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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)(x2−1)]nfor...(A)Alln(B)Oddn(C)Evenn(D)n⩾3(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)(x2−1))nwemusthavep(i)=0andp(−i)=0butp(i)=4n−(2(−2))n=4n−(−4)nbutwelookthatp(−i)=p(i)=0⇒n≡0[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)(x2−1)]n=(x4+2x2+1−2x2+2)n−[(x2+1+2)(x2−1)]n=[(x2+1)2−2(x2−1)]n−[(x2+1)(x2−1)+2(x2−1)]n=(...)+(−2)n(x2−1)n−(...)−2n(x2−1)n=(...)+[(−2)n−2n](x2−1)n(...)=termswithfactor(x2+1)weseethatthelasttermiszeroifniseven.⇒AnswerC

Commented by Joel578 last updated on 03/Mar/18

thank you very much

thankyouverymuch

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