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Question Number 31125 by Joel578 last updated on 02/Mar/18

Let n be a positive integer. Then x^2  + 1   is a factor of (x^4  + 3)^n  − [(x^2  + 3)(x^2  − 1)]^n   for ...  (A) All n  (B) Odd n  (C) Even n  (D) n ≥ 3  (E) None of these options

Letnbeapositiveinteger.Thenx2+1isafactorof(x4+3)n[(x2+3)(x21)]nfor...(A)Alln(B)Oddn(C)Evenn(D)n3(E)Noneoftheseoptions

Commented by prof Abdo imad last updated on 04/Mar/18

if x^2  +1 divide p(x)=(x^4  +3)^(n ) ((x^2  +3)(x^2 −1))^n   we must have p(i)=0 and p(−i)=0 but  p(i)=4^n  −(2(−2))^n =4^n  −(−4)^n  but we look  that p(−i)=p(i)=0 ⇒n≡0[2] ⇒n even.

ifx2+1dividep(x)=(x4+3)n((x2+3)(x21))nwemusthavep(i)=0andp(i)=0butp(i)=4n(2(2))n=4n(4)nbutwelookthatp(i)=p(i)=0n0[2]neven.

Answered by supungamage001 last updated on 03/Mar/18

(c)even

(c)even

Answered by mrW2 last updated on 03/Mar/18

(x^4  + 3)^n  − [(x^2  + 3)(x^2  − 1)]^n   =(x^4 +2x^2  +1−2x^2  +2)^n  − [(x^2 +1 +2)(x^2  − 1)]^n   =[(x^2  +1)^2 −2(x^2  −1)]^n  −[(x^2 +1)(x^2 −1)+2(x^2  −1)]^n   =(...)+(−2)^n (x^2 −1)^n −(...)−2^n (x^2 −1)^n   =(...)+[(−2)^n −2^n ](x^2 −1)^n   (...)=terms with factor (x^2 +1)  we see that the last term is zero if n is even.  ⇒Answer C

(x4+3)n[(x2+3)(x21)]n=(x4+2x2+12x2+2)n[(x2+1+2)(x21)]n=[(x2+1)22(x21)]n[(x2+1)(x21)+2(x21)]n=(...)+(2)n(x21)n(...)2n(x21)n=(...)+[(2)n2n](x21)n(...)=termswithfactor(x2+1)weseethatthelasttermiszeroifniseven.AnswerC

Commented by Joel578 last updated on 03/Mar/18

thank you very much

thankyouverymuch

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