Find the cubic polynomial whose zeroes …
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Sia ? 6 years, 5 months ago
Let the cubic polynomial be ax3 + bx2 +cx + d
and its zeroes be {tex}\alpha ,\beta {/tex} and {tex}\gamma{/tex}.
Then, {tex}\alpha + \beta + \gamma= 2 = \frac { - b } { a }{/tex}
{tex}\alpha \beta + \beta \gamma + \gamma \alpha = - 7 = \frac { C } { a }{/tex}
and {tex}\alpha \beta \gamma = - 14 = \frac { - d } { a }{/tex}
If a = 1, then b = -2, c = -7, d = 14
So, one cubic polynomial which fits the given
conditions is x3 - 2x2 - 7x + 14.
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