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Question Number 64508 by lalitchand last updated on 18/Jul/19
factorize (x+1)(x+3)(x+5)(x+7)+16
$$\mathrm{factorize}\:\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}+\mathrm{7}\right)+\mathrm{16} \\ $$
Commented by Prithwish sen last updated on 18/Jul/19
(x+1)(x+7)(x+3)(x+5)+16  =(x^2 +8x+7)(x^2 +8x+15)+16  = (a+7)(a+15)+16 putting x^2 +8x = a  = a^2 +22a+121  =(a+11)^2   =(x^2 +8x+11)(x^2 +8x+11)
$$\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{7}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)+\mathrm{16} \\ $$$$=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{7}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{15}\right)+\mathrm{16} \\ $$$$=\:\left(\mathrm{a}+\mathrm{7}\right)\left(\mathrm{a}+\mathrm{15}\right)+\mathrm{16}\:\mathrm{putting}\:\mathrm{x}^{\mathrm{2}} +\mathrm{8x}\:=\:\mathrm{a} \\ $$$$=\:\mathrm{a}^{\mathrm{2}} +\mathrm{22a}+\mathrm{121} \\ $$$$=\left(\mathrm{a}+\mathrm{11}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{11}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{11}\right) \\ $$
Answered by behi83417@gmail.com last updated on 18/Jul/19
(x+1)(x+7)=x^2 +8x+7  (x+3)(x+5)=x^2 +8x+15  x^2 +8x=p⇒(p+7)(p+15)+16=  =p^2 +22p+121=(p+11)^2   ⇒(x+1)(x+3)(x+5)(x+7)+16=         =(x^2 +8x+11)^2    .■
$$\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{7}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{7} \\ $$$$\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{15} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{8x}=\mathrm{p}\Rightarrow\left(\mathrm{p}+\mathrm{7}\right)\left(\mathrm{p}+\mathrm{15}\right)+\mathrm{16}= \\ $$$$=\mathrm{p}^{\mathrm{2}} +\mathrm{22p}+\mathrm{121}=\left(\mathrm{p}+\mathrm{11}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}+\mathrm{7}\right)+\mathrm{16}= \\ $$$$\:\:\:\:\:\:\:=\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{x}}+\mathrm{11}\right)^{\mathrm{2}} \:\:\:.\blacksquare \\ $$
Answered by MJS last updated on 18/Jul/19
put x=t−4  (t−3)(t−1)(t+1)(t+3)+16  t^4 −10t^2 +25  (t^2 −5)^2   put t=x+4  ((x+4)^2 −5)^2   (x+4−(√5))^2 (x+4+(√5))^2   or  (x^2 +8x+11)^2
$$\mathrm{put}\:{x}={t}−\mathrm{4} \\ $$$$\left({t}−\mathrm{3}\right)\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)\left({t}+\mathrm{3}\right)+\mathrm{16} \\ $$$${t}^{\mathrm{4}} −\mathrm{10}{t}^{\mathrm{2}} +\mathrm{25} \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{put}\:{t}={x}+\mathrm{4} \\ $$$$\left(\left({x}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{4}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \left({x}+\mathrm{4}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\mathrm{or} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{11}\right)^{\mathrm{2}} \\ $$
Commented by Prithwish sen last updated on 18/Jul/19
great idea sir.
$$\mathrm{great}\:\mathrm{idea}\:\mathrm{sir}. \\ $$

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