Answer
$a.\quad n(n+1)$
$b.\quad $Always even.
Work Step by Step
a.
Find the GCF
$n^{2}=n\cdot n$
$n=n$
GCF=$n$.
$n^{2}+n\qquad...$factor out the GCF
$=n(n)+n(1)\qquad...$apply the Distributive Property.
$=n(n+1)$
b.
When n is even, both $n^{2}$ and $ n $ are even, so their sum is even.
When n is odd, both $n^{2}$ and $n $ are odd, so their sum is even.
Answer:Always even.
