Question

Find the cubic polynomial whose three zeros are 3 , -1 and 1/3

Answer 1

The required answer is in the attachment.

Answer 2

$Given\: \alpha = 3 , \: \beta = -1 \:and \\ \gamma = \frac{1}{3} \: are \: zeroes \:of \: a \\cubic \: polynomial$

$i) \alpha +\beta + \gamma\\ = 3 + (-1)+ \frac{1}{3} \\= 3 - 1 + \frac{1}{3} \\= 2 + \frac{1}{3} \\= \frac{6+1}{3} \\=\frac{7}{3} \: --(1)$

$ii ) \alpha \beta + \beta \gamma + \beta \alpha \\= 3\times (-1) + (-1) \times \frac{1}{3} + \frac{1}{3} \times 3 \\= -3 -\frac{1}{3} + 1 \\= -2 -\frac{1}{3} \\= \frac{-6- 1}{3} \\= \frac{-7}{3} \: --(2)$

$iii ) \alpha \beta \gamma\\= 3 \times (-1) \times \frac{1}{3} \\= -1 \:---(3)$

$\underline{ \blue { Form \: of \:a \:cubic \: polynomial :}}$

$\boxed { \pink { k[x^{3} -(\alpha +\beta + \gamma)x^{2} + (\alpha \beta + \beta \gamma + \beta \alpha )x - \alpha \beta \gamma] }}$

$k[ x^{3} - \frac{7}{3}x^{2} + \frac{-7}{3}x+ (-1) ]$

$= k[ x^{3} - \frac{7}{3}x^{2} - \frac{7}{3}x-1]$

$We\:can \:put \: different \: values \:of \:k .$

$When \: k = 3 , \: the\: Cubic \: polynomial \\will \:be \: \orange { 3x^{2} - 7x^{2} - 7x - 3 }$

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