Question

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

Answer 1

since that is a factor....
so put X =1,-1
u get
a+b+c+d+e=0 // at x=1
or
a-b+c-d+e=0 //at x=-1
so a+c+e=b+d

Answer 2

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