(x-(i-1))*(x-(i+1))

(x-(i-1))*(x-(i+1))

$$\small{\text{
$
(x-(i-1))*(x-(i+1)) =[(x-i)+1] [(x-i)-1]= (x-i)^2-1
$
}}\\
\small{\text{
$
=x^2-2xi+i^2-1 \quad | \quad i^2=-1$
}}\\
\small{\text{
$
=x^2-2xi-2 = (x^2-2) - 2xi $
}}$$

(x-(i-1))*(x-(i+1))

$$\\(x-(i-1))*(x-(i+1))\\
=x^2-x(i+1)-x(i-1)+(i^2-1)\\
=x^2-xi-x-xi+x+(-1-1)\\
=x^2-2xi-2\\
=x^2-2-2xi$$