How to prove that $4n^2+4n+8$ is even?

Question

I'm trying to prove that $4n^2+4n+8$ is even. I tried dividing the polynomial by $2n$ to get a remainder of $8$. Is this correct? how do I proceed ?

Answer 1

An even number looks like $2n$ for some $n \in \mathbb{N}$. Can't you see a way to do that?

$$4n^2 + 4n + 8 = 2(2n^2 + 2n + 2*2)$$ You wrote the polynomial as $2$ times a natural number. So, it is even.

Answer 2

Any number is said to even when it is divisible by $2$

Given that $$4n^2+4n+8=2(2n^2+2n+4)=2\lambda $$ where, $\lambda$ is some integer

Above, number $2\lambda$ is even i.e. $4n^2+4n+8$ is even number.