Question:
If $\alpha, \beta$ are the zeros of polynomial $f(x)=x^{2}-p(x+1)-c$, then $(\alpha+1)(\beta+1)=$
(a) $c-1$
(b) $1-c$
(c) $C$
(d) $1+c$
Solution:
Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial
$f(x)=x^{2}-p(x+1)-c$
$=x^{2}-p x-p-c$
$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$
$=-\left(\frac{-p}{1}\right)$
$=p$
$\alpha \times \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$
$=\frac{-p-c}{1}$
$=-p-c$
We have
$(\alpha+1)(\beta+1)$
$=\alpha \beta+\beta+\alpha+1$
$=\alpha \beta+(\alpha+\beta)+1$
$=-p-c+(p)+1$
$=-c+1$
$=1-c$
The value of $(\alpha+1)(\beta+1)$ is $1-c$
Hence, the correct choice is $(b)$