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(B) Let the zeroes of the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ be $\alpha, \beta,$ and $\gamma$.Given that two zeroes are $0$,let $\alpha =

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$\frac{-b}{a}$

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(B) Let the zeroes of the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ be $\alpha, \beta,$ and $\gamma$.Given that two zeroes are $0$,let $\alpha = 0$ and $\beta = 0$.According to the relationship between zeroes and coefficients of a cubic polynomial,the sum of the product of zeroes taken two at a time is given by $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.Substituting the values: $(0)(0) + (0)(\gamma) + (\gamma)(0) = 0 = \frac{c}{a}$. This implies $c = 0$.The product of the zeroes is given by $\alpha\beta\gamma = -\frac{d}{a}$.Substituting the known values: $(0)(0)(\gamma) = 0 = -\frac{d}{a}$. This implies $d = 0$.The sum of the zeroes is given by $\alpha + \beta + \gamma = -\frac{b}{a}$.Substituting $\alpha = 0$ and $\beta = 0$: $0 + 0 + \gamma = -\frac{b}{a}$.Therefore,the third zero $\gamma = -\frac{b}{a}$.

Given that two of the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$ are $0$,the third zero is

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