A) \[-\frac{c}{a}\]
B) \[\frac{c}{a}\]
C) \[0\]
D) \[-\frac{b}{a}\]
Correct Answer: B
Solution :
| [b] Let \[p(x)=a{{x}^{3}}+b{{x}^{2}}+cx+d\] |
| Let \[\alpha ,\beta \] and \[\gamma \] be the zeroes of \[p(x),\] where \[\alpha =0.\]. |
| We know that sum of product of zeroes taken two at a time |
| \[=\frac{c}{a}=\alpha \beta +\beta \gamma +\gamma \alpha \] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,0\times \beta +\beta \gamma +\gamma \times 0=\frac{c}{a}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\beta \gamma =\frac{c}{a}\] |
| Hence, product of other two zeroes \[=\frac{c}{a}\]. |
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