Question Number 179327 by greougoury555 last updated on 28/Oct/22
$$\:{Find}\:{polynomial}\:{u},{v}\:\in{Q}\left[{x}\right]\:{such} \\ $$$$\:\:{that}\:\left({x}^{\mathrm{4}} −\mathrm{1}\right){u}\left({x}\right)+\left({x}^{\mathrm{7}} −\mathrm{1}\right){v}\left({x}\right)=\left({x}−\mathrm{1}\right) \\ $$
Answered by manxsol last updated on 28/Oct/22
$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right){u}\left({x}\right)+\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{6}} +{x}^{\mathrm{5}} +{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right){v}\left({x}\right)={x}−\mathrm{1} \\ $$$${then} \\ $$$${u}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)} \\ $$$$\mathrm{v}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{6}} +\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)} \\ $$