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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
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The AC-method reduces to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $\color{#c00}{=1})$ as follows

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\!\!\rm\: \color{#c00}{X^2} + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\overbrace{ac}^{\rm\qquad\ \ \ \ \ {\bf AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\ \end{eqnarray}$$

For example

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

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: $\rm\ X = a\:x$

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\!\!\rm\: X^2 + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\smash[t]{\overbrace{ac}^{\rm\qquad\ \ \ \ \ {\bf AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\ \\ \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\!\!\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c \end{eqnarray}$$ 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 $\rm\ a^{n-1}$ 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).

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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

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

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  • $\begingroup$ I was taught a different method, where you seek $m$ and $n$ that add to $b$ and multiply to $ac$. (However, the next step is not to write it in the form $a(x+m)(x+n)$, since your post shows that this doesn't work.) $\endgroup$ – Akiva Weinberger Mar 18 '15 at 23:23
  • $\begingroup$ If you have $2x^2+17x+21$, you want them to sum to 17 and multiply to 42; 14 and 3 work nicely. So you split it into $(2x^2+14x)+(3x+21)$. Factoring gives you $2x(x+7)+3(x+7)$, and factoring again gives you $(2x+3)(x+7)$. $\endgroup$ – Akiva Weinberger Mar 18 '15 at 23:27
  • $\begingroup$ It seems a litte tedious. What I presented is known as the Newton-Girard formulae. Both ways are, of course, the same. $\endgroup$ – Aaron Maroja Mar 18 '15 at 23:29
  • $\begingroup$ But then you have to deal with fractions. (By the way, rewriting it as $(2x^2+3x)+(14x+21)$ also works, if you were wondering.) $\endgroup$ – Akiva Weinberger Mar 18 '15 at 23:31
  • $\begingroup$ Yes, this may be a downpoint. But you see this again in generalized in Galois Theory. So I thought it was worth mentioning. $\endgroup$ – Aaron Maroja Mar 18 '15 at 23:33

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