Question

$a{x}^4+b{x}^3+c{x}^2+dx+e$ is exactly divisible by $x^2-1$, when

• A. $a+b+c+d+e=0$
• B. $a+c+e=0$
• C. $b+d=0$
• D. $a+c+e=b+d=1$

Answer 1

Answer: (A), (B), (C)

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

$x^2-1=(x-1)(x+1)$

By factor theorem, $f\left(1\right)=0$ and $f(-1)=0$

$\therefore a+b+c+d+e=0$ and $a+c+e=b+d$