Question

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

Answer 1

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