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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?

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Ah! I... didn't see that. Somehow. I'm eager to see how to extract such though. – Zol Tun Kul Jul 3 '12 at 1:49
I wrote out a little something about extracting. – André Nicolas Jul 3 '12 at 1:59
up vote 3 down vote accepted

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:

$$\begin{eqnarray}\rm 2f\, &=&\rm\ \ 4n^2\ +\ 13\cdot 2n\, +\, 2\cdot 18 \\ &=&\rm\ (2n)^2 + 13\,(2n) + 2\cdot 18 \\ &=&\rm\ \ N^2\ +\ 13\, N\ +\ 36\quad for\quad N = 2n \\ &=&\rm\ (N\ +\ 4)\,(N\ +\ 9) \\ &=&\rm\ (2n\, +\, 4)\,(2n\, +\, 9) \\ \rm f\, &=&\rm\ (\ n\ +\ 2)\,(2n\ +\ 9) \end{eqnarray}$$

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I'm enlightened. Thank you. – Zol Tun Kul Jul 3 '12 at 3:58

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.

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