Question

If x^2-1 is a factor of ax^4+bx^3+cx^2+dx+e then find a+c+e=b+d=0

Answer 1

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

Answer 2

Since x2 - 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 , 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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