## Algebra and Trigonometry 10th Edition

The property was proved for $n=1$ The property is correct if $n$ is changed by $n+1$
Let's prove the property for $n=1$: $n^3+3n^2+2n=1^3+3(1)^2+2(1)=1+3+2=6=2(3)$ 3 is a factor. It is correct! Suppose that the property is correct, that is: $n^3+3n^2+2n$ has 3 as a factor for all integers $n\geq1$ Now, let's prove the property for $n+1$: $(n+1)^3+3(n+1)^2+2(n+1)=n^3+3n^2+3n+1+3n^2+6n+3+2n+2=(n^3+3n^2+2n)+(3n^2+3n+1+6n+3+2)=(n^3+3n^2+2n)+3(n^2+3n+2)$ It is a sum of two terms that have 3 as a factor.