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(D) Let the quadratic polynomial be $p(x) = k(x^2 - (	ext{sum of zeroes})x + (	ext{product of zeroes}))$, where $k$ is a non-zero constant.Given zer

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$\frac{x^{2}}{2}-\frac{x}{2}-6$

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(D) Let the quadratic polynomial be $p(x) = k(x^2 - (	ext{sum of zeroes})x + (	ext{product of zeroes}))$, where $k$ is a non-zero constant.Given zeroes are $\alpha = -3$ and $\beta = 4$.Sum of zeroes $(\alpha + \beta) = -3 + 4 = 1$.Product of zeroes $(\alpha \cdot \beta) = -3 	imes 4 = -12$.Substituting these values into the formula, we get $p(x) = k(x^2 - 1x + (-12)) = k(x^2 - x - 12)$.If we choose $k = \frac{1}{2}$, the polynomial becomes $\frac{1}{2}(x^2 - x - 12) = \frac{x^2}{2} - \frac{x}{2} - 6$.Since the zeroes of a polynomial remain unchanged when multiplied by a non-zero constant, option $D$ is a valid quadratic polynomial with the given zeroes.

$A$ quadratic polynomial, whose zeroes are $-3$ and $4,$ is

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