Question:
If two of the zeros of the cubic polynomial $a x^{3}+b x^{2}+c x+d$ are each equal to zero, then the third zero is
(a) $\frac{-d}{a}$
(b) $\frac{c}{a}$
(c) $\frac{-b}{a}$
(d) $\frac{b}{a}$
Solution:
Let $\alpha=0, \beta=0$ and $\gamma$ be the zeros of the polynomial
$f(x)=a x^{3}+b x^{2}+c x+d$
Therefore
$\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$
$=-\left(\frac{b}{a}\right)$
$\alpha+\beta+\gamma=-\frac{b}{a}$
$0+0+\gamma=-\frac{b}{a}$
$\gamma=-\frac{b}{a}$
The value of $\gamma=-\frac{b}{a}$
Hence, the correct choice is $(c)$
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