Question

If $x^2-1$ is a factor of $x^4+a{x}^3+3x-b$, then

• A. $a=3,b=-1$
• B. $a=-3,b=1$
• C. $a=3,b=1$
• D. none of these

Answer 1

Answer: (B)

$a=-3,b=1$

Let $f\left(x\right)=x^4+a{x}^3+3x-b$
As $x^2-1=(x-1)(x+1)$ are factors
Then
$f\left(1\right)={\left(1\right)}^4+a{\left(1\right)}^3+3\left(1\right)-b=0$
$\Rightarrow 1+a+3-b=0\Rightarrow a-b=-4$ ...(1)
And
$f(-1)={(-1)}^4+a{(-1)}^3+3(-1)-b=0$
$\Rightarrow 1-a-3-b=0\Rightarrow a+b=-2$ ...(2)
Solving (1) and (2)
$a=-3,b=1$
Hence, option 'B' is correct.