Question:
If one of the zeros of the cubic polynomial $a x^{3}+b \times 2+c x+d$ is 0 , then the product of the other two zeros is
(a) $\frac{-c}{a}$
(b) $\frac{c}{a}$
(c) 0
(d) $\frac{-b}{a}$
Solution:
(b) $\frac{c}{a}$
Let $\alpha, \beta$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d$.
Then, sum of the products of zeroes taking two at at a time is given by
$(\alpha \beta+\beta \times 0+\alpha \times 0)=\frac{c}{a}$
$=>\alpha \beta=\frac{c}{a}$
$\therefore$ The product of the other two zeroes is $\frac{c}{a}$.
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