Question

If $x^2-1$ is a factor of $a{x}^4+b{x}^3+c{x}^2+dx+e$, then prove that
(a)$a+b+c+d+e=0$
(b)$a+c+e=b+d$

Answer 1

Given : $x^2-1$ is a factor of $a{x}^4+b{x}^3+c{x}^2+dx+e$

$\Rightarrow (x-1)(x+1)$ divides $a{x}^4+b{x}^3+c{x}^2+dx+e$

$\Rightarrow$ When $x=1$, then $a+b+c+d+e=0$ ----- $\left(i\right)$

When $x=-1$, then $a-b+c-d+e=0$ ----- $\left(ii\right)$

$\left(ii\right)\Rightarrow a+c+e=b+d$