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The normal way to factor ax^2+bx+c=0 is to look for t,u,v,w such that:

(tx+u)(vx+w) = 0

so that tv=a, uw=c, and uv+wt=b.

This can be tricky, since there can be several possibilities for t,u,v,w.

Ashley, a bright student, showed me a shortcut that we couldn't quite work out 100%. Instead of factoring a into tv, we do the following:

I. rs=c (equivalence of constant term) 
II. r+sa=b (equivalence of 'x' term) 
III. sa=b-r (subtract II by r) 
IV. rsa=ac (multiply I by a) 
V. r(b-r)=ac (substitute III into IV) 
VI. r^2-br+ac=0 (rearrange V) 

We now have a quadratic with a leading term of 1, so we simply need to find factors of ac that add to -b. Call these p and q. Thus,


and r=p, while s=c/p. Thus


and x=-p/a, x=-c/p

Of course, this is too complex to be considered a shortcut in the form above.

The shortcut would be something like:

  • Given ax^2+bx+c=0, find factors (p and q) of ac that add to -b

  • The solutions are then -p/a and -c/p

However, Ashley believes the technique was even simpler.

Is this a well-known/named technique? Is there a simpler way to state it?

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1 Answer 1

up vote 2 down vote accepted

This is sometimes called the $\rm\color{#C00}{AC}$ method and it works for higher degree polynomials too. Namely, 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 leading coefficient $\rm\:a\:$ then changing variables $\rm\: X = a\:x$

$$\begin{eqnarray} \rm\: a\:f(x)\, =\, a\:(a\:x^2 + b\:x + c)\, &=&\rm\, X^2 + b\:X + \color{#C00}{ac}\, =\, g(X),\ \ \ X = a\:x \\ \\ \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c)\, &=&\rm\, X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c \end{eqnarray}$$ 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 the constant $\rm\:a^{n-1}\:$ must divide into the factors $\rm\,g_i\:$of $\rm\:g\:$ by Gauss' Lemma, i.e. primes 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 technique also works for multivariate polynomial factorization, e.g. it applies to this question.

Remark $\ $ Those who know university algebra might be interested to know that this works not only over UFDs and GCD domains but also for integrally-closed domains satisfying the

Primal Divisor Property $\rm\ \ c\ |\ AB\ \Rightarrow\ c = ab,\ \ a\ |\: A,\ \ b\ |\ B$

Such domains are called Schreier rings by Paul Cohn (or Riesz domains, because they satisfy a divisibility form of the Riesz interpolation property). See this answer for further discussion.

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