When t ^ 2 - 1 exactly divides the polynomial P(t)=a 1 t^ 4 +a 2 t^ 3 +a 3 t^ 2 +a 4 t+a 5 then prove a 1 +a 3 +a 5 =a 2 +a 4 =0
Answers
Given : t² - 1 exactly divides the polynomial P(t)=a₁t⁴ +a₂ t³ +a₃ t² +a₄ t +a₅
To Find : prove a₁ +a₃ +a₅ =a₂ +a₄ =0
Solution:
P(t)=a₁t⁴ +a₂ t³ +a₃ t² +a₄ t +a₅
t² - 1 = (t + 1)(t - 1)
Hence 1 , -1 are zeroes
P(-1) = 0
a₁(-1)⁴ +a₂(-1)³ +a₃ (-1)² +a₄(-1) +a₅
=> a₁ - a₂ +a₃ - a₄ +a₅ = 0
=> a₁ +a₃ +a₅ = a₂ +a₄
P(1) = 0
=> a₁ +a₂ +a₃ +a₄ +a₅ = 0
=> (a₁ +a₃ +a₅ )+ (a₂ +a₄ ) = 0
=> (a₂ +a₄ ) + (a₂ +a₄ ) = 0
=> 2 (a₂ +a₄ ) = 0
=> (a₂ +a₄ ) = 0
a₁ +a₃ +a₅ = a₂ +a₄ = 0
QED
Hence proved
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