Question Number 58103 by smiak8742 last updated on 17/Apr/19
$${f}\left({x}\right)=−{x}^{\mathrm{6}} +\mathrm{3}\:{x}^{\mathrm{4}} \:+\:\mathrm{4}{x}^{\mathrm{2}} {find}\:{the}\:{zeros} \\ $$
Answered by MJS last updated on 17/Apr/19
$${x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$${x}=\sqrt{{t}} \\ $$$${t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{4}=\mathrm{0} \\ $$$${t}_{\mathrm{1}} =−\mathrm{1};\:{t}_{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\:{x}_{\mathrm{3}} =−\mathrm{i};\:{x}_{\mathrm{4}} =\mathrm{i};\:{x}_{\mathrm{5}} =−\mathrm{2};\:{x}_{\mathrm{6}} =\mathrm{2} \\ $$