Factoring a polynomium with no pattern Along the way of deriving the formula for a sum of integer squares on this webpage (it is about proving the pyramid volume formula), I noticed this step:
$$2n^2+7n+6=(n+2)(2n+3)$$
From right-to-left, you simply multiply the terms into each other's brackets. 
But from left-to-right, as is done in the proof, I can't figure out the method.
Is there are formula for such factoring or is it simply a "good guess"?
 A: The polynomial $2n^2+7n+6$ is a simple quadratic with discriminant $\Delta = 49-48=1$ and roots $\cfrac{-7 \pm \sqrt{1}}{4}=-2,-\frac{3}{2}\,$, therefore it is divisible by $(n+2)(n+\frac{3}{2})$. The dominant coefficient is $2\,$, so the full factorization is $\,2(n+2)(n+\frac{3}{2}) = (n+2)(2n+3)\,$.
A: The trick as always when dealing with polynomials with integer coefficients is to use the rational root theorem.
In your case, if $p/q$ is a root of $2X^2+7X+6$, then $q\vert 2$ and $p\vert 6$. 
A: Use the box method. What two numbers do you multiply to get $12$ and add to get $7$?
We have the tuple (3, 4). Draw a square box on your paper and divide it into four quadrants. Fit $2x^2$ on the top-left, and $6$ on the lower-right. Then fit $3x$ on the top-right box and $4x$ in the lower-left box.
Find the greatest common factors. In the first row, $x$ is the common factor. In the second row, $2$ is the common factor.
In the first column, $2x$ is the common factor. In the second column, $3$ is the common factor.
Therefore, you get $(x+2)(2x+3)$. You may be able to find more helpful examples here demonstrating the use of this method.
