Question Number 11024 by ABD last updated on 07/Mar/17
$${P}\left({x}+\mathrm{1}\right)×{P}\left({x}−\mathrm{1}\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}+{a}−\mathrm{5} \\ $$$${a}=? \\ $$
Answered by ajfour last updated on 07/Mar/17
$${let}\:{P}\left({x}\right)\:=\:{bx}+{c} \\ $$$${Then}\:{P}\left({x}+\mathrm{1}\right)×{P}\left({x}−\mathrm{1}\right) \\ $$$$=\left({bx}+{b}+{c}\right)\left({bx}−{b}+{c}\right) \\ $$$$=\left({bx}+{c}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} \:\:\:=\:\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{a}−\mathrm{9} \\ $$$${so},\:{b}={c}=\pm\mathrm{2}.\:{and}\:{a}−\mathrm{9}=−{b}^{\mathrm{2}} \\ $$$${thus}\:\:{a}−\mathrm{9}\:=\:−\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{a}=\mathrm{5}\:\:. \\ $$