I suppose you mean:
x^2-1 is a factor for
X^2-1 is a factor means that x^2-1 zero are the function's zeros too.
==> x1= 1 x2=-1
Now by substituting in the function:
x=1 ==> a(1)+b(1)+c(1)+d(1)+e=0
==> a+b+c+d+e =0
2) a+c+e = b+d
To prove that we need to substitute with x=-1 (the other root)
==> a+c+e = b+d
Supposing that you mean that (x^2 - 1) is a factor of ax^4+bx^3+cx^2+dx+e, that means also that the roots of the factor x^2-1 are the roots of the polynomial ax^4+bx^3+cx^2+dx+e, too.
So, let's find the roots of x^2-1=0
We notice that is a difference of squares:
a^2 - b^2 = (a-b)(a+b)
x^2-1 = (x-1)(x+1) = 0
Now we'll put each factor zero:
x-1 = 0
x+1 = 0
x = -1
So, as we've said before, the roots of x^2-1 are the roots of the given polynomial, meaning that:
f(-1) = 0, where f(x) = ax^4+bx^3+cx^2+dx+e
f(-1) = a-b+c-d+e = 0 (1)
f(1) = a+b+c+d+e = 0 (2)
From the relation (1), we'll get:
a+c+e = b+d
a+b+c+d+e = 0
To prove if x^2-1 is factor of f(x) = ax^4+bx^3+cx^2+dx+e.....(1) then
a+b+c+d = 0 and a+c+e = b+d.
If x^2-1 is a factor, then (x-1) and (x+1) are also the factors of the given expression at (1)
So for x=1 , f(1) = 0 and f(-1) = 0. This gives: For x=1,a*1^4+b*1^3 + c*1^2+d*1+e = 0. Or
a+b+c+d+e = 1
Eor x = -1, f(-1) = 0.This gives: a*(-1)^4+b*(-1)^3+c*(-1)^2+ d*(-1)+e = 0. Or a-b+c-d+e = 0. Or
1+c+e+ = b+d.