Factorising quadratics - coefficient of $x^2$ is greater than $1$ In factoring quadratics where the coefficient of $x^2$ is greater than $1$, I use the grouping method where we multiply the coefficient and constant together and then factor.
My question is can someone explain the math behind that?
Example:
$5x^2+11x+2,\quad  5\cdot2=10$
$5x^2+10x+x+2$
$5x(x+2)+1(x+2)$
$(5x+1)(x+2)$
 A: This doesn't seem like a very reliable method. It seems to only catch factorizations of the form $$(ax+1)(x+b)$$
for integers $a, b$.
The reason it works for these cases is that when you multiply out you get
$$(ax+1)(x+c) = ax^2 + (1+ac)x + c$$
The first coefficient is just $a$, the third is $c$, and the second is the product of them, plus one. 

More generally, for four integers $a, b, c$, and $d$, we have
$$(ax+b)(cx+d) = acx^2 + (ad+bc)x + bd$$
The method you described would lead to us multiplying $ac$ and $bd$ together, to get $$abcd$$
You could factor this number $abcd$ (which, to be clear, is the product of the four numbers $a, b, c$ and $d$), and then try to write out all of the possible factors and test them. This isn't very efficient.

The reason the grouping method "works" (which it rarely does) is really just that it uses the expansion of $(ax+b)(cx+d)$.
In the end, the assumption of an integer root isn't that great, and it's easier/better to complete the square.
A: Consider the factorization of $9x^2 - 18x - 16$.  To split the linear term, we must find two numbers with product $9 \cdot -16 = -144$ and sum $-18$.  They are $-24$ and $6$.  Hence,
\begin{align*}
9x^2 - 32x - 16 & = 9x^2 - 24x + 6x - 16 && \text{split the linear term}\\
                & = 3x(3x - 8) + 2(3x - 8) && \text{factor by grouping}\\
                & = (3x + 2)(3x - 8) && \text{extract the common factor}
\end{align*}
Observe that when you split the linear term, the product of the quadratic and constant coefficients is equal to the product to the two linear coefficients, that is, $9 \cdot -16 = -24 \cdot 6 = -144$.  
Suppose that $a, b, c, k, l, m, n$ are integers such that 
$$ax^2 + bx + c = (kx + l)(mx + n) \tag{1}$$
First, we show that 
\begin{align*}
a & = km \tag{2}\\
b & = kn + lm \tag{3}\\
c & = ln \tag{4}
\end{align*}
Since equation 1 is an algebraic identity, it holds for each value of the variable.  In particular, it holds for $x = 0, 1, -1$.  
If we substitute $0$ for $x$ in equation 1, we obtain
$$c = ln$$
Thus, equation 4 holds.  If we substitute $1$ for $x$ in equation 1, we obtain
\begin{align*}
a + b + c & = (k + l)(m + n)\\
          & = k(m + n) + l(m + n)\\
          & = km + kn + lm + ln
\end{align*}  
Since $c = ln$, we can cancel $c$ from the left hand side and $ln$ from the right hand side of the equation to obtain
$$a + b = km + kn + lm \tag{5}$$
If we substitute $-1$ for $x$ in equation 1, we obtain
\begin{align*}
a - b + c & = (-k + l)(-m + n)\\
          & = -k(-m + n) + l(-m + n)\\
          & = km - kn - lm + ln
\end{align*}
Since $c = ln$, we cancel $c$ from the left hand side and $ln$ from the right hand side of the equation to obtain
$$a - b = km - kn - ln \tag{6}$$
Adding equations 5 and 6 yields
$$2a = 2km$$
Dividing each side of the equation by $2$ yields
$$a = km$$
so equation 2 holds.  Since $a = km$, we can cancel $a$ from the left hand side and $km$ from the right hand side of equation 5 to obtain
$$b = kn + ln$$
so equation 3 holds.
Thus, if $a, b, c, k, l, m, n$ are integers such that equation 1 holds, then 
$$ac = (km)(ln) = klmn = (kn)(lm)$$
where $b = kn + ln$, so the product of the quadratic and constant coefficients is equal to the product of the two coefficients into which the linear coefficient splits.  
We then obtain the factorization
\begin{align*}
ax^2 + bx + c & = kmx^2 + (kn + lm)x + ln && \text{by equations 2, 3, and 4}\\
              & = kmx^2 + knx + lmx + ln && \text{split the linear term}\\
              & = kx(mx + n) + l(mx + n) && \text{factor by grouping}\\
              & = (kx + l)(mx + n) && \text{extract the common factor}
\end{align*}
Note that to check the factorization is correct, we multiply $(kx + l)(mx + n)$, in which we perform the steps of the factorization by splitting the linear term and grouping in reverse order. 
A: Sometimes, instead of using symbols, it can help to just use numbers to keep track of patterns (and then generalize with symbols later).
To understand this grouping method, it is instructive to work the problem backwards, and to expand, instead of factor.
For example
$$(2x+3)(5x+7)\text{ the constants are all distinct primes. How fortunate.}$$
$$2\cdot5x^2 + 2\cdot7x+3\cdot5x+3\cdot7$$
$$\color{green}{2\cdot5}x^2 + (\color{blue}{2\cdot7}+\color{blue}{3\cdot5})x+\color{green}{3\cdot7}$$
Notice how we can partition the middle coefficient ($2\cdot7+3\cdot5$) into $2\cdot7$ and $3\cdot5$. The product of these partitions is $(2\cdot7)(3\cdot5)=2\cdot3\cdot5\cdot7$, which is equal to the product of the outside cofficients $(2\cdot5)(3\cdot7)=2\cdot3\cdot5\cdot7$.
In a real problem, however, we don't know the components (2, 3, 5, and 7). We only know the coefficients 10, 29, and 21 for the quadratic $$10x^2+29x+21$$
Our first step is to partition the middle coefficient into two parts such that the sum of the parts is $29$ and the product is $10\cdot21=210$. After analyzing the factors of $210$, we settle on the partitions  $14$ and $15$.
Looking above at the quadratic with the colored coefficients, notice that neither of the factors of the $x^2$ term appear in both partitions (2 is a factor of the first partition, and 5 is a factor of the second partition). The same property holds true for the target factors of the constant term (the 3 and 7 appear in different partitions). This property will allow us to factor by grouping, thus discovering numbers 2,3,5, and 7 from the numbers 10, 14, 15, 21.
$$(10x^2+14x)+(15x+21)=2x(5x+7)+3(5x+7)=(2x+3)(5x+7)$$
Also take note that it doesn't matter what partition we associate with the $10x^2$ term. We just get a different common factor, depending upon whether we choose $14x$ or $15x$.
$$(10x^2+15x)+(14x+21)=5x(2x+3)+7(2x+3)=(5x+7)(2x+3)$$
As an exercise, you may want to restate the preceding argument for the quadratic
$$(px+r)(qx+s)$$
