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Question

If all the zeroes of cubic polynomial x3 + ax2 – bx + c are negative then a, b and c all have _______ sign.

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Solution

Let f(x) = x3 + ax2 – bx + c $\mathrm{Let}\mathrm{the}\mathrm{zeroes}\mathrm{of}f\left(x\right)\mathrm{be}\alpha ,\beta ,\gamma ,\mathrm{where}\mathrm{all}\mathrm{these}\mathrm{zeroes}\mathrm{are}\mathrm{negative}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}\alpha +\beta +\gamma =-a\phantom{\rule{0ex}{0ex}}\mathrm{Thus},a\mathrm{is}\mathrm{positive}\left(\because \alpha ,\beta ,\gamma \mathrm{are}\mathrm{negative}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\alpha \beta +\beta \gamma +\gamma \alpha =-b\phantom{\rule{0ex}{0ex}}\mathrm{Thus},b\mathrm{is}\mathrm{positive}\left(\because \alpha ,\beta ,\gamma \mathrm{are}\mathrm{negative}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\alpha \beta \gamma =-c\phantom{\rule{0ex}{0ex}}\mathrm{Thus},c\mathrm{is}\mathrm{positive}\left(\because \alpha ,\beta ,\gamma \mathrm{are}\mathrm{negative}\right)$ Hence, if all the zeroes of cubic polynomial x3 + ax2 – bx + c are negative then a, b and c all have positive sign.

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