Question

State True or False.

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

• A. True
• B. False

Answer 1

The correct option is A True

Let $p\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$
Given, $x^2-1$ is factor of $p\left(x\right).$
$i.e(x-1)(x+1)$ is factor of $p\left(x\right)$
$\Rightarrow (x-1),(x+1)$ are factor of $p\left(x\right)$
$\therefore p\left(1\right)=0andp(-1)=0$
$\Rightarrow p\left(1\right)=a+b+c+d+e=0$ ...(i)
$\Rightarrow p(-1)=a-b+c-d+e=0$ ...(ii)
From (ii), we get
$a+c+e=b+d$
Putting this value in (i),
$\Rightarrow (a+c+e)+b+d=0$
$\Rightarrow (b+d)+b+d=0$
$\Rightarrow 2(b+d)=0$
$\Rightarrow b+d=0$
$\therefore a+c+e=b+d=0$