Question Number 176846 by Ar Brandon last updated on 27/Sep/22
$${m}\in\mathbb{Z} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{leading}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\mathrm{P}\left({x}\right)=\mathrm{4}{x}^{\frac{\mathrm{13}}{{m}−\mathrm{5}}} −\mathrm{6}{x}^{\mathrm{25}−\mathrm{2}{m}} +\mathrm{4}{x}^{\mathrm{13}} +\mathrm{5}{x}^{\mathrm{10}} −\mathrm{4}\:? \\ $$
Answered by mr W last updated on 27/Sep/22
$$\frac{\mathrm{13}}{{m}−\mathrm{5}}=+{integer}\:\Rightarrow{m}=\mathrm{18}\:{or}\:\mathrm{6} \\ $$$$\mathrm{25}−\mathrm{2}{m}=+{integer}\:\Rightarrow{m}=\mathrm{6} \\ $$$$\mathrm{P}\left({x}\right)=\mathrm{4}{x}^{\mathrm{13}} −\mathrm{6}{x}^{\mathrm{13}} +\mathrm{4}{x}^{\mathrm{13}} +\mathrm{5}{x}^{\mathrm{10}} −\mathrm{4} \\ $$$$\mathrm{P}\left({x}\right)=\mathrm{2}{x}^{\mathrm{13}} +\mathrm{5}{x}^{\mathrm{10}} −\mathrm{4} \\ $$$$\Rightarrow{leading}\:{coeff}.\:{is}\:\mathrm{2}. \\ $$
Commented by Ar Brandon last updated on 28/Sep/22
Thanks Sir