# If α, β, γ are the zeros of the polynomial f(x) = ax3 + bx2 + cx + d,

Question:

If $\alpha, \beta, y$ are the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$, then $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=$

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

(b) $\frac{c}{d}$

(c) $-\frac{c}{d}$

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

Solution:

We have to find the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$

Given $\alpha, \beta, \gamma$ be the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$

We know that

$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{3}}$

$=\frac{c}{a}$

$\alpha \beta \gamma=\frac{-\text { Constant term }}{\text { Coefficient of } x^{3}}$

$=\frac{-d}{a}$

So

$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}$

$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\frac{c}{a}}{-\frac{d}{a}}$

$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{c}{a} \times\left(-\frac{a}{d}\right)$

$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=-\frac{c}{d}$

Hence, the correct choice is (c)