How do I factorize a polynomial $ax^2 + bx + c$ where $a \neq 1$ How do I factorize a polynomial $ax^2 + bx + c$ where $a \neq 1$?
E.g. I know that $6x^2 + 5x + 1$ will factor to $(3x + 1)(2x + 1)$, but is there a recipe or an algorithm that will allow me to factorize this?
 A: The roots of a quadratic polynomial $ax^2+bx+c$ are $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$. Then the factorization is just $a(x-x_1)(x-x_2)$.
In your example, $a=6,b=5,c=1$, so $x_1=\frac{-5+\sqrt{5^2-4\cdot 6\cdot1}}{2\cdot 6}=-\frac13$ and $x_2=\frac{-5-\sqrt{5^2-4\cdot 6\cdot1}}{2\cdot 6}=-\frac12$. The factorization is then $6(x+\frac13)(x+\frac12)$, or $(3x+1)(2x+1)$.
A: If you are lucky enough to find "by guesswork" values $x_1$ and $x_2$ so that
$$x_1 + x_2 = -\frac{b}{a}$$
and
$$x_1 x_2 = \frac{c}{a}$$
Then you'll be able to factor $ax^2 + bx + c$ to $a(x-x_1)(x-x_2)$.
Of course it will be easy to guess if $x_1$ and $x_2$ are integers.
A: That AC-method reduces to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $\color{#c00}{=1})$ as follows
$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c)  &\,=\,&\!\!\rm\: \color{#c00}{X^2} + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\overbrace{ac,}^{\rm\qquad\ \ \ \ \ {\bf\large\ \  AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \  X = a\:x \\
\end{eqnarray}$$ In your case $$ {\begin{eqnarray}
f \, &\,=\,& \ \ \,  6 x^2+\ 5\ x\,\ +\ \ 1\\
\Rightarrow\,\ 6f\, &\,=\,&\!\,\ (6x)^2\! +5(6x)+6\\
         &\,=\,& \ \ \   \color{#c00}{X^2}+\, 5\  X\,\ +\ 6,\,\ \ X\, =\, 6x\\
         &\,=\,&  \ \ (X+2)\ (X+\,3)\\
         &\,=\,& \  (6x+2)\,(6x+3)\\
\Rightarrow\ \  f\:=\: \color{#0a0}{6^{-1}}\,(6f)\,  &\,=\,&   \,  (3x+ 1)\ (2x+1)\\
\end{eqnarray}}$$
In the final step we cancel $\,\color{#0a0}6\,$  by cancelling $\,2\,$ from the first factor, and $\,3\,$ from the second.
If we denote our factoring algorithm by $\,\cal F,\,$ then the above transformation is simply
$$\cal F f\,  = a^{-1}\cal F\, a\,f\quad\,$$
Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. The same idea also works for higher degree polynomials, see this answer, which also gives links to closely-related ring-theoretic topics. 
A: Hint: another possibility is to use the vieta's formulas
https://en.wikipedia.org/wiki/Vieta%27s_formulas
but it's more a "guess and proof strategy"^^
A: I actually learned this one neat trick in middle school. Take your polynomial 
$$6x^2+5x+1$$
Take out the leading coefficient, and multiply it into the constant term to get
$$x^2+5x+6$$
Now factor like usual 
$$(x+2)(x+3)$$
Divide both of those constant terms by the original leading coefficient, and reduce the fractions
$$\left(x+\frac{2}{6}\right)\left(x+\frac{3}{6}\right) = \left(x+\frac{1}{3}\right)\left(x+\frac{1}{2}\right)$$
Now take the denominators and put them in front of the $x$ terms to get
$$(3x+1)(2x+1)$$
(this process essentially only works for polynomials with integer coefficients, and it assumes that there are no constant factors in the original polynomial)
A: If you know how to factor polynomials for $a=1$, then simply writing
$$ax^2+bx+c=a(x^2+\frac bax+\frac ca )$$
makes the task immediate.
