Chapter 3 - Quadratic Functions - 3.1 Introduction to the Family of Quadratic Functions - Exercises and Problems for Section 3.1 - Exercises and Problems - Page 121: 3

$w(n)=3 n^2+6 n+0$ $a=3, b=6, c=0$

We have $$ \begin{aligned} w(n) & =n(n-3)(n-2)-n^2(n-8) \\ & =n\left(n^2-5 n+6\right)-\left(n^3-8 n^2\right) \\ & =n^3-5 n^2+6 n-n^3+8 n^2 \\ & =3 n^2+6 n+0, \end{aligned} $$ so $a=3, b=6, c=0$