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Let's solve these exercises one by one:

### 1. Determine \( m \in \mathbb{R} \) such that the polynomial \( p(x) = (m^2 - 2)x^4 + 6x^3 - 4x + 2 \) has degree 4.

**Solution:**  
The degree of a polynomial is determined by the highest power of \( x \) with a non-zero coefficient. Here, the term with the highest power is \( x^4 \) with the coefficient \( (m^2 - 2) \).

To ensure the polynomial has degree 4, the coefficient of \( x^4 \) must be non-zero:

\[
m^2 - 2 \neq 0
\]

Solving for \( m \):

\[
m^2 \neq 2 \quad \Rightarrow \quad m \neq \pm\sqrt{2}
\]

Thus, \( m \) must be any real number except \( \pm\sqrt{2} \).

### 2. For which real values of \( k \) does the polynomial expression \( (2k^2 - 8)x^3 - 2x^2 + 5x - 1 \) have degree 2?

**Solution:**  
For the polynomial to have degree 2, the coefficient of the \( x^3 \) term must be zero:

\[
2k^2 - 8 = 0
\]

Solving for \( k \):

\[
2k^2 = 8 \quad \Rightarrow \quad k^2 = 4 \quad \Rightarrow \quad k = \pm 2
\]

Thus, the polynomial has degree 2 when \( k = \pm 2 \).

### 3. The polynomial \( p(x) = ax + (b - 2) \) is zero. What are the values of \( a \) and \( b \)?

**Solution:**  
For the polynomial to be zero for all \( x \):

\[
ax + (b - 2) = 0
\]

This implies:

- The coefficient of \( x \), \( a \), must be 0.
- The constant term \( b - 2 \) must also be 0.

So,

\[
a = 0 \quad \text{and} \quad b - 2 = 0 \quad \Rightarrow \quad b = 2
\]

Thus, \( a = 0 \) and \( b = 2 \).

### 4. Determine the values of \( a, b, c, d \) such that the polynomial \( (a - 1)x^3 + (2a - b)x^2 + (b - c)x + (c - 2d) \) is zero.

**Solution:**  
For the polynomial to be zero for all \( x \):

1. \( a - 1 = 0 \quad \Rightarrow \quad a = 1 \)
2. \( 2a - b = 0 \quad \Rightarrow \quad 2(1) - b = 0 \quad \Rightarrow \quad b = 2 \)
3. \( b - c = 0 \quad \Rightarrow \quad 2 - c = 0 \quad \Rightarrow \quad c = 2 \)
4. \( c - 2d = 0 \quad \Rightarrow \quad 2 - 2d = 0 \quad \Rightarrow \quad d = 1 \)

Thus, \( a = 1 \), \( b = 2 \), \( c = 2 \), and \( d = 1 \).

### 5. Determine \( m \in \mathbb{R} \) such that \( -1 \) is a root of the polynomial \( x^2 - 4x + (m + 4) \).

**Solution:**  
If \( -1 \) is a root of the polynomial, then substituting \( x = -1 \) into the polynomial should give 0:

\[
(-1)^2 - 4(-1) + (m + 4) = 0
\]

Simplifying:

\[
1 + 4 + (m + 4) = 0 \quad \Rightarrow \quad 9 + m = 0 \quad \Rightarrow \quad m = -9
\]

Thus, \( m = -9 \).

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Do you need more details on any of the solutions, or have any questions?

### Related Questions:

1. How can you find the roots of a polynomial of degree 3?
2. What is the relationship between the roots and coefficients of a quadratic polynomial?
3. How do you determine the degree of a polynomial given in factored form?
4. Can a polynomial of degree 2 have more than 2 roots? Why or why not?
5. What is the effect of the coefficient of the highest degree term on the graph of the polynomial?
6. How can the discriminant be used to determine the nature of roots for a quadratic equation?
7. What are the properties of a polynomial with all zero coefficients except for the constant term?
8. How do you solve a system of linear equations derived from polynomial conditions?

### Tip:
When solving polynomial equations, always check the highest degree term first to determine the overall degree and nature of the polynomial.

Explore and solve 5 exercises in polynomial equations, covering topics such as determining polynomial degrees, finding roots of quadratic equations, and solving polynomial expressions. Detailed step-by-step solutions provided.

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