If the zeros of the cubic polynomial p(x) = a | vedclass

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(B) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d,$ the relationships between zeros and coefficients are:$1$. Sum of zeros: $\alpha + \beta + \g

Vedclass · Accepted Answer

$-\frac{c}{d}$

Vedclass · Answer

(B) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d,$ the relationships between zeros and coefficients are:$1$. Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$$2$. Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$$3$. Product of zeros: $\alpha\beta\gamma = -\frac{d}{a}$Now,we need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}.$Taking the least common multiple $(LCM)$ of the denominators:$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}$Substituting the values from the relationships:$= \frac{c/a}{-d/a} = \frac{c}{a} 	imes \left(-\frac{a}{d}ight) = -\frac{c}{d}.$

If the zeros of the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ $(a \neq 0)$ are $\alpha, \beta,$ and $\gamma,$ then $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \dots$

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