If `x^2-1` is `a` factor of `a x^4+b x^3+c x^2+dx+e ,` show that `a+c+e=b+d`

Explanation

Since` x^2 - 1 = (x - 1)` is a factor of `p(x) = ax^4 + bx^3 + cx^2 + 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 , 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`