Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 

The general form of a quadratic polynomial when the sum and product of its zeroes are given is:

$x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})$

Now, applying this formula for each given pair:

1) Sum = $\frac{1}{4}$, Product = $-1$
$x^2 - \frac{1}{4}x - 1$

2) Sum = $\sqrt{2}$, Product = $\frac{1}{3}$
$x^2 - \sqrt{2}x + \frac{1}{3}$

3) Sum = $0$, Product = $\sqrt{5}$
$x^2 + \sqrt{5}$

4) Sum = $1$, Product = $1$
$x^2 - x + 1$

5) Sum = $- \frac{1}{4}$, Product = $\frac{1}{4}$
$x^2 + \frac{1}{4}x + \frac{1}{4}$

6) Sum = $4$, Product = $1$
$x^2 - 4x + 1$

Thus, these are the required quadratic polynomials. 

use the formula $x^2-(\text{sum of zeroes})x+(\text{product of zeroes})=0$. Here, 

$\text{A quadratic polynomial with sum }S\text{ and product }P\text{ of its zeroes is given by:}$

$x^2-Sx+P$ $\text{For each given sum and product:}$

$\text{3. Given }S=0,P=\sqrt{5}$ $x^2-(0)x+\sqrt{5}=x^2+\sqrt{5}$

$\text{4. Given }S=1,P=1$ $x^2-(1)x+1=x^2-x+1$

$\text{5. Given }S=-\frac{1}{4},P=\frac{1}{4}$ $x^2-(-\frac{1}{4})x+\frac{1}{4}=x^2+\frac{1}{4}x+\frac{1}{4}$

$\text{6. Given }S=4,P=1$ $x^2-(4)x+1=x^2-4x+1$