Question Number 164135 by mathlove last updated on 14/Jan/22
$${p}\left({x}\right)+{xp}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${faind}\:\:\:{p}\left(\mathrm{2}\right)=? \\ $$
Answered by mr W last updated on 14/Jan/22
$${p}\left({x}\right)+{xp}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1}\:\:\:...\left({i}\right) \\ $$$${p}\left(−{x}\right)−{xp}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${xp}\left(−{x}\right)−{x}^{\mathrm{2}} {p}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right){x}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){p}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{x}\right) \\ $$$$\Rightarrow{p}\left({x}\right)=\mathrm{1}−{x} \\ $$$$\Rightarrow{p}\left(\mathrm{2}\right)=\mathrm{1}−\mathrm{2}=−\mathrm{1} \\ $$
Commented by mathlove last updated on 14/Jan/22
$${thanks}\:\:\:{thear}\:{mr}\:{W} \\ $$$$ \\ $$
Commented by mathlove last updated on 14/Jan/22
Commented by mr W last updated on 14/Jan/22
$${look}\:{above}\:{again}! \\ $$
Commented by mathlove last updated on 14/Jan/22
$${thank}\:\:\:{I}\:{know} \\ $$