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

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.