If α and β are the zeros of the quadratic polynomial f(x)

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-p(x+1)-c$

Then

$x^{2}-p(x+1)-c$

$x^{2}-p x-p-c$

$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-(-p)}{1}$

= p

$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{-p-c}{1}$

$=-p-c$

We have to prove that $(\alpha+1)(\beta+1)=1-c$

$(\alpha+1)(\beta+1)=1-c$

$(\alpha+1) \beta+(\alpha+1)(1)=1-c$

$\alpha \beta+\beta+\alpha+1=1-c$

$\alpha \beta+(\alpha+\beta)+1=1-c$'

Substituting $\alpha+\beta=p$ and $\alpha \beta=-p-c$ we get,

$-p-c+p+1=1-c$

$1-c=1-c$

Hence, it is shown that $(\alpha+1)(\beta+1)=1-c$.