Sum of the zeros of the cubic polynomial
P(x)= $3 x^{3} - 5 x^{2} - 11 x - 3$ is:

\frac{5}{3}

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Solution

The correct option is D $\frac{5}{3}$
We know in a cubic polynomial

Sum of zeroes $= \frac{- b}{a}$

Here $b = - 5$ and $a = 3$

Hence,sum of zeroes $= \frac{- (- 5)}{3}$ $= \frac{5}{3}$

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Similar questions

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

3 x^{3} - 5 x^{2} - 11 x - 3

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

- \frac{1}{3}

f (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

P (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

\frac{- 1}{3}

a, - 1, - \frac{1}{a}

p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

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