Quadratic Polynomial factorization

This could be primary school stuff. But I want to ask it.

In factoring $x^2+bx+c$ (i.e. $a = 1$ in $ax^2+bx+c$), we find $m$ and $n$ such that $m+n = b$ and $mn=c$. We can reason this well as follows:

In the expansion of $(x+m)(x+n)$, $mx$ and $nx$ should add up to $bx$ and $mn$ should result in $c$.

However I am not able to come up with the logic for following:
For factoring $qx^2+bx+c$ (i.e. $a \neq 1$ in $ax^2+bx+c$), we find $m$ and $n$ such that $m+n=b$ and $mn = ac$.

  • I am not able to realize the logic behind why it is $mn = ac$ and
  • how doing this result in correct factors

Solutions Collecting From Web of "Quadratic Polynomial factorization"

The 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 AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\

For example

$$ {\begin{eqnarray}
f \, &\,=\,& \ \, 2\ x^2-\ 3\ x\,\ +\ \ 1\\
\Rightarrow\ 2f\, &\,=\,&\!\ (2x)^2\! -3(2x)+2\\
&\,=\,& \ \ \ \color{#c00}{X^2}-\, 3\ X\,\ +\,\ 2,\,\ \ X\, =\, 2x\\
&\,=\,& \ \ (X-2)\ (X-\,1)\\
&\,=\,& \ (2x-2)\,(2x-1)\\
\Rightarrow\ \ f\:=\: 2^{-1}\,(2f)\, &\,=\,& \ \ \, (x- 1)\ (2x\,-1)\\

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.
This is sometimes called the AC method (cf. below). It works for higher degree polynomials too: we can reduce the problem of factoring a non-monic polynomial to that of factoring a monic polynomial by scaling by a $ $ power of the lead coefficient $\rm\:a\:$ then changing variables: $\rm\ X = a\:x$

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\!\!\rm\: X^2 + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\smash[t]{\overbrace{ac}^{\rm\qquad\ \ \ \ \ {\bf AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\
\rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\!\!\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c
After factoring the monic $\rm\,g(X)\, =\, a^{n-1}f(x),\,$ we are guaranteed that the transformation reverses to yield a factorization of $\rm\:f,\ $ since $\rm\ a^{n-1}$ must divide into the factors of $\rm\ g\ $ by Gauss’ Lemma, i.e. primes $\,p\in\rm\mathbb Z\,$ remain prime in $\rm\,\mathbb Z[X],\,$ so $\rm\ p\ |\ g_1(x)\:g_2(x)\,$ $\Rightarrow$ $\,\rm\:p\:|\:g_1(x)\:$ or $\rm\:p\:|\:g_2(x).$

This method also works for multivariate polynomial factorization, e.g. it applies to this question.

See this answer for more on the ring-theoretic concepts that lie at the heart of the matter (primal elements, Schreier refinement, Riesz interpolation, etc).

If $m,n $ are the roots of $ax^2 + bx +c$ then we may write

$$ax^2 + bx +c = \color{red} d\ (x- m)(x-n) = dx^2 -d(m+ n)x + dmn$$

Using the equality of polynomials we have taht

$$d = a ; d(m+n) = -b ; mnd = c $$

that is

$$a = d ; m+n = \color {red}{-\frac{b}{a}} ; mn = \color{red}{\frac{c}{a}}$$