x^2-1 is a factor of ax^4+bx^3+cx^2+dx+e, show that a+c+e=b+d=0
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answer for the given problem is given
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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
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