Question

the number of polynomials having zeroes as 4and 7 is
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Answer 1

Answer:

${ {x}^{2} - 11x}^{} + 28 = 0$

Step-by-step explanation:

$\alpha + \beta = 4 + 7 \\ = 11 \\ \\ \alpha \beta = 4 \times 7 \\ = 28 \\ \\ required \: equation \: \: \: \: \: \: \: \: {x}^{2} - (\alpha + \beta )x + \alpha \beta = 0 \\ \\ \\ { {x}^{2} - 11x}^{} + 28 = 0$