(i) Obtain all other zeros of `2x^4 + 7x^3 - 19x^2 - 14x + 30`, if two of its zeros are `sqrt(2)` and `-sqrt(2)`.
(ii) Obtain all other zeros of `2x^(3) + x^2 - 6x - 3`, if two of its zeros are `-sqrt(3)` and `sqrt(3)` .

Answer

Let's solve the given problems step by step.

### Part (i)

**Given Polynomial:**
\[ P(x) = 2x^4 + 7x^3 - 19x^2 - 14x + 30 \]
**Known Zeros:**
\[ \sqrt{2}, -\sqrt{2} \]

#### Step 1: Identify the factors from the known zeros
Since \( \sqrt{2} \) and \( -\sqrt{2} \) are zeros, we can form a factor from these zeros:
\[ (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \]

#### Step 2: Divide the polynomial by the factor
Now, we will divide \( P(x) \) by \( x^2 - 2 \) using polynomial long division.

1. Divide \( 2x^4 \) by \( x^2 \) to get \( 2x^2 \).
2. Multiply \( 2x^2 \) by \( x^2 - 2 \) to get \( 2x^4 - 4x^2 \).
3. Subtract this from \( P(x) \):
\[ (2x^4 + 7x^3 - 19x^2 - 14x + 30) - (2x^4 - 4x^2) = 7x^3 - 15x^2 - 14x + 30 \]

4. Divide \( 7x^3 \) by \( x^2 \) to get \( 7x \).
5. Multiply \( 7x \) by \( x^2 - 2 \) to get \( 7x^3 - 14x \).
6. Subtract this:
\[ (7x^3 - 15x^2 - 14x + 30) - (7x^3 - 14x) = -15x^2 + 30 \]

7. Divide \( -15x^2 \) by \( x^2 \) to get \( -15 \).
8. Multiply \( -15 \) by \( x^2 - 2 \) to get \( -15x^2 + 30 \).
9. Subtract this:
\[ (-15x^2 + 30) - (-15x^2 + 30) = 0 \]

Thus, we have:
\[ P(x) = (x^2 - 2)(2x^2 + 7x - 15) \]

#### Step 3: Factor the quadratic \( 2x^2 + 7x - 15 \)
Now we need to factor \( 2x^2 + 7x - 15 \).

1. Look for two numbers that multiply to \( 2 \times -15 = -30 \) and add to \( 7 \). The numbers are \( 10 \) and \( -3 \).
2. Rewrite the middle term:
\[ 2x^2 + 10x - 3x - 15 \]
3. Factor by grouping:
\[ 2x(x + 5) - 3(x + 5) = (2x - 3)(x + 5) \]

#### Step 4: Find the remaining zeros
Now we have:
\[ P(x) = (x^2 - 2)(2x - 3)(x + 5) \]

Setting each factor to zero:
1. \( x^2 - 2 = 0 \) gives \( x = \sqrt{2}, -\sqrt{2} \) (already known).
2. \( 2x - 3 = 0 \) gives \( x = \frac{3}{2} \).
3. \( x + 5 = 0 \) gives \( x = -5 \).

**Remaining Zeros:**
\[ \frac{3}{2}, -5 \]

### Part (ii)

**Given Polynomial:**
\[ P(x) = 2x^3 + x^2 - 6x - 3 \]
**Known Zeros:**
\[ -\sqrt{3}, \sqrt{3} \]

#### Step 1: Identify the factors from the known zeros
Since \( -\sqrt{3} \) and \( \sqrt{3} \) are zeros, we can form a factor:
\[ (x - \sqrt{3})(x + \sqrt{3}) = x^2 - 3 \]

#### Step 2: Divide the polynomial by the factor
Now, we will divide \( P(x) \) by \( x^2 - 3 \).

1. Divide \( 2x^3 \) by \( x^2 \) to get \( 2x \).
2. Multiply \( 2x \) by \( x^2 - 3 \) to get \( 2x^3 - 6x \).
3. Subtract this from \( P(x) \):
\[ (2x^3 + x^2 - 6x - 3) - (2x^3 - 6x) = x^2 - 3 \]

4. Divide \( x^2 \) by \( x^2 \) to get \( 1 \).
5. Multiply \( 1 \) by \( x^2 - 3 \) to get \( x^2 - 3 \).
6. Subtract this:
\[ (x^2 - 3) - (x^2 - 3) = 0 \]

Thus, we have:
\[ P(x) = (x^2 - 3)(2x + 1) \]

#### Step 3: Find the remaining zeros
Now we have:
\[ P(x) = (x^2 - 3)(2x + 1) \]

Setting each factor to zero:
1. \( x^2 - 3 = 0 \) gives \( x = \sqrt{3}, -\sqrt{3} \) (already known).
2. \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \).

**Remaining Zero:**
\[ -\frac{1}{2} \]

### Summary of Solutions:
- For part (i), the remaining zeros are \( \frac{3}{2}, -5 \).
- For part (ii), the remaining zero is \( -\frac{1}{2} \).