If two zeroes of the polynomial `x^4+3x^3-20x^2-6x+36` are `sqrt2 and -sqrt2` . find the other zeroes of polynomial

Explanation

To find the other zeroes of the polynomial \( p(x) = x^4 + 3x^3 - 20x^2 - 6x + 36 \) given that two of its zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \), we can follow these steps:

### Step 1: Write the polynomial in terms of its known roots
Since \( \sqrt{2} \) and \( -\sqrt{2} \) are roots, we can express the polynomial as:
\[ p(x) = (x - \sqrt{2})(x + \sqrt{2}) \cdot q(x) \]
where \( q(x) \) is a quadratic polynomial. The product \( (x - \sqrt{2})(x + \sqrt{2}) \) can be simplified using the difference of squares:
\[ (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \]
Thus, we can rewrite the polynomial as:
\[ p(x) = (x^2 - 2) \cdot q(x) \]

### Step 2: Perform polynomial long division
To find \( q(x) \), we need to divide \( p(x) \) by \( x^2 - 2 \).

1. **Divide the leading term**: Divide \( x^4 \) by \( x^2 \) to get \( x^2 \).
2. **Multiply**: Multiply \( x^2 \) by \( x^2 - 2 \) to get \( x^4 - 2x^2 \).
3. **Subtract**: Subtract \( (x^4 - 2x^2) \) from \( p(x) \):
\[ (x^4 + 3x^3 - 20x^2 - 6x + 36) - (x^4 - 2x^2) = 3x^3 - 18x^2 - 6x + 36 \]

4. **Repeat the process**: Now, divide \( 3x^3 \) by \( x^2 \) to get \( 3x \).
5. **Multiply**: Multiply \( 3x \) by \( x^2 - 2 \) to get \( 3x^3 - 6x \).
6. **Subtract**: Subtract \( (3x^3 - 6x) \):
\[ (3x^3 - 18x^2 - 6x + 36) - (3x^3 - 6x) = -18x^2 + 36 \]

7. **Divide again**: Divide \( -18x^2 \) by \( x^2 \) to get \( -18 \).
8. **Multiply**: Multiply \( -18 \) by \( x^2 - 2 \) to get \( -18x^2 + 36 \).
9. **Subtract**: Subtract \( (-18x^2 + 36) \):
\[ (-18x^2 + 36) - (-18x^2 + 36) = 0 \]

Thus, we find that:
\[ q(x) = x^2 + 3x - 18 \]

### Step 3: Factor the quadratic polynomial
Now we need to factor \( q(x) = x^2 + 3x - 18 \).

To factor, we look for two numbers that multiply to \(-18\) and add to \(3\). The numbers \(6\) and \(-3\) work:
\[ q(x) = (x + 6)(x - 3) \]

### Step 4: Find the other zeroes
Setting each factor equal to zero gives us the other zeroes:
1. \( x + 6 = 0 \) → \( x = -6 \)
2. \( x - 3 = 0 \) → \( x = 3 \)

### Final Answer
The other zeroes of the polynomial \( p(x) = x^4 + 3x^3 - 20x^2 - 6x + 36 \) are \( 3 \) and \( -6 \).