Question:
If two of the zeros of the cubic polynomial $a x^{3}+b x^{2}+c x+d$ is 0 , then the third zeros is
(a) $\frac{-b}{a}$
(b) $\frac{b}{a}$
(C) $\frac{c}{a}$
(d) $\frac{-d}{a}$
Solution:
(a) $\frac{-b}{a}$
Let $\alpha, 0$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d=0$.
Then sum of the zeroes $=\frac{-b}{a}$
$=>\alpha+0+0=\frac{-b}{a}$
$\Rightarrow \alpha=\frac{-b}{a}$
Hence, the third zero is $\frac{-b}{a}$.
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