For `x+a` to be a factor of `p(x)`, `-a` should be a zero of `p(x)`.
(i) `p(x) = x^3+x^2+x+1`
For `x+1` to be a factor of `p(x)`, `-1` should be a zero of `p(x)`.
`:. p(-1)`should be `0.`
`p(-1) = (-1)^3+(-1)^2+(-1)+1 = -1+1-1+1 = 0`
`=>p(-1) = 0`
`:. x+1` is a factor of `p(x)`.
(ii)`p(x) = x^4+x^3+x^2+x+1`
For `x+1` to be a factor of `p(x)`, `-1` should be a zero of `p(x)`.
`:. p(-1)`should be `0.`
`p(-1) = (-1)^4+(-1)^3+(-1)^2+(-1)+1 = 1-1+1-1+1 = 1`
`=>p(-1) = 1`
As `p(-1) !=0`, so `x+1` is not a factor of `p(x)`.