\(x^2-1=x^2-1^2=(x-1)(x+1)\)
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.
\(p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\
p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\
p(1)=a+b+c+d+e \\
p(-1)=a-b+c-d+e \\ \\
p(1)=0 \\
p(-1)=0 \\ \\ \hbox{add both equations:} \\
a+b+c+d+e=0 \\
\underline{a-b+c-d+e=0} \\
2a+2c+2e=0 \\
2(a+c+e)=0 \\
a+c+e=0 \\ \\
\hbox{substitute 0 for a+c+e in the first equation:} \\
a+b+c+d+e=0 \\
(a+c+e)+b+d=0 \\
0+b+d=0 \\
b+d=0 \\ \\
\boxed{a+c+e=b+d=0} \\
\hbox{proved } \checkmark\)
