Simplifying to a desired expression structure My book has this expression:
\begin{align}
((n(n+1)(2n+7))/6)+(n+1)(n+3)
\end{align}
And then the book simplified it, and ended up with the desired expression:
\begin{align}
((n+1)(n+2)(2n+9))/6
\end{align}
I tried to do such simplification. But I ended up with this:
\begin{align}
(2n^3+15n^2+31n+18)/6
\end{align}
Even though the "value" in my expression is the same as the desired result's, I need the structure of my expression to be the same as well. But I don't understand how to do such.
Here are the full steps my book shows. Of course, I understand that the steps make sense, but I don't understand why did my book take that approach? Is that the only approach possible to reach the desired expression? How was I supposed to know that? I mean, I know how to simplify, but clearly my book is using a different "style" or "path"
\begin{align}
((n(n+1)(2n+7))/6)+(n+1)(n+3)\\
(n(n+1)(2n+7)+6(n+1)(n+3))/6\\
((n+1)[n(2n+7)+6(n+3)])/6\\
((n+1)(2n^2 +7n + 6n + 18))/6\\
((n+1)(n+2)(2n+9))/6
\end{align}
So yes, how should I simplify to get a desired expression structure?
 A: Below I explain in detail the solution given in your book.
$\begin{eqnarray}
 &&\rm\ \ \,n(n\!+\!1)(2n\!+\!7)/\color{#C00}6+(n\!+\!1)(n\!+\!3)\\
\rm put\ all\ over\ a\ common\ denominator = \color{#C00}6:\quad &= &\rm\ (n(\color{#0A0}{n\!+\!1})(2n\!+\!7)\,+\,\color{#C00}6\,(\color{#0A0}{n\!+\!1})(n\!+\!3))/\color{#C00}6\\
\rm pull\ out\ the\ common\ factor\ \color{#0A0}{n\!+\!1}:\quad &=&\rm\ (\color{#0A0}{n\!+\!1})\,(n(2n\!+\!7)+6(n\!+\!3))/6\\
\rm apply\ the\ distributive\ law:\quad &=&\rm\ (n\!+\!1)\,(2n^2 + 13n + 18)/6\\
\rm factor\ the\ quadratic,\ see\ below:\quad &=&\rm\ (n\!+\!1)\,(n\!+\!2)\,(2n\!+\!9)/6
\end{eqnarray}$
In order to factor the quadratic $\rm\:f = 2n^2+13n+18\:$ one can apply the AC-method as follows:
$$\begin{eqnarray}\rm 2f\, &=&\rm\ \ 4n^2\ +\ 13\cdot 2n\, +\, 2\cdot 18 \\
&=&\rm\  (2n)^2 + 13\,(2n) + 2\cdot 18 \\
&=&\rm\ \  N^2\ +\ 13\, N\ +\ 36\quad for\quad N = 2n \\
&=&\rm\ (N\ +\ 4)\,(N\ +\ 9) \\
&=&\rm\ (2n\, +\, 4)\,(2n\, +\, 9) \\
\rm f\, &=&\rm\ (\ n\ +\ 2)\,(2n\ +\ 9) \end{eqnarray}$$
A: Your expression is 
$$\frac{2n^3+15n^2+31n+18}{6}.$$
Let's forget about the $6$. Also, I would like to change the $n$ to $x$, for no good reason except familiarity.
So we want to factor
$$2x^3+15x^2+31x+18.$$
To do this, we look for rational roots of the polynomial $2x^3+15x^2+31x+18$. By the Rational root Theorem, such roots must have shape $a/b$, where $a$ divides $18$ and $b$ divides $2$. (There may be no rational roots, though in this case there are three.)
It is easy to spot the root $x=-1$. So $x-(-1)$, that is, $x+1$, divides our polynomial. Do the division, using the ordinary division process for polynomials, which is much like "long division." We get
$$2x^3+15x^2+31x+18=(x+1)(2x^2+13x+18).$$
Now factor the quadratic as one did in school. Or else note that $x=-2$ is a root of $2x^2+13x+18$.
Remark: The book grabbed the obvious common factor of $n+1$. The rest turned out to be a quadratic that factors nicely. You multiplied out, meaning you buried the $n+1$ term. It can be extracted from your expression, and the above calculation shows how, but why bury and then extract?  A factored or partly factored expression is often more useful. Anyway, "taking out" common factors usually simplifies calculations. 
