Given that one of the zeroes of the cubic polynomial $a x^{3} + b x^{2} + c x + d$ is zero, the product of the other two zeroes is

- \frac{c}{a}

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\frac{c}{a}

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- \frac{b}{a}

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Solution

The correct option is B $\frac{c}{a}$
For the given equation, $a x^{3} + b x^{2} + c x + d$ , 0 is the root of this equation implies d=0
Hence, the equation is $x (a x^{2} + b x + c)$
The other two roots will be from $a x^{2} + b x + c$
Product of the other roots = $\frac{c}{a}$

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Similar questions

Given that one of the zeroes of the cubic polynomial $a x^{3} + b x^{2} + c x + d$ is zero, the product of the other two zeroes is

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

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

\frac{- c}{a}

\frac{c}{a}

\frac{- b}{a}

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

\frac{- c}{a}

\frac{c}{a}

\frac{- b}{a}

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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