Question:
If one of the zeros of the cubic polynomial $x^{3}+a x^{2}+b x+c$ is $-1$, then the product of the other two zeros is
(a) a − b − 1
(b) b − a − 1
(c) 1 − a + b
(d) 1 + a − b
Solution:
(c) $1-a+b$
Since $-1$ is a zero of $x^{3}+a x^{2}+b x+c$, we have:
$(-1)^{3}+a \times(-1)^{2}+b \times(-1)+c=0$
$=>a-b+c-1=0$
$=>c=1-a+b$
Also, product of all zeroes is given by
$\alpha \beta \times(-1)=-c$
$=>\alpha \beta=c$
$=>\alpha \beta=1-a+b$