Find the zeroes of the polynomial `x^(3)+6x^(2)+11x+6` which are integers .

Answer

To find the integer zeroes of the polynomial \( f(x) = x^3 + 6x^2 + 11x + 6 \), we will use the trial and error method (also known as the Rational Root Theorem) to test possible integer factors.

### Step 1: Identify possible integer factors
The possible integer factors of the constant term (6) are \( \pm 1, \pm 2, \pm 3, \pm 6 \).

### Step 2: Test each factor
We will substitute each possible integer into the polynomial to see if it results in zero.

1. **Testing \( x = 1 \)**:
\[ f(1) = 1^3 + 6(1^2) + 11(1) + 6 = 1 + 6 + 11 + 6 = 24 \quad (\text{not a zero}) \]

2. **Testing \( x = -1 \)**:
\[ f(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0 \quad (\text{is a zero}) \]

Since \( x = -1 \) is a zero, we can factor the polynomial using \( x + 1 \).

### Step 3: Factor the polynomial
We can perform polynomial long division or synthetic division to divide \( f(x) \) by \( x + 1 \).

#### Synthetic Division:
\[ \begin{array}{r|rrrr} -1 & 1 & 6 & 11 & 6 \\ & & -1 & -5 & -6 \\ \hline & 1 & 5 & 6 & 0 \\ \end{array} \]
The result is \( x^2 + 5x + 6 \).

### Step 4: Factor the quadratic
Now we need to factor \( x^2 + 5x + 6 \):
\[ x^2 + 5x + 6 = (x + 2)(x + 3) \]

### Step 5: Write the complete factorization
Thus, the complete factorization of the polynomial is:
\[ f(x) = (x + 1)(x + 2)(x + 3) \]

### Step 6: Identify all zeroes
Setting each factor to zero gives us the zeroes:
1. \( x + 1 = 0 \) → \( x = -1 \)
2. \( x + 2 = 0 \) → \( x = -2 \)
3. \( x + 3 = 0 \) → \( x = -3 \)

### Conclusion
The integer zeroes of the polynomial \( x^3 + 6x^2 + 11x + 6 \) are \( -1, -2, \) and \( -3 \).

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