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

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the zeroes of the polynomial \( q(x) = x^3 + 2x^2 + a \) are also the zeroes of the polynomial \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \). We will also identify which zeroes of \( p(x) \) are not zeroes of \( q(x) \).

### Step 1: Set up the problem
We have:
- \( q(x) = x^3 + 2x^2 + a \)
- \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \)

We want to find \( a \) and \( b \) such that the zeroes of \( q(x) \) are also zeroes of \( p(x) \).

### Step 2: Perform polynomial long division
We will divide \( p(x) \) by \( q(x) \) to find the quotient and the remainder.

1. Divide the leading term of \( p(x) \) by the leading term of \( q(x) \):
\[ \frac{x^5}{x^3} = x^2 \]
Multiply \( q(x) \) by \( x^2 \):
\[ x^2 \cdot (x^3 + 2x^2 + a) = x^5 + 2x^4 + ax^2 \]
Subtract this from \( p(x) \):
\[ p(x) - (x^5 + 2x^4 + ax^2) = -x^4 - 4x^3 + 3x^2 + 3x + b - (2x^4 + ax^2) \]
This simplifies to:
\[ -3x^4 - 4x^3 + (3 - a)x^2 + 3x + b \]

2. Divide the leading term of the new polynomial by the leading term of \( q(x) \):
\[ \frac{-3x^4}{x^3} = -3x \]
Multiply \( q(x) \) by \( -3x \):
\[ -3x \cdot (x^3 + 2x^2 + a) = -3x^4 - 6x^3 - 3ax \]
Subtract this from the previous result:
\[ (-3x^4 - 4x^3 + (3 - a)x^2 + 3x + b) - (-3x^4 - 6x^3 - 3ax) = 2x^3 + (3 - a)x^2 + (3 + 3a)x + b \]

3. Divide the leading term again:
\[ \frac{2x^3}{x^3} = 2 \]
Multiply \( q(x) \) by \( 2 \):
\[ 2 \cdot (x^3 + 2x^2 + a) = 2x^3 + 4x^2 + 2a \]
Subtract this from the previous result:
\[ (2x^3 + (3 - a)x^2 + (3 + 3a)x + b) - (2x^3 + 4x^2 + 2a) = (3 - a - 4)x^2 + (3 + 3a)x + (b - 2a) \]
This simplifies to:
\[ (3 - a - 4)x^2 + (3 + 3a)x + (b - 2a) \]

### Step 3: Set the remainder to zero
For \( q(x) \) to be a factor of \( p(x) \), the remainder must be zero. Thus, we set the coefficients of the polynomial equal to zero:

1. Coefficient of \( x^2 \):
\[ 3 - a - 4 = 0 \implies a = -1 \]

2. Coefficient of \( x \):
\[ 3 + 3a = 0 \implies 3 + 3(-1) = 0 \implies 0 = 0 \quad \text{(no new information)} \]

3. Constant term:
\[ b - 2a = 0 \implies b - 2(-1) = 0 \implies b + 2 = 0 \implies b = -2 \]

### Step 4: Final values
Thus, we have:
\[ a = -1, \quad b = -2 \]

### Step 5: Identify zeroes of \( p(x) \) not in \( q(x) \)
Now we need to find the zeroes of both polynomials:
- \( q(x) = x^3 + 2x^2 - 1 \)
- \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x - 2 \)

To find the zeroes of \( p(x) \), we can factor it or use numerical methods. We can check for rational roots using the Rational Root Theorem or synthetic division.

After finding the roots of \( p(x) \), we can compare them with the roots of \( q(x) \) to determine which roots are not common.

### Conclusion
The values of \( a \) and \( b \) are:
\[ \boxed{a = -1, b = -2} \]
The zeroes of \( p(x) \) that are not zeroes of \( q(x) \) can be determined after finding the roots of both polynomials.