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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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