Question Number 175976 by rexford last updated on 10/Sep/22
Commented by Rasheed.Sindhi last updated on 10/Sep/22
$${Not}\:{clear}\:{enough}. \\ $$
Answered by floor(10²Eta[1]) last updated on 10/Sep/22
$$\mathrm{x}^{\mathrm{4}} =\mathrm{x}^{\mathrm{4}} \mathrm{k}_{\mathrm{4}} +\left(\mathrm{4k}_{\mathrm{4}} +\mathrm{k}_{\mathrm{3}} \right)\mathrm{x}^{\mathrm{3}} +\left(\mathrm{6k}_{\mathrm{4}} +\mathrm{3k}_{\mathrm{3}} +\mathrm{k}_{\mathrm{2}} \right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{4k}_{\mathrm{4}} +\mathrm{3k}_{\mathrm{3}} +\mathrm{2k}_{\mathrm{2}} +\mathrm{k}_{\mathrm{1}} \right)\mathrm{x}+\mathrm{k}_{\mathrm{4}} +\mathrm{k}_{\mathrm{3}} +\mathrm{k}_{\mathrm{2}} +\mathrm{k}_{\mathrm{1}} +\mathrm{k}_{\mathrm{0}} \\ $$$$\Rightarrow\mathrm{k}_{\mathrm{4}} =\mathrm{1},\:\mathrm{4k}_{\mathrm{4}} +\mathrm{k}_{\mathrm{3}} =\mathrm{0}\Rightarrow\mathrm{k}_{\mathrm{3}} =−\mathrm{4} \\ $$$$ \\ $$
Commented by rexford last updated on 11/Sep/22
$${i}\:{get}\:{it} \\ $$$${thank}\:{you}\:{very}\:{much}\:{for}\:{your}\:{time} \\ $$