Question

If $(x+1)$ and $(x-1)$ are the factors of $x^4+a{x}^3-3x^2+2x+b$ then find the values of a and b.

Answer 1

Let $p\left(x\right)=x^4+a{x}^3-3x^2+2x+b$
$\because (x+1)$ is a factor of p(x)
$\therefore p(-1)=0$
$\Rightarrow (-1{)}^4+a(-1{)}^3-3(-1{)}^2+2(-1)+b=0$
$\Rightarrow 1-a-3-2+b=0$
$\Rightarrow -a+b=4$ .........(i)
Also $(x-1)$ is a factor of p(x)
$\therefore p\left(1\right)=0$
$\Rightarrow (1{)}^4+a(1{)}^3-3(1{)}^2+2(1)+b=0$
$\Rightarrow 1+a-3+2+6=0$
$\Rightarrow a+b=0$ ............(ii)
Solving (i) and (ii), we get
$2b=4$

$\Rightarrow b=2$
Substituting $b=2$ in (ii), we get
$a+2=0\Rightarrow a=-2$
Hence, $a=-2,b=2$