13. find all zeroes of `2x^4-3x^3-3x^2+6x-2` , if you know that two of its zeroes are ` sqrt2 and -(sqrt2) `

Explanation

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

### Step 1: Identify the known roots
The given roots are \(\sqrt{2}\) and \(-\sqrt{2}\). We can express these roots in terms of factors of the polynomial:
\[ (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \]

### Step 2: Divide the polynomial by the factor
Now, we will divide the polynomial \(2x^4 - 3x^3 - 3x^2 + 6x - 2\) by \(x^2 - 2\) using polynomial long division.

### Step 3: Perform polynomial long division
1. Divide the leading term \(2x^4\) 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 the original polynomial:
\[ (2x^4 - 3x^3 - 3x^2 + 6x - 2) - (2x^4 - 4x^2) = -3x^3 + x^2 + 6x - 2 \]

4. Now, divide \(-3x^3\) by \(x^2\) to get \(-3x\).
5. Multiply \(-3x\) by \(x^2 - 2\) to get \(-3x^3 + 6x\).
6. Subtract this from the current polynomial:
\[ (-3x^3 + x^2 + 6x - 2) - (-3x^3 + 6x) = x^2 - 2 \]

7. Finally, divide \(x^2\) by \(x^2\) to get \(1\).
8. Multiply \(1\) by \(x^2 - 2\) to get \(x^2 - 2\).
9. Subtract this from the current polynomial:
\[ (x^2 - 2) - (x^2 - 2) = 0 \]

The result of the division gives us:
\[ 2x^2 - 3x + 1 \]

### Step 4: Factor the quotient polynomial
Now we need to factor \(2x^2 - 3x + 1\):
1. We can use the factorization method:
\[ 2x^2 - 2x - x + 1 = 0 \]
Grouping gives:
\[ 2x(x - 1) - 1(x - 1) = 0 \]
This can be factored as:
\[ (2x - 1)(x - 1) = 0 \]

### Step 5: Solve for the remaining roots
Setting each factor to zero gives us the remaining roots:
1. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
2. \(x - 1 = 0 \Rightarrow x = 1\)

### Step 6: Compile all the roots
Thus, the complete set of zeroes of the polynomial \(2x^4 - 3x^3 - 3x^2 + 6x - 2\) are:
\[ \sqrt{2}, -\sqrt{2}, \frac{1}{2}, 1 \]