If (x² – 1) is a factor of ax⁴ + bx³ + cx² + dx + e, show that a + c + e = b + d = 0

Let f(x) = ax⁴ + bx³ + cx² + dx + e

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

Also we can write, (x² – 1) = (x + 1)(x – 1)

Since (x² – 1) is a factor of f(x), this means (x + 1) and (x – 1) both are factors of f(x).

So, if (x – 1) is a factor of f(x)

⇒ f(1) = 0

⇒ a(1)⁴ + b(1)³ + c(1)² + d(1) + e = 0

⇒ a + b + c + d + e = 0 ----- (A)

Also as (x + 1) is also a factor,

⇒ f(–1) = 0

⇒ a(–1)⁴ + b(–1)³ + c(–1)² + d(–1) + e = 0

⇒ a – b + c – d + e = 0

⇒ a + c + e = b + d ---- (B)

On solving equations (A) and (B), we get,

a + c + e = b + d = 0