Question:
If α, β, γ are the zeros of the polynomial x3 − 6x2 − x + 30, then the value of (αβ + βγ + γα) is
(a) −1
(b) 1
(c) −5
(d) 30
Solution:
(a) −1
Here, $p(x)=x^{3}-6 x^{2}-x+3$
Comparing the given polynomial with $x^{3}-(\alpha+\beta+\gamma) x^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma$, we get:
$(\alpha \beta+\beta \gamma+\gamma \alpha)=-1$