Given a polynomial P(x) = x^3 - (a +1)x^2 +ax -b is divided by x -2, the remainder is n1, and when it is divided by x + 3, the remainder is n2. Given -1 iss a zero of P(x) and n2 = -14n1, find the values of n1, n2, a and b. Hence, find the other zeroes

Given the polynomial \( P(x) = x^3 - (a + 1)x^2 + ax - b \), and the conditions provided:

1. The remainder when \( P(x) \) is divided by \( x - 2 \) is \( n_1 \).
2. The remainder when \( P(x) \) is divided by \( x + 3 \) is \( n_2 \).
3. \( -1 \) is a zero of \( P(x) \), meaning \( P(-1) = 0 \).
4. \( n_2 = -14n_1 \).

### Step 1: Use the Remainder Theorem
According to the Remainder Theorem:
- \( P(2) = n_1 \)
- \( P(-3) = n_2 = -14n_1 \)

### Step 2: Express \( P(2) \) and \( P(-3) \)
Let's calculate \( P(2) \) and \( P(-3) \) using the polynomial \( P(x) = x^3 - (a + 1)x^2 + ax - b \).

- \( P(2) = 2^3 - (a + 1)2^2 + 2a - b = 8 - 4(a + 1) + 2a - b \)
  \[
  P(2) = 8 - 4a - 4 + 2a - b = 4 - 2a - b
  \]
  Therefore, \( n_1 = P(2) = 4 - 2a - b \).

- \( P(-3) = (-3)^3 - (a + 1)(-3)^2 + (-3)a - b = -27 - 9(a + 1) - 3a - b \)
  \[
  P(-3) = -27 - 9a - 9 - 3a - b = -36 - 12a - b
  \]
  Therefore, \( n_2 = P(-3) = -36 - 12a - b \).

### Step 3: Relate \( n_1 \) and \( n_2 \)
We are given \( n_2 = -14n_1 \). Substituting the expressions for \( n_1 \) and \( n_2 \):

\[
-36 - 12a - b = -14(4 - 2a - b)
\]

Expand and simplify:

\[
-36 - 12a - b = -56 + 28a + 14b
\]

Collecting like terms:

\[
20 - 40a - 15b = 0
\]

This simplifies to:

\[
4a + 3b = 4 \quad \text{(Equation 1)}
\]

### Step 4: Use the fact that \( -1 \) is a root
Given that \( -1 \) is a root:

\[
P(-1) = (-1)^3 - (a + 1)(-1)^2 + (-1)a - b = -1 - (a + 1) - a - b
\]
\[
P(-1) = -1 - a - 1 - a - b = -2 - 2a - b = 0
\]

This simplifies to:

\[
2a + b = -2 \quad \text{(Equation 2)}
\]

### Step 5: Solve the system of equations
We have two simultaneous equations:

1. \( 4a + 3b = 4 \)
2. \( 2a + b = -2 \)

Solve Equation 2 for \( b \):

\[
b = -2 - 2a
\]

Substitute this into Equation 1:

\[
4a + 3(-2 - 2a) = 4
\]

Expand and simplify:

\[
4a - 6 - 6a = 4
\]
\[
-2a - 6 = 4
\]
\[
-2a = 10 \implies a = -5
\]

Substitute \( a = -5 \) back into Equation 2:

\[
2(-5) + b = -2
\]
\[
-10 + b = -2 \implies b = 8
\]

### Step 6: Find \( n_1 \) and \( n_2 \)
With \( a = -5 \) and \( b = 8 \):

\[
n_1 = 4 - 2(-5) - 8 = 4 + 10 - 8 = 6
\]
\[
n_2 = -36 - 12(-5) - 8 = -36 + 60 - 8 = 16
\]

Given \( n_2 = -14n_1 \), and \( n_1 = 6 \):

\[
n_2 = -14 \times 6 = -84
\]

### Step 7: Find the other zeroes
Since \( -1 \) is a zero, we know one root. To find the others, we factor the polynomial:

\[
P(x) = x^3 - (-5 + 1)x^2 - 5x - 8 = x^3 - 4x^2 - 5x - 8
\]

Given \( -1 \) is a root, factor \( P(x) \) as:

\[
P(x) = (x + 1)(x^2 - 5x - 8)
\]

To find the roots of the quadratic \( x^2 - 5x - 8 = 0 \):

\[
x = \frac{5 \pm \sqrt{25 + 32}}{2} = \frac{5 \pm \sqrt{57}}{2}
\]

Thus, the other roots are \( \frac{5 + \sqrt{57}}{2} \) and \( \frac{5 - \sqrt{57}}{2} \).

### Final Answer:
- \( n_1 = 6 \)
- \( n_2 = -84 \)
- \( a = -5 \)
- \( b = 8 \)
- Other roots: \( \frac{5 + \sqrt{57}}{2} \) and \( \frac{5 - \sqrt{57}}{2} \).

Would you like further details or have any questions? Here are five related questions for further exploration:

1. How does the Remainder Theorem help in finding remainders of polynomials?
2. How do we solve a quadratic equation to find its roots?
3. Can you find the sum and product of the roots for \( P(x) \)?
4. What is the significance of knowing one root when factoring polynomials?
5. How can we verify if the found roots satisfy the original polynomial?

**Tip:** When solving for roots of polynomials, always check for possible factorization methods first to simplify the process.

Learn how to solve a polynomial problem involving remainder theorem, roots of polynomials, and quadratic equations. Detailed step-by-step solution provided with specific values of n1, n2, a, b, and additional roots found.

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