Question

If $(x+p-1)$ is a factor of the expression $x^2+px+1-p$, then the roots of the equation $x^2+px+1=p$ are

• A. $0,1$
• B. $-1,1$
• C. $0,-1$
• D. $-1,2$

Answer 1

The correct option is C $0,-1$

Let $x=1-p$
Hence, $f\left(x\right)=x^2+px+1-p$
Now, $f(1-p)=0$ ... since it is a factor.
$\Rightarrow (1-p{)}^2+p(1-p)+1-p=0$
$\Rightarrow (1-p)[1-p+p+1]=0$
$\Rightarrow 2(1-p)=0$
$\Rightarrow p=1$
Now the second expression becomes
$x^2+x+1=1$
$\Rightarrow x^2+x=0$
$\Rightarrow x(x+1)=0$
$\Rightarrow x=0$ and $x=-1$
Hence, the required roots are $0,-1$.