Question

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

• A. $a=3,b=1$
• B. None of these
• C. $a=3,b=–1$
• D. $a=–3,b=1$

Answer 1

The correct option is D $a=–3,b=1$

Given: $(x^2-1)\text{is a factor of}(x^4+a{x}^3+3x-b)$
$\text{Let}f\left(x\right)=x^4+a{x}^3+3x-b$
Using factor theorem, $x=\pm 1$ are the zeros of $f\left(x\right)$.
$\therefore f\left(1\right)=0\text{and}f(-1)=0$

Taking $f\left(1\right)=0$
$\Rightarrow a-b+4=0$ . . . . .(i)

And $f(–1)=0$
$\Rightarrow a+b+2=0$ . . . . .(ii)

On solving $\left(i\right)$ and $\left(ii\right)$,we get
$a=–3,b=1$