The zeros of the quadratic polynomial are -3 | vedclass

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Question

(A) quadratic polynomial with zeros $\alpha$ and $\beta$ is given by the formula: $p(x) = k[x^{2} - (\alpha + \beta)x + (\alpha \cdot \beta)]$,where $

Vedclass · Accepted Answer

$p(x) = x^{2} - x - 12$

Vedclass · Answer

(A) quadratic polynomial with zeros $\alpha$ and $\beta$ is given by the formula: $p(x) = k[x^{2} - (\alpha + \beta)x + (\alpha \cdot \beta)]$,where $k$ is a constant.Given zeros are $\alpha = -3$ and $\beta = 4$.Sum of zeros: $\alpha + \beta = -3 + 4 = 1$.Product of zeros: $\alpha \cdot \beta = (-3) 	imes 4 = -12$.Substituting these values into the formula (taking $k=1$): $p(x) = x^{2} - (1)x + (-12) = x^{2} - x - 12$.Therefore,the correct polynomial is $p(x) = x^{2} - x - 12$.

The zeros of the quadratic polynomial are $-3$ and $4$. Which of the following represents this polynomial?

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