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

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

Answer 2

Answer:

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