Question

If alpha and beta are the zeros of th polynomia l f(x) =x^2-p(x+1)-c, show that (alpha+beta)(beta+1)=1-c

Answer 1

I hope it is
$( \alpha + 1)( \beta + 1) = 1 - c$
for that the solution is here

${x}^{2} - p(x + 1) - c \\ {x}^{2} - px - p - c \\ {x}^{2} - px - (p + c) \\ \alpha + \beta = p \\ \alpha \beta = - (p + c) \\ ( \alpha + 1 )( \beta + 1) = \alpha \beta + \alpha + { \beta }+ 1 \\ = - (p + c) + p + 1 \\ = - p + p - c + 1 \\ = 1 - c$