer coefficients
- Cannot be factored further over the integers

So, this is the complete factorization into irreducible polynomials.

---

### Step 4: Evaluate Each Factor at $x = 1$

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

- $p_1(x) = x - 1 \Rightarrow p_1(1) = 0$
- $p_2(x) = x + 1 \Rightarrow p_2(1) = 2$
- $p_3(x) = x^2 + 1 \Rightarrow p_3(1) = 1 + 1 = 2$
- $p_4(x) = x^2 + 2x + 2 \Rightarrow p_4(1) = 1 + 2 + 2 = 5$
- $p_5(x) = x^2 - 2x + 2 \Rightarrow p_5(1) = 1 - 2 + 2 = 1$

---

### Step 5: Compute the Sum

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

---

### 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 polynomial evaluated at 1. Let me start by trying to factor this polynomial step by step.

First, I notice that the polynomial is of degree 8, and it has only even powers of x: $ x^8 $, $ x^4 $, and the constant term. That makes me think of substitution. Maybe I can let $ y = x^4 $, so the polynomial becomes $ y^2 + 3y - 4 $. Let me try that.

So substituting $ y = x^4 $, we have:

$ y^2 + 3y - 4 $.

Now, let me factor this quadratic in y. To factor $ y^2 + 3y - 4 $, I need two numbers that multiply to -4 and add to 3. Let's see: 4 and -1. Yes, 4 * (-1) = -4 and 4 + (-1) = 3. Therefore, the quadratic factors as:

$ (y + 4)(y - 1) $.

So 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^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) $. Then, $ x^2 - 1 $ is also a difference of squares: $ (x - 1)(x + 1) $. Therefore, $ x^4 - 1 $ factors into $ (x - 1)(x + 1)(x^2 + 1) $.

Now, the other factor is $ x^4 + 4 $. Hmm, this might be more challenging. Let me recall that $ x^4 + 4 $ can be factored as a sum of squares, but over the integers? Wait, sum of squares can't be factored over the real numbers unless we use complex numbers. However, there might be a way to factor it into quadratics with integer coefficients.

Let me try to factor $ x^4 + 4 $. I remember something about factoring expressions of the form $ x^4 + a^4 $. There's a formula for that, isn't there? Let me recall. It's called the Sophie Germain identity:

$ x^4 + 4y^4 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2) $.

Yes, that's right. So if I apply that here, where $ y = 1 $, since $ x^4 + 4 = x^4 + 4(1)^4 $. Therefore, using the identity:

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

Let me check that. Multiplying those two quadratics:

First, $ (x^2 + 2x + 2)(x^2 - 2x + 2) $. Multiply them out:

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

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

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

Now, combine all these terms:

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

So indeed, the product is $ x^4 + 4 $. Perfect! So that works. Therefore, $ x^4 + 4 $ factors into $ (x^2 + 2x + 2)(x^2 - 2x + 2) $.

Therefore, putting it all together, the original polynomial factors as:

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

Now, I need to check if each of these factors can be factored further over the integers. Let me check each one:

1. $ x - 1 $: linear, so 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 $ 0 - 4 = -4 $), so it can't be factored into linear terms with integer coefficients.
4. $ x^2 + 2x + 2 $: discriminant is $ 4 - 8 = -4 $, so also irreducible over integers.
5. $ x^2 - 2x + 2 $: discriminant is $ 4 - 8 = -4 $, also irreducible over integers.

Therefore, all factors are irreducible over integers. Therefore, the complete factorization is:

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

So the polynomials $ p_1, p_2, ..., p_k $ are:

1. $ x - 1 $
2. $ x + 1 $
3. $ x^2 + 1 $
4. $ x^2 + 2x + 2 $
5. $ x^2 - 2x + 2 $

Now, the problem says each non-constant polynomial is monic with integer coefficients, which they are. So now, I need to compute the sum $ p_1(1) + p_2(1) + \dots + p_k(1) $.

Let me compute each polynomial evaluated at 1:

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 $

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

Wait, but let me check again. Let me recompute each value step by step to make sure I didn't make any mistakes.

First polynomial: $ x - 1 $. At x=1: 1 -1 = 0. Correct.

Second polynomial: x + 1. At x=1: 1 + 1 = 2. Correct.

Third polynomial: x² + 1. At x=1: 1 + 1 = 2. Correct.

Fourth polynomial: x² + 2x + 2. At x=1: 1 + 2 + 2 = 5. Correct.

Fifth polynomial: x² - 2x + 2. At x=1: 1 - 2 + 2 = 1. Correct.

Adding them up: 0 + 2 = 2; 2 + 2 = 4; 4 + 5 = 9; 9 + 1 = 10. So total is 10.

But let me think again. Is there a possibility that the problem considers the factors in a different order, or maybe I missed some factors?

Wait, let me check the original factorization again. The original polynomial was factored as:

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

Yes, that's five factors. So there are five polynomials. So the sum is 0 + 2 + 2 + 5 + 1 = 10. Therefore, the answer is 10?

But wait, let me check if I need to consider something else. Wait, 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 factorization is correct, since all factors are irreducible. Therefore, the answer should be 10.

Wait, but let me check if there is another possible factorization. For example, maybe there's a different way to factor the original polynomial into more factors? But since we have already factored it completely, into linear and irreducible quadratic factors, so I think that's the unique factorization.

Alternatively, let me check the original polynomial. Let me compute the value of the original polynomial at x=1 to see if the sum of the p_i(1) equals something else? Wait, the original polynomial is x^8 + 3x^4 -4. At x=1, that's 1 + 3 -4 = 0. However, the product of all p_i(1) should equal 0, since the original polynomial evaluated at 1 is 0. Let me check that.

Product of p_i(1) is 0 * 2 * 2 * 5 * 1 = 0, which matches the original polynomial evaluated at 1. So that seems correct.

But the question is not asking for the product, but the sum. So sum is 10. Let me check again, maybe my arithmetic is wrong. Let me write down each term:

p1(1) = 0

p2(1) = 2

p3(1) = 2

p4(1) = 5

p5(1) = 1

Total: 0 + 2 + 2 + 5 + 1 = 10. Yes, that's correct.

Wait, but maybe there's a mistake in the factorization? Let me check if the original polynomial equals the product of the factors. Let me compute (x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2). Let me compute 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 this:

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 altogether: x^6 + 2x^5 + 2x^4 - x^2 - 2x - 2.

Then, multiply by (x^2 - 2x + 2). Let me denote that as:

Let me call the previous result as A = x^6 + 2x^5 + 2x^4 - x^2 - 2x - 2.

Multiply A by (x^2 - 2x + 2). This seems tedious, but maybe there's a smarter way. Alternatively, note that earlier steps showed that x^4 + 4 factors into (x^2 + 2x + 2)(x^2 - 2x + 2), and x^4 - 1 factors into (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1). Therefore, multiplying them together gives the original polynomial. Therefore, the product is indeed x^8 + 3x^4 -4. Therefore, the factorization is correct.

Thus, the sum of p_i(1) is 10. Therefore, the answer should be 10. But let me check again with another approach.

Alternatively, if I have the factorization into p1, p2,..., pk, then p1(1) + ... + pk(1) is equal to the sum of each factor evaluated at 1. Alternatively, since the original polynomial is the product of the p_i(x), then the product of p_i(1) is the original polynomial evaluated at 1, which is 0, as we saw.

But the question is about the sum. Is there a way to relate the sum of p_i(1) with something else? Not directly that I can think of. So I think 10 is the correct answer. Wait, but maybe I made a mistake in the factorization? Let me check with specific values.

Take x = 1: original polynomial is 1 + 3 -4 = 0. The product of p_i(1) is 0 * 2 * 2 * 5 * 1 = 0. Correct.

Take x = 0: original polynomial is 0 + 0 -4 = -4. Let me compute the product of p_i(0):

p1(0) = -1, p2(0) = 1, p3(0) = 0 + 1 = 1, p4(0) = 0 + 0 + 2 = 2, p5(0) = 0 - 0 + 2 = 2. Then product is (-1)(1)(1)(2)(2) = -4, which matches the original polynomial evaluated at 0. So that's correct.

Another check: x = 2. Original polynomial: 2^8 + 3*2^4 -4 = 256 + 3*16 -4 = 256 + 48 -4 = 256 + 44 = 300.

Now compute product of p_i(2):

p1(2) = 2 - 1 = 1

p2(2) = 2 + 1 = 3

p3(2) = 4 + 1 = 5

p4(2) = 4 + 4 + 2 = 10

p5(2) = 4 - 4 + 2 = 2

Multiply them: 1 * 3 = 3; 3 * 5 = 15; 15 * 10 = 150; 150 * 2 = 300. Correct, which matches the original polynomial evaluated at 2. Therefore, the factorization is correct. Hence, my previous calculation of the sum of p_i(1) as 10 is correct.

But wait, just to be thorough, let me check if there's another possible factorization. For example, could x^4 + 4 be factored differently?

Well, we used the Sophie Germain identity, which is standard for such cases. Also, trying to factor x^4 + 4 into two quadratics with integer coefficients. Suppose we attempt to factor it as (x^2 + ax + b)(x^2 + cx + d). Then expanding:

x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd.

This should equal x^4 + 0x^3 + 0x^2 + 0x + 4. Therefore, we have the equations:

1. a + c = 0
2. ac + b + d = 0
3. ad + bc = 0
4. bd = 4

From equation 1: c = -a.

Substitute into equation 3: ad + b(-a) = 0 => a(d - b) = 0.

So either a = 0 or d = b.

Case 1: a = 0. Then c = 0. Then equation 2: 0 + b + d = 0 => b + d = 0. Equation 4: bd = 4. So solving b + d = 0 and bd = 4. Then, d = -b, so b(-b) = -b² = 4 => b² = -4, which has no real solutions, let alone integer ones. Therefore, Case 1 invalid.

Case 2: d = b. Then equation 4: b*d = b² = 4. Therefore, b = ±2. Then, equation 2: ac + b + d = (-a²) + b + b = -a² + 2b = 0. Therefore, -a² + 2b = 0 => a² = 2b.

Since b = ±2, then:

If b = 2: a² = 4 => a = ±2.

If b = -2: a² = -4, which is impossible.

Therefore, b = 2, and a = ±2. Therefore, possible solutions:

a = 2, then c = -2, and d = b = 2.

Therefore, the factors would be (x² + 2x + 2)(x² - 2x + 2), which is exactly what we found earlier. Similarly, if a = -2, then c = 2, and the factors would be (x² - 2x + 2)(x² + 2x + 2), same thing. Therefore, no other factorizations are possible. Therefore, our factorization of x^4 + 4 is correct and unique over integers.

Therefore, the factorization is correct, and the sum is indeed 10. So, I think the answer is 10.

But hold on, let me check once more if I have the correct number of factors. The original polynomial factors into (x - 1)(x + 1)(x^2 + 1)(x^2 + 2x + 2)(x^2 - 2x + 2). So that's five factors. So when we evaluate each at 1, we get 0, 2, 2, 5, 1. Summing to 10. Therefore, unless there's a miscalculation here, which I don't see, the answer is 10.

Alternatively, maybe the problem expects more factors? For example, if there is a different factorization? Let me think again. Suppose someone tried to factor x^4 -1 further, but we already did that. x^4 -1 factors into (x^2 -1)(x^2 + 1), and x^2 -1 factors into (x -1)(x +1). So that's all.

Alternatively, is there a way to factor the original polynomial into more factors? For example, if we factor x^2 + 1 further? But it's irreducible over integers. Similarly for the quadratics.

Therefore, I think the factorization is complete. Therefore, the answer is 10.

But just to make sure, let me check if the problem says "non-constant polynomial p_i(x)", so we have five factors, all non-constant. Therefore, the sum is 10. So I think that's the answer. Therefore, the answer is \boxed{10}.

**Final Answer**
\boxed{10}

To solve the problem, we start with the given polynomial:

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

---

### Step 1: Factor the Polynomial

We notice the polynomial only includes even powers of $x$, suggesting we can use substitution. Let:

$$
y = x^4
$$

Then the expression becomes:

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

This is a quadratic in $y$. Factoring, we find:

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

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

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

---

### Step 2: Factor Each Component

Now, we factor each term further:

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

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

- $x^2 - 1$ is again a difference of squares:

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

So:

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

Now, we handle $x^4 + 4$. This appears irreducible over the integers, but we can factor it using the **Sophie Germain identity**:

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

---

### Step 3: Final Factorization

Putting everything together, we have:

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

All of these polynomials are:

- Monic
- Have integer coefficients
- Cannot be factored further over the integers

So, this is the complete factorization into irreducible polynomials.

---

### Step 4: Evaluate Each Factor at $x = 1$

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

- $p_1(x) = x - 1 \Rightarrow p_1(1) = 0$
- $p_2(x) = x + 1 \Rightarrow p_2(1) = 2$
- $p_3(x) = x^2 + 1 \Rightarrow p_3(1) = 1 + 1 = 2$
- $p_4(x) = x^2 + 2x + 2 \Rightarrow p_4(1) = 1 + 2 + 2 = 5$
- $p_5(x) = x^2 - 2x + 2 \Rightarrow p_5(1) = 1 - 2 + 2 = 1$

---

### Step 5: Compute the Sum

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

---

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