Question Number 96904 by pticantor last updated on 05/Jun/20
Commented by pticantor last updated on 05/Jun/20
$$\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{there}}\:\boldsymbol{{are}}\:\mathrm{4}\:\boldsymbol{{real}}\:\boldsymbol{{numbers}}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\:\boldsymbol{{and}}\:\boldsymbol{{d}}\:\:\:\:\:\:\:\:\: \\ $$$$\boldsymbol{{which}}\:\boldsymbol{{will}}\:\boldsymbol{{be}}\:\boldsymbol{{specified}}\:\boldsymbol{{such}}\:\boldsymbol{{that}} \\ $$$$\left(\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}+\mathrm{5}\right)^{\mathrm{2}} +\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} =\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{ax}}+\boldsymbol{{b}}\right)\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{cx}}+\boldsymbol{{d}}\right) \\ $$
Commented by john santu last updated on 05/Jun/20
$$\Rightarrow{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{27}{x}^{\mathrm{2}} −\mathrm{38}{x}+\mathrm{26}\:=\: \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{13}\right)\: \\ $$$$\begin{cases}{{a}=−\mathrm{2},\:{b}\:=\:\mathrm{2}\:,\:{c}\:=\:−\mathrm{6}\:,\:{d}\:=\:\mathrm{13}\:{or}}\\{{a}=−\mathrm{6},\:{b}\:=\:\mathrm{13}\:,\:{c}\:=\:−\mathrm{2}\:,\:{d}\:=\:\mathrm{2}}\end{cases} \\ $$
Commented by pticantor last updated on 05/Jun/20
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}} \\ $$