Question

If $p,q$ and $r$ are the zeros of the polynomial $f\left(x\right)=a{x}^3+b{x}^2+cx+d$, find the value of $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$

• A. $\frac{-b}{a}$
• B. $\frac{c}{a}$
• C. $-\frac{c}{d}$
• D. $\frac{c}{d}$

Answer 1

The correct option is C $-\frac{c}{d}$

Given that

$p\left(x\right)=a{x}^3+b{x}^2+cx+d$ .........(1) where $a\ne 0$ is a cubic polynomial

And $p,q,r$ are the zeroes of the polynomial $p\left(x\right)$

We all know that

$p+q+r=\frac{-b}{a}=$ Sum of the roots

$pq+qr+rp=\frac{c}{a}=$ Product of roots taken two at a time

$pqr=\frac{-d}{a}=$ product of roots

In the Question

$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$

$=\frac{pq+qr+rp}{pqr}$

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

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

Option $C$ is correct