For which values of a and b are zeroes of q(X)`=x^(3)+2x+a` also zeroes of the polynomial p(x)`=x^(5)-x^(4)-4x^(3)+3x^(2)+3x+b`? Which zeroes of p(x) are not zeroes of p(x)?

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) such that the zeros of the polynomial \( q(x) = x^3 + 2x^2 + a \) are also zeros of the polynomial \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \).

### Step 1: Understand the problem
We need to determine the values of \( a \) and \( b \) such that every root (zero) of \( q(x) \) is also a root of \( p(x) \). This means that \( q(x) \) must be a factor of \( p(x) \).

### Step 2: Perform polynomial long division
To check if \( q(x) \) is a factor of \( p(x) \), we can use polynomial long division. We will divide \( p(x) \) by \( q(x) \).

1. **Divide the leading term**: The leading term of \( p(x) \) is \( x^5 \) and the leading term of \( q(x) \) is \( x^3 \). We divide \( x^5 \) by \( x^3 \) to get \( x^2 \).

2. **Multiply \( q(x) \) by \( x^2 \)**:
\[ x^2 \cdot (x^3 + 2x^2 + a) = x^5 + 2x^4 + ax^2 \]

3. **Subtract from \( p(x) \)**:
\[ p(x) - (x^5 + 2x^4 + ax^2) = (-1 - 2)x^4 + (-4 - a)x^3 + 3x^2 + 3x + b \]
This simplifies to:
\[ -3x^4 + (-4 - a)x^3 + 3x^2 + 3x + b \]

### Step 3: Continue the division
4. **Divide the leading term again**: The leading term is now \( -3x^4 \). Divide by \( x^3 \) to get \( -3x \).

5. **Multiply \( q(x) \) by \( -3x \)**:
\[ -3x(x^3 + 2x^2 + a) = -3x^4 - 6x^3 - 3ax \]

6. **Subtract again**:
\[ (-4 - a + 6)x^3 + (3 + 3a)x + b \]
This simplifies to:
\[ (2 - a)x^3 + (3 + 3a)x + b \]

### Step 4: Final division step
7. **Divide the leading term**: The leading term is \( (2 - a)x^3 \). Divide by \( x^3 \) to get \( 2 - a \).

8. **Multiply \( q(x) \) by \( (2 - a) \)**:
\[ (2 - a)(x^3 + 2x^2 + a) = (2 - a)x^3 + 2(2 - a)x^2 + (2 - a)a \]

9. **Subtract**:
\[ (3 + 3a - 2(2 - a))x^2 + (b - (2 - a)a) \]
This gives:
\[ (3 + 3a - 4 + 2a)x^2 + (b - (2a - a^2)) \]
Simplifying gives:
\[ (5a - 1)x^2 + (b - 2a + a^2) \]

### Step 5: Set the remainder to zero
For \( q(x) \) to be a factor of \( p(x) \), the remainder must be zero:
1. Set \( 5a - 1 = 0 \) → \( a = \frac{1}{5} \)
2. Set \( b - 2a + a^2 = 0 \):
\[ b = 2a - a^2 \]
Substituting \( a = \frac{1}{5} \):
\[ b = 2 \cdot \frac{1}{5} - \left(\frac{1}{5}\right)^2 = \frac{2}{5} - \frac{1}{25} = \frac{10}{25} - \frac{1}{25} = \frac{9}{25} \]

### Final values
Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{1}{5}, \quad b = \frac{9}{25} \]

### Step 6: Identify zeros of \( p(x) \) that are not zeros of \( q(x) \)
To find the zeros of \( p(x) \) and \( q(x) \):
- The zeros of \( q(x) \) are the solutions to \( x^3 + 2x^2 + \frac{1}{5} = 0 \).
- The zeros of \( p(x) \) are the solutions to \( x^5 - x^4 - 4x^3 + 3x^2 + 3x + \frac{9}{25} = 0 \).

After solving both equations, we can identify which zeros of \( p(x) \) are not zeros of \( q(x) \).