Find a cubic polynomial whose zeroes …
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Posted by Himanshu Pandey 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let {tex}\alpha{/tex} = 3, {tex}\beta{/tex} = {tex}\frac{1}{2}{/tex} and {tex}\gamma{/tex} = -1. Then,
{tex}( \alpha + \beta + \gamma ) = \left( 3 + \frac { 1 } { 2 } - 1 \right) = \frac { 5 } { 2 }{/tex},
{tex}( \alpha \beta + \beta \gamma + \gamma \alpha ) = \left( \frac { 3 } { 2 } - \frac { 1 } { 2 } - 3 \right) = \frac { - 4 } { 2 }{/tex} = -2
and {tex}\alpha \beta y = \left\{ 3 \times \frac { 1 } { 2 } \times ( - 1 ) \right\} = \frac { - 3 } { 2 }{/tex}
The polynomial with zeros α,β and {tex}\gamma{/tex} is:
{tex}\mathrm x^3-(\mathrm\alpha+\mathrm\beta+\mathrm\gamma)\mathrm x^2+(\mathrm{αβ}+\mathrm{βγ}+\mathrm{γα})\mathrm x-\mathrm{αβγ}{/tex}
{tex}=\mathrm x^3-\frac52\mathrm x^2-2\mathrm x+\frac32{/tex}
Thus, 2x3- 5x2- 4x + 3 is the desired polynomial.
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