Question

If x raised to 2 - 1 is a factor of ax raise to 4 + bx raise to 3 + cx raise to 2 + dx + e, Prove that a + c +e = b + d = 0​

Answer 1

Answer:

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

Solution:

Given:

• x² -1 is a factor of P(x) = ax⁴ +bx³ + cx²+dx +e

if x² -1 is a factor of P(x) then P(±1) = 0

x² -1 = 0

or, x² = 1

or, x = ± 1

substituting value of x in P(X)

P(x) = ax⁴ +bx³ + cx²+dx +e

or, P(1) : a(1)⁴ + b(1)³ + c (1)²+dx + e = 0

or, P(1) : a + b +c+d+e = 0 ------- equ(1)

Simlarly,

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

or, P(-1): a -b +c-d + e = 0

or, P(-1): a + c+ e = b +d ------- equ(2)

From equ (1) & (2) , We get

a + c +e = b + d = 0