If two zeroes of a cubic polynomial $a x^{3} + b x^{2} + c x + d$ are each equal to zero , then the third zero is

- \frac{d}{a}

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\frac{c}{a}

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\frac{- b}{a}

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\frac{b}{9}

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Solution

The correct option is B $\frac{- b}{a}$
Given two zeroes of a cubic polynomial $a x^{3} + b x^{2} + c x + d$

are zero, Let third zero be $y$

then sum of zeroes $= \frac{- b}{a}$

$\Rightarrow 0 + 0 + y = \frac{- b}{a}$

$\Rightarrow y = \frac{- b}{a}$

Suggest Corrections

\frac{- d}{a}

\frac{c}{a}

\frac{- b}{a}

\frac{b}{a}

α, β, γ

a x^{3} + b x^{2} + c x + d

If one of the zeros of the cubic polynomial $a x^{3} + b x^{2} + c x + d$ is 0 then the product of the other two zeros is

(a) $\frac{- c}{a}$ (b) $\frac{c}{a}$ (c) 0 (d) $\frac{- b}{a}$

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