If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e, show that a + c + e = b + d = 0.
Since x2 - 1 = (x + 1)(x - 1) is a factor of 
          p(x) = ax4 + bx3 + cx2 + dx + e
... p(x) is divisible by (x + 1) and (x - 1) separately
⇒ p(1) = 0 and p(- 1) = 0
   p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0
⇒ a + b + c + d + e = 0 ……(i)
     
Similarly, p(- 1) = a(- 1)4 + b(- 1)3 + c(- 1)2+ d(- 1) + e = 0
⇒ a - b + c - d + e = 0
           ⇒ a + c + e = b + d ……(ii)
Putting the value of a + c + e in eqn (i), we get 
    a + b + c + d + e = 0
⇒ a + c + e + b + d = 0
     ⇒ b + d + b + d = 0
             ⇒ 2(b + d) = 0
                 ⇒ b + d = 0 ……(iii)
Comparing equations (ii) and (iii), we get
     a + c + e = b + d = 0
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