Let a and B be the two rOot of eQn-- given, a+B = 1/4aB= -a quadratic polynomial caN be written in the form::::p(x) = (sum of root) x + product of rootp(x) x^2 - (a+B ) x + aB = 0 p(x) = x^2 - 1/4 - 1/8p(x) = 8x^2 - 2x -1 cheer""" provedhope it'll be help U Harshita

**Finding the Quadratic Polynomial with Given Zeroes**

To find a quadratic polynomial with given zeroes, we will use the fact that the sum of the zeroes of a quadratic polynomial is equal to the negation of the coefficient of the linear term, and the product of the zeroes is equal to the constant term divided by the coefficient of the quadratic term.

Let's denote the quadratic polynomial as ax² + bx + c, where a, b, and c are constants.

**Finding the Sum of the Zeroes**

Given that the zeroes are -3 and 2, the sum of the zeroes can be found as follows:

Sum of Zeroes = -3 + 2 = -1

Therefore, the coefficient of the linear term (-b) is equal to -1.

**Finding the Product of the Zeroes**

Given that the zeroes are -3 and 2, the product of the zeroes can be found as follows:

Product of Zeroes = -3 * 2 = -6

Therefore, the constant term (c) is equal to -6.

**Plugging in the Values**

Now that we have determined the values of two of the coefficients, we can write the quadratic polynomial as follows:

ax² + bx + c = a(x² + bx/a) + c

Substituting the values we found earlier, the quadratic polynomial becomes:

ax² + bx + c = a(x² - x/a) - 6

**Simplifying the Polynomial**

To simplify the polynomial further, we can multiply a into the expression inside the parentheses:

ax² + bx + c = ax² - x + (-6)

Combining like terms, we have:

ax² + bx + c = ax² - x - 6

Therefore, the quadratic polynomial with zeroes -3 and 2 is given by:

ax² + bx + c = ax² - x - 6

**Conclusion**

In conclusion, the quadratic polynomial with zeroes -3 and 2 is ax² - x - 6. The sum of the zeroes is -1, and the product of the zeroes is -6.