# 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

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