If $- \frac{1}{3}$ is the zero of the cubic polynomial $f (x) = 3 x^{3} - 5 x^{2} - 11 x - 3$ the other zeros are

- 3, - 1

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1, 3

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3, - 1

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- 3, 1

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Solution

The correct option is D $3, - 1$
Let $a, b$ be the other zeros of the cubic polynomial, then,
$\frac{- 1}{3} + a + b = \frac{- b}{a} = - \frac{(- 5)}{3} = \frac{5}{3}$

$=> a + b = \frac{5}{3} + \frac{- 1}{3}$

$=> a + b = \frac{6}{3}$
$=> a + b = 2$

Also,
$(\frac{- 1}{3}) a b = \frac{- d}{a} = \frac{- (- 3)}{3} = \frac{3}{3} = 1$
$=> a b = - 3$
Thus, either of $a$ and $b$ is negative and the larger number is positive.
Thus, $a = 3$ and $b = - 1$

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Similar questions

3, - 1, \frac{- 1}{3}

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One of the zeroes of cubic polynomial $(3 x^{3} - 5 x^{2} - 11 x - 3)$ is $\frac{- 1}{3}$ . Find the other zeroes.

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