Answer
$n^3-5n^2+10n-12$.
Work Step by Step
The given expression is
$=(n-3)(n^2-2n+4)$
Distribute $(n^2-2n+4)$ to each term of $(n-3)$.
$=(n)(n^2-2n+4)-3(n^2-2n+4)$
Use distributive property.
$=n(n^2)+n(-2n)+n(4)-3(n^2)-3(-2n)-3(4)$
Multiply.
$=n^3-2n^2+4n-3n^2+6n-12$
Combine like terms.
$=n^3-5n^2+10n-12$
Hence, the product is $n^3-5n^2+10n-12$.
