Put in i + 2 for 'x'
(i+2)(i+2-1)^2
(i+2)(i+1)^2
(i+2) (i^2 + 2i + 1)
(i+2)(-1+2i+1)
(i+2)(2i)
2i^2 + 4i = 4i -2
Expand the following: f(i+2) = (i-1+2)^2 (i+2) Add like terms. 2-1 = 1: f(i+2) = (i+2) (i+1)^2 (i+1) (i+1) = (i) (i) + (i) (1) + (1) (i) + (1) (1) = i^2+i+i+1 = i^2+2 i+1: f(i+2) = i^2+2 i+1 (i+2) | | | | i | + | 2 | | i^2 | + | 2 i | + | 1 | | | | i | + | 2 | | 2 i^2 | + | 4 i | + | 0 i^3 | + | 2 i^2 | + | 0 i | + | 0 i^3 | + | 4 i^2 | + | 5 i | + | 2: Answer: | | f(i+2) = i^3+4i^2+5i+2
