Factorization Of Polynomials — Maths STD 9 — Question

If $x^2-1$ is factor of $p(x)=a x^4+b x^3+c x^2+d x+e$.
Then $(x-1)$ and $(x+1)$ will also be factors of $p(x)$.
Because $x^2-1=(x-1)(x+1)$
Then, at $x=1$ and $x=-1, p(x)=0$
$\Rightarrow p(1)=0 \text { and } p(-1)=0$
$\Rightarrow a+b+c+d+e=0 \ldots \text { (1) }$
And
$\Rightarrow a-b+c-d+e=0 \ldots \text { (2) }$
Adding equations $(1)$ and $(2)$.
$2 a+2 c+2 e=0$
$\Rightarrow a+c+e=0 \ldots \text { (3) }$
Substracting equation $(2)$ from $(1)$
$2 b+2 d=0$
$\Rightarrow b+d=0 \ldots \text { (4) }$
From equations $(3)$ and $(4)$, we get
$a+c+e=b+d$