Question Number 12255 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17
Answered by mrW1 last updated on 16/Apr/17
![(x+1)(x+2)(x+3)(x+4)+1 =(x+(5/2)−(3/2))(x+(5/2)−(1/2))(x+(5/2)+(1/2))(x+(5/2)+(3/2))+1 =[(x+(5/2))^2 −((3/2))^2 ][(x+(5/2))^2 −((1/2))^2 ]+1 =(x+(5/2))^4 −[((1/2))^2 +((3/2))^2 ](x+(5/2))^2 +((1/2))^2 ((3/2))^2 +1 =(x+(5/2))^4 −2×(5/4)×(x+(5/2))^2 +((5/4))^2 =[(x+(5/2))^2 −(5/4)]^2](https://www.tinkutara.com/question/Q12256.png)
$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{4}\right)+\mathrm{1} \\ $$$$=\left({x}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{1} \\ $$$$=\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \right]\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \right]+\mathrm{1} \\ $$$$=\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{4}} −\left[\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \right]\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$=\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{4}} −\mathrm{2}×\frac{\mathrm{5}}{\mathrm{4}}×\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$=\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right]^{\mathrm{2}} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17
$$=\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}\right)^{\mathrm{2}} \\ $$$${nice}.{thank}\:{you}\:{so}\:{much}. \\ $$
Answered by ajfour last updated on 17/Apr/17
$${f}\left({x}\right)=\:\left({x}+\mathrm{1}\right)\left({x}+\mathrm{4}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)+\mathrm{1} \\ $$$$=\:\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}\right)+\mathrm{1} \\ $$$${let}\:{t}\:=\:{x}^{\mathrm{2}} +\mathrm{5}{x} \\ $$$${f}\left({x}\right)\:=\:\left({t}+\mathrm{4}\right)\left({t}+\mathrm{6}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{t}^{\mathrm{2}} +\mathrm{10}{t}+\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left({t}+\mathrm{5}\right)^{\mathrm{2}} \:=\:\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}\right)^{\mathrm{2}} \:. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Apr/17
$${so}\:{beautiful}.{thanks}\:{a}\:{lot}. \\ $$