Question

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

Answer 1

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

Then,

$(x^2-1)$ is a factor of $p\left(x\right)$

$(x-1)(x+1)$ is a factor of $p\left(x\right)$

$(x-1)$ and $(x+1)$ are factors of $p\left(x\right)$

$p\left(1\right)=0$ and $p(-1)=0$

$a+b+c+d+e=0$ and $a-b+c-d+e=0$

Adding and subtracting these two equations, we get,

$2(a+c+e)=0$ and $2(b+d)=0$

$⟹a+c+e=0$ and $b+d=0⟹a+c+e=b+d=0$