Why does $n(n+1)(2n+1)+6(n+1)^2=(n+1)(n(2n+1)+6(n+1))$?

Question

I'm not sure how my professor got from: $$n(n+1)(2n+1)+6(n+1)^2 ~\text{to}~ (n+1)(n(2n+1)+6(n+1))$$

I have a feeling it's an algebra thing but I just can't seem to get it. Would really appreciate the help.

Answer 1

It is an algebra thing.

$\color{blue}{n}\color{green}{(n+1)}\color{blue}{(2n+1)}$ and $\color{red}6\color{green}{(n+1)}^{\color{red}2}$ both have $\color{green}{(n+1)}$ as a common factor. So the common factor can be factored out:

$\color{blue}{n}\color{green}{(n+1)}\color{blue}{(2n+1)} + \color{red}6\color{green}{(n+1)}^{\color{red}2} = \color{green}{(n+1)}[\color{blue}{n}\color{blue}{(2n+1)} + \color{red}6\color{green}{(n+1)}^{\color{red}1}]$

Answer 2

You can expand your equation to :

$$n(n+1)(2n+1)+6(n+1)^2 = n(n+1)(2n+1)+6(n+1)(n+1)$$

On the right hand side you will notice that $n+1$ occurs in each group of terms and this can be factored out as follows:

$$n(n+1)(2n+1)+6(n+1)(n+1) = (n+1)\cdot\left[n(2n+1)+6(n+1)\right]$$

Answer 3

One way is to start with $$n(n+1)(2n+1)+6(n+1)^2$$

and factor $(n+1)$ to get $$ (n+1)(n(2n+1)+6(n+1))$$

The other way which is lengthy is to multiply both sides and see if you get the same expressions. ( not recommended)