If(X2-1) is a factor of ax4+bx3+cx2+dx+e,show that a+c+e=b+d=0

Given : x 2 – 1 is a factor of f(x) = ax 4 + bx 3 + cx 2 + dx + e

Now (x 2 – 1) can be written as (x + 1) (x – 1)

⇒ (x – 1) and (x + 1) are factors 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 + c + e = b + d ..... (1)

and f (1) = 0

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

a + b + c + d + e = 0

b + d + b + d = 0 ( from (1) )

⇒ 2 (b + d) = 0

b + d = 0 ..... (2)

from (1) and (2) we get

a + c + e = b + d = 0

  • 129
What are you looking for?