Question Number 7775 by sandy_suhendra last updated on 14/Sep/16
Commented by sou1618 last updated on 15/Sep/16
$${x}=\mathrm{4}−\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{18}{x}+\mathrm{23}=? \\ $$$$ \\ $$$$\left({x}−\mathrm{4}+\sqrt{\mathrm{3}}\right)\left({x}−\mathrm{4}−\sqrt{\mathrm{3}}\right)=\mathrm{0}\:\:\:\:\left(\because{x}=\mathrm{4}−\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{13}=\mathrm{0} \\ $$$$ \\ $$$$\left({x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{18}{x}+\mathrm{23}\right)\boldsymbol{\div}\left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{13}\right) \\ $$$$=\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)+\frac{\mathrm{10}}{{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{13}} \\ $$$$ \\ $$$${so} \\ $$$${x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{18}{x}+\mathrm{23}=\left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{13}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)+\mathrm{10} \\ $$$$=\mathrm{10} \\ $$$$ \\ $$
Commented by sandy_suhendra last updated on 15/Sep/16
$${Nice},\:{thank}'{s}\:{for}\:{your}\:{answer} \\ $$