Question

answer the following question by step by step​

Answer 1

Step-by-step explanation:

Given p(x)=x²-p(x+1)-c

=>x²-px-p-c

=>x²-px-(p+c)

hence on comparing it with ax²+bx+c

we get

a=1

b=-p

c=-(p+c)

we know that

@+B=-b/a

=>-(-p)/1

=>p

and

@B=c/a

=>-(p+c)/1

=>-(p+c)

now it's given that

(@+1)(B+1)=0

=>@B+1+@+B=0

=>@B+@+B=-1

=>-(p+c)+p=-1

=>-c=-1

=>c=1 (Answer)