Question

11. If a +c+e = 0 and b + d = 0, then find the zeroes of the polynomials
axt + bx² + cx² + dx te​

Answer 1

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

Answer 2

Answer:

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