Question

Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time and product of its zeros as 3 , -1 and 3 respectively.


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Answer 1

hope it help more in future so

Answer 2

$\mathtt\green{GENERAL \: FORM \: OF \: CUBIC \: EQUATION}$

$\mathtt { {ax}^{3} + {bx}^{2} + cx + d}$

$\mathtt \green{SOLUTION}$

Sum of zeros

$\mathtt{ \alpha + \beta + \gamma = \frac{ - b}{a} }$

$\mathtt{ 3= \frac{ - b}{a} }$

$\mathtt{ - 3= \frac{ b}{a} }$

$\mathtt \purple{ \alpha \beta + \beta \gamma + \gamma \alpha = \frac{ c}{a} }$

$\mathtt\purple{ - 1 = \frac{c}{a} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

Product of zeros

$\mathtt{ \alpha \beta \gamma = \frac{ - d}{a} }$

$\mathtt{ 3 = \frac{ - d}{a} }$

$\mathtt{ - 3 = \frac{ d}{a} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\mathtt \purple{ \frac{ {ax}^{3} + {bx}^{2} + cx + d }{a} = { \frac{0}{a}} }$

$\mathtt \purple{ {ax}^{3} + \frac{b}{a} {x}^{2} + \frac{c}{a} {x}^{2} + \frac{d}{a} = 0 }$

$\mathtt \purple{ {x}^{3} - {3x}^{2} - x - 3 = 0 }$