Obtain other zeroes of the polynomial `f(x)=2x^4+3x^3-5x^2-9x-3` if two of its zeroes are `-sqrt3 and sqrt3`

Answer

To find the other zeroes of the polynomial \( f(x) = 2x^4 + 3x^3 - 5x^2 - 9x - 3 \) given that two of its zeroes are \( -\sqrt{3} \) and \( \sqrt{3} \), we can follow these steps:

### Step 1: Identify the known zeroes
We know two zeroes of the polynomial:
- \( r_1 = -\sqrt{3} \)
- \( r_2 = \sqrt{3} \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas, for a polynomial of the form \( ax^n + bx^{n-1} + cx^{n-2} + \ldots + k = 0 \):
- The sum of the roots \( r_1 + r_2 + r_3 + r_4 = -\frac{b}{a} \)
- The product of the roots \( r_1 \cdot r_2 \cdot r_3 \cdot r_4 = (-1)^n \frac{k}{a} \)

Here, \( a = 2 \) and \( b = 3 \).

### Step 3: Calculate the sum of the roots
Using Vieta's formula for the sum of the roots:
\[ r_1 + r_2 + r_3 + r_4 = -\frac{b}{a} = -\frac{3}{2} \]
Substituting the known roots:
\[ -\sqrt{3} + \sqrt{3} + r_3 + r_4 = -\frac{3}{2} \]
This simplifies to:
\[ r_3 + r_4 = -\frac{3}{2} \]

### Step 4: Calculate the product of the roots
Using Vieta's formula for the product of the roots:
\[ r_1 \cdot r_2 \cdot r_3 \cdot r_4 = (-1)^4 \frac{k}{a} = \frac{-3}{2} \]
Calculating the product of the known roots:
\[ (-\sqrt{3}) \cdot (\sqrt{3}) \cdot r_3 \cdot r_4 = 3 \cdot r_3 \cdot r_4 \]
Setting this equal to the product from Vieta's:
\[ 3 \cdot r_3 \cdot r_4 = -\frac{3}{2} \]
Thus:
\[ r_3 \cdot r_4 = -\frac{1}{2} \]

### Step 5: Set up equations for \( r_3 \) and \( r_4 \)
Now we have a system of equations:
1. \( r_3 + r_4 = -\frac{3}{2} \)
2. \( r_3 \cdot r_4 = -\frac{1}{2} \)

Let \( r_3 = x \) and \( r_4 = y \). Then we can rewrite the equations as:
1. \( x + y = -\frac{3}{2} \)
2. \( xy = -\frac{1}{2} \)

### Step 6: Substitute \( y \) in terms of \( x \)
From the first equation, we can express \( y \):
\[ y = -\frac{3}{2} - x \]
Substituting into the second equation:
\[ x\left(-\frac{3}{2} - x\right) = -\frac{1}{2} \]
Expanding this gives:
\[ -\frac{3}{2}x - x^2 = -\frac{1}{2} \]
Rearranging leads to:
\[ x^2 + \frac{3}{2}x - \frac{1}{2} = 0 \]

### Step 7: Solve the quadratic equation
Multiplying through by 2 to eliminate the fraction:
\[ 2x^2 + 3x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 3, c = -1 \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ x = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4} \]

### Step 8: Find the values of \( r_3 \) and \( r_4 \)
Thus, the roots are:
\[ r_3 = \frac{-3 + \sqrt{17}}{4}, \quad r_4 = \frac{-3 - \sqrt{17}}{4} \]

### Conclusion
The other zeroes of the polynomial \( f(x) = 2x^4 + 3x^3 - 5x^2 - 9x - 3 \) are:
\[ r_3 = \frac{-3 + \sqrt{17}}{4}, \quad r_4 = \frac{-3 - \sqrt{17}}{4} \]