HCF

Asked by  | 14th Apr, 2008, 09:45: AM

Expert Answer:

x32-1 = (x16)2 - (1)2 = (x16 - 1)(x16 + 1) = [(x8)2 - (1)2](x16 + 1) = (x8 - 1)(x8 + 1)(x16 + 1)

                                   = [(x4)2 - (1)2](x8 + 1)(x16 + 1) = (x4 - 1)(x4 + 1)(x8 + 1)(x16 + 1)

                                  = [(x2)2 - (1)2(x4 + 1)(x8 + 1)(x16 + 1)  = (x2 - 1)(x2 + 1)(x4 + 1)(x8 + 1)(x16 + 1)

                                  = (x-1)(x +1)(x2 + 1)(x4 + 1)(x8 + 1)(x16 + 1)                        ...(i)

And

X22-1 = (x11)2 - 12 = (x11 -1)(x11 +1)                                                                     ...(ii)

If we put x = 1 in (x11 -1) of  X22-1 then it wil becomes zero

 x - 1 will be the factor of ( X22-1).

  from (i) and (ii)

 HCF of (x32-1) and (X22-1)  is (x - 1)

Answered by  | 17th Apr, 2008, 04:39: PM

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