Question

Find a cubic polynomial whose zeros are 2,_√7 and√7.also verify the relationship between the zero and the coefficient of the polynomial

Answer 1

$Let \: \alpha ,\beta \:and \:\gamma\: are$

$zeroes \:of \: a \: cubic \: polynomial$

$\alpha = 2, \: \beta = -\sqrt{7}\:and \:\gamma = \sqrt{7}$

$The \: polynomial \:can \: be \: written \: as$

$= (x - 2)(x+\sqrt{7})(x-\sqrt{7})$

$= (x-2)[x^{2} - (\sqrt{7})^{2}]$

$= (x-2)(x^{2}-7)$

$= x^{3} - 7x - 2x^{2} + 14$

$\green {= x^{3}-2x^{2} - 7x + 14 }$

Verification :

$i) \alpha + \beta + \gamma$

$= 2 - \sqrt{7} + \sqrt{7}$

$= 2$

$= \frac{2}{1}$

$= \frac{- x^{2} \: coefficient }{x^{3} \: coefficient }$

$ii) \alpha \beta + \beta \gamma+\alpha \gamma$

$= 2\times (-\sqrt{7}) + (-\sqrt{7})(\sqrt{7}) + (\sqrt{7}) \times 2$

$= -2\sqrt{7} - 7 + 2\sqrt{7}$

$= -7$

$= \frac{-7}{1}$

$= \frac{x \: coefficient }{x^{3} \: coefficient }$

$iii) \alpha \beta \gamma$

$= 2 \times (- \sqrt{7} )(\sqrt{7})$

$= 2 \times (-7)$

$= \frac{-14}{1}$

$= \frac{- constant \:term}{x^{3} \: coefficient }$

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