Question

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

Answer 1

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

Then $(x-1)$ and $x+1$ are also factors of $a{x}^4+b{x}^3+c{x}^2+dx+e$.

Let, $f\left(x\right)=$$a{x}^4+b{x}^3+c{x}^2+dx+e$

Since $(x-1)$ is a factor of $f\left(x\right)$.

Then $f\left(1\right)=0$. [Using Remainder theorem]

or, $a+b+c+d+e=0$........(1).

Again since $(x+1)$ is a factor of $f\left(x\right)$.

Then $f(-1)=0$. [Using Remainder theorem]

or, $a-b+c-d+e=0$........(2).

Now adding (1) and (2) we get,

$2(a+c+e)=0$

or, $a+c+e=0$

Using this from (2) we get,

$b+c=0$

So $a+c+e=b+d=0$