# Factoring quadratics where the coefficient on the $x^2$ term does not equal 1

so we are working through various methods of factoring quadratic equations and the students seem comfortable factoring basic quadratics such as:

$$x^2 - 7x + 12 = 0$$

by finding the factors of $12$ that add up to $-7$. I feel like this is intuitive because it is just reasoning through "anti-distribution" and the students have picked it up very well.

However, we have gotten to the point where the coefficient on the $x^2$ term is not $1$ such as:

$$3x^2 -14x - 5 = 0$$

and I feel like finding all of the factors and essentially guessing and checking is much less intuitive. This is the way that I learned to factor these types of equations but it took a very long time for me to get comfortable and develop an intuition as to which factors to use. My feeling that this is a less than ideal instructional strategy is backed up by the fact that many of the students who are great with $a=1$ are totally lost when the $a$ term is changed. what are some intuitive ways to teach how to factor an equation like this other than finding all of the factors and guessing and checking?

• Look at the ac method for factoring. – Dave L Renfro Dec 15 '14 at 21:08
• @Benjamin Dickman: I'm too busy with work right now to write anything (and probably will be for the next day or two), so if anyone wants to write about it, that would be fine with me. For anyone whose interested, a couple of places I've written about it are this 28 April 2006 post and this 31 July 2007 post at Math Forum. – Dave L Renfro Dec 15 '14 at 21:49
• @DaveLRenfro Okay; I have written up a brief bit on the ac method. – Benjamin Dickman Dec 15 '14 at 23:05
• There are couple of other interesting bits on Patterns in Practice (see also that post's follow-up). – Benjamin Dickman Dec 16 '14 at 2:27
• Why not just teach them the quadratic formula? I've never understood the emphasis on factoring polynomials with integer coefficients whose roots happen to be integers; those cases just don't come up in practice. – Ben Crowell Dec 16 '14 at 16:28

$$3x^2 -14x - 5 = 0$$

Multiply through by A or here, 3

$$9x^2 -42x - 15 = 0$$

Now, use substitution, u=$3x$ (3X is the square root of this first term, and by using the u substitution, we now have an 'a' of 1. )

$$u^2 -14u - 15 = 0$$

factor to

$$(u-15)(u+1)$$

Substitute back u=3x

$$(3x-15)(3x+1)$$

last, divide out that 3 we multiplied by -

$$(x-5)(3x+1)$$

If you look at the number of steps, you'll realize it's consistent, 5 steps every time. Compare this to the guess and bang head method, and you'll find this one is a bit better. Students who get this are happy to learn and use it. But the u substitution isn't for everyone. I use this with high schoolers at multiple levels who need to factor just as you asked, with a non-zero A.

• Beautiful, intuitive, and works! How have I never seen this? Also a good introduction to substitution for use in calculus. – Richard Dec 16 '14 at 0:35
• But this is still guess and head-bang because your polynomial $u^2 - 14u -15$ is not factored, merely put into monic form. In this particular case since both $a$ and $c$ are prime it greatly reduces the possible forms of the factorization and it's not hard to guess $(3x+1)(x-5)$ right off the bat without all the extra work...especially on a test. For a method it's hard to beat completing the square since it generalizes to other situations such as when wanting to put various equations of conic sections into standard forms. – azdahak Dec 16 '14 at 4:46
• @azdahak - In my opinion, and to the point of the OPs question, the struggle with factoring seems to be with a quadratic when A isn't 1. I agree that the 3/15 example might not be tough, but the method I showed works well when A and C both have a few prime factors and the guess & check method becomes quite cumbersome. – JTP - Apologise to Monica Dec 16 '14 at 13:06
• @BenjaminDickman: your constant term in all equations but your first should be $ac$, not $c$, so you end up with $u^2+ub+ac=0$. Otherwise, a good clarification. – Rory Daulton Dec 16 '14 at 19:51
• @JoeTaxpayer I can certainly appreciate the transformation, but I would be concerned why the students are struggling with the factoring in the first place. Perhaps because of the focus on the special monic case they get trained to expect sums on the $x$ and products on the constant and then the rug gets pulled out from under them. But you never really get anything for free -- transforming the polynomial in this way just shifts all the possible products onto the constant term making more work there. – azdahak Dec 16 '14 at 20:42

Factoring non-monic quadratic polynomials can be done by factoring with respect to a particular constraint. More precisely, DL Renfro points to the ac Method of Factoring which can be summarized roughly as follows:

Given a quadratic $ax^2 + bx + c$, the polynomial can be factored iff there is a factor pair for $ac$ whose sum is $b$; here, I denote by "factor pair" a pair of integers whose product is the target integer.

The aforelinked works out a few examples, and also suggests pulling out common factors of $a,b,c$ beforehand (when relevant).

Applying the approach to the OP's quadratic of $3x^2−14x−5$:

Observe that $3(-5) = -15$ has a factor pair, $(-15, 1)$, whose sum is $-15 + 1 = -14$.

Now we know the answer is yes, it can be factored; rather than providing the general justification, one can work it out as per the linked examples (I include an optional $1$ coefficient for clarity):

$$3x^2 - 14x - 5 = 3x^2 - 15x + 1x - 5 = 3x(x - 5) + 1(x - 5) = (3x+1)(x-5)$$

Do note the role of the factor pair $(-15, 1)$ in re-writing the original expression; more generally, the factor pair for $ac$ indicates precisely how to re-write the expression in a manner that makes the polynomial factorization more transparent.

I will "leave as an exercise" why this approach works, but I would expect an explanation to be covered at some point in a relevant course (whether before or after the algorithm is presented may depend on constraints outside the scope of this question).

Given the above discussion, I would like to make one additional comment:

Viewing the task at hand as consisting of "factoring with respect to a particular constraint," it is important that students be able to factor integers more generally. But when do students have the opportunity to practice factoring integers? Unfortunately, the answer seems to be: Not often enough. To this end, here are some brief bits from a recent article about factoring (written in the context of working with pre-service teachers, but certainly applicable here). The citation is:

Feldman, Z. (2014). Rethinking Factors. Mathematics Teaching in the Middle School, 20(4), 230-236.

First, a bit on when factoring is (or should be) covered in the United States:

Second, and perhaps surprising to some, students can struggle to work with a number's (prime) factorization; again, from the same MTMS Nov. 2014 article:

Without overgeneralizing, the risk indicated in my reading of this work is students (who are unlikely to encounter an actual number theory course unless they are college mathematics majors) may lack a great deal of fluency in working with factoring integers. So, before delving too deeply into topics around factoring polynomials, it may be worthwhile to (re)visit integer factorization to ensure that students are not lacking a more foundational component.

• From a practical teaching perspective: can this "ac method" be generalised to real number factorisation? If it is not generalisable, what advantage does it have over completing the square? Unlike the factoring method for monic quadratics, it does not feel direct and intuitive. My concern is that I don't want to teach my students a 'trick' that works in one special case, when I have to teach them completing the square in the next year anyway. – Richard Dec 16 '14 at 0:31
• I'm not surprised that many students don't realize the connection between factoring and divisibility. I'd guess that many of the same students aren't aware that division is the inverse operation of multiplication — a fact that's obvious if you understand what division and multiplication are, but not obvious if you think of multiplication and division merely as the numbers you get by following certain procedures. Easy way to check this: ask students to compute something like $(1728 \cdot 23)/23$. – Daniel Hast Dec 16 '14 at 1:54
• Your write up answers the OP very well, and it is a cool trechnique. Just for myself, in a crammed curriculum I have to be selective. I like to present multiple approaches, but they need to add to an intuitive understanding or introduce generally useful techniques. The intermediate discovery of a1b2 and a2b1 makes the "ac approach" indirect and therefore less intuitive - even as you say "I will leave as an exercise why this works". It is also not generally useful outside of a Year 10 textbook as Year 11 goes to real coefficients. Other techniques presented here do not share those limitations. – Richard Dec 16 '14 at 2:47
• (If the down-voter would clarify their reasoning, then I would gladly try to improve this response!) – Benjamin Dickman Dec 18 '14 at 21:40
• (See also Bill Dubuque's nice write-up on the ac Method in MESE 7918.) – Benjamin Dickman Jun 7 '15 at 6:38

The QF yields that $-{\frac13}$ and $5$ are roots. So $$3x^2-14x-5=c\left(x+\frac13\right)(x-5)$$ Comparing leading coefficients, $c$ must be $3$: \begin{align}3x^2-14x-5&=3\left(x+\frac13\right)(x-5)\\&=(3x+1)(x-5)\end{align}

## Use Parabola Vertex Form (deterministic---no guess and check about it)

The $x$-coordinate of the vertex of the parabola $y=3x^2-14x-5$ is $-{\frac{b}{2a}}=-{\frac{-14}{2\cdot3}}={\frac73}$. The $y$-coordinate is $3\left(\frac73\right)^2-14\left(\frac73\right)-5=\frac{49}{3}-\frac{2\cdot49}{3}-5=-{\frac{49}{3}}-\frac{15}{3}=-{\frac{64}{3}}$.

So $y=c\left(x-\frac73\right)^2-\frac{64}{3}$. Comparing leading coefficients, $c=3$, so \begin{align} y &=3\left(x-\frac73\right)^2-\frac{64}{3}\\ &=\frac{1}{3}\left(9\left(x-\frac73\right)^2-64\right)\\ &=\frac{1}{3}\left(3\left(x-\frac73\right)-8\right)\left(3\left(x-\frac73\right)+8\right)\\ &=\frac{1}{3}\left(3x-15\right)\left(3x+1\right)\\ &=\left(x-5\right)\left(3x+1\right) \end{align}

## Complete the Square (deterministic---no guess and check about it)

Starting with $3x^2-14x-5$, always multiply and divide by $4a$ to avoid fractions: \begin{align} &3x^2-14x-5\\ &=\frac{4\cdot3}{4\cdot3}\left(3x^2-14x-5\right)\\ &=\frac{1}{12}\left(36x^2-12\cdot14x-60\right)\\ &=\frac{1}{12}\left(\left(6x\right)^2-2(6x)(14)-60\right)\\ &=\frac{1}{12}\left(\left(6x\right)^2-2(6x)(14)+14^2-14^2-60\right)\\ &=\frac{1}{12}\left((6x-14)^2-196-60\right)\\ &=\frac{1}{12}\left((6x-14)^2-256\right)\\ &=\frac{1}{12}(6x-14-16)(6x-14+16)\\ &=\frac{1}{6\cdot2}(6x-30)(6x+2)\\ &=(x-5)(3x+1)\\ \end{align}

## AC Method (involves integer factorization and a list of things to inspect)

$$3x^2-14x-5$$

Take $3\cdot(-5)=-15$. List pairs that multiply to $-15$:

$$(-15,1),(-5,3),(-3,5),(-1,15)$$

We could have stopped at the first pair, because $-15+1=-14$, the middle coefficient. Use this to replace the $-14$:

$$3x^2-15x+x-5$$

Group two terms at a time and factor out the GCF:

$$3x(x-5)+1(x-5)$$ $$(3x+1)(x-5)$$

## Prime Factor what you can version 1 (involves integer factorization and a list of things to inspect)

If $3x^2-14x-5$ factors, then prime factoring $3$, it factors as $$(3x+?)(x+??)$$ And $(?)(??)=-5$. There are only four possibilities. $(?,??)$ is one of $$(1,-5),(-1,5),(5,-1),(-5,1)$$ Multiplying out $(3x+?)(x+??)$ for each of the four cases reveals $3x^2-14x-5=(3x+1)(x-5)$.

## Rational Root Theorem (involves integer factorization and a list of things to inspect)

If $3x^2-14x-5$ factors, there are rational roots. They must be of the form $\pm\frac{a}{b}$ where $a\mid5$ and $b\mid3$. The only options are $\pm5,\pm{\frac53},\pm1,\pm{\frac13}$. Check these eight inputs to $3x^2-14x-5$ and find that $-{\frac13}$ and $5$ are roots. So $$3x^2-14x-5=c(x+1/3)(x-5)$$ Comparing leading coefficients, $c$ must be $3$.

## Prime Factor what you can version 2 (using Rational Root Theorem to speed up version 1)

If $3x^2-14x-5$ factors, then prime factoring $3$, it factors as $$(3x+?)(x+??)$$ The latter factor reveals that if the thing factors at all, one of its roots is an integer. Considering the RRT, check if any of $\pm5,\pm1$ are roots, and discover that $5$ is. Conclude $$(3x+?)(x-5)$$ and then conclude $$(3x+1)(x-5)$$

## Graphing to improve efficiency ot Rational Root Theorem method

Using the vertex formula again, locate the vertex at $\left(\frac73,-{\frac{64}{3}}\right)$. Since $a=3$, consider the sequence $\{3\cdot1,3\cdot3,3\cdot5,3\cdot7,\ldots\}$. Extend horizontally outward from the vertex by $1$ in each direction, move up $3$ and plot a point. Extend horizontally outward again by $1$, move up $9$ and plot a point. Continue until you've plotted points that cross over the $x$-axis.

Now you have a rough idea where the roots are. Returning to the rational root theorem approach, you can eliminate many of the potential roots now from the initial list, speeding up that approach.

• That is quite a collection of methods. However, some of them usually take longer to do than factoring. And just about all high school students hate completing the square or anything like it (such as vertex form). But for those who are bad at factoring or at guessing. one of those would do. – Rory Daulton Jan 19 '15 at 22:36
• @RoryDaulton You say "some of them usually take longer to do than factoring", but each of these is a method of factoring. It sounds like maybe you have a method in mind that is not on this list that you are comparing to? – alex.jordan Jan 19 '15 at 22:40
• I am referring to the method stated in the OP: "finding all of the factors and essentially guessing and checking." – Rory Daulton Jan 19 '15 at 22:48

Teach them how to complete the square. This is probably the simplest systematic method for factoring quadratic polynomials, and it's also very geometrically intuitive (you can literally visualize it in terms of a square). Guessing and checking usually works for very simple examples — monic quadratic polynomials with small integer factors — but it's not effective for anything much more complicated than that. (I suspect your students would also have trouble factoring polynomials with non-integer roots in this way.)

Completing the square is also essential for understanding the quadratic formula, which is derived using this technique. Since roots of a polynomial correspond to linear factors ($r$ is a root if and only if $x - r$ is a factor), this is the same thing in slightly different language.

Also, one of the most important mathematical skills is being able to reduce more complicated problems to simpler ones. Given a quadratic polynomial $ax^2 + bx + c$, we can write this as $a(x^2 + \frac{b}{a} x + \frac{c}{a})$, and now study the monic quadratic polynomial $x^2 + \frac{b}{a}x + \frac{c}{a}$, which has exactly the same roots as $ax^2 + bx + c$. So, if your students learn how to factor monic quadratics (by completing the square or by another method), then they can factor all quadratics by reducing in this way.

You can motivate completing the square as coming from a series of such reductions to simpler problems, starting from the easiest cases. An equation of the form $x^2 = c$ is easily solved, and $(x - d)^2 - c = 0$ isn't much harder. But $(x - d)^2 - c = x^2 - 2dx + d^2 - c$, so this is general enough to encompass all monic quadratics.

Guessing and checking might lead to an ad hoc way of completing the square anyway. Factoring a polynomial $x^2 + rx + s$ means finding numbers whose sum is $r$ and product is $s$, so one might start looking around $\frac{r}{2}$ and work outwards until numbers with the appropriate product are found. For instance, starting with $x^2 + 8x + 12$, one could guess $4$ and $4$ (whose product is too big), then $3$ and $5$ (still too big), then $2$ and $6$ (which is correct). This corresponds to the fact that $x^2 + 8x + 12 = (x + 4)^2 - 2^2$, whose roots are $4 \pm 2$. (Of course, this needs to be understood more explicitly to handle something like $x^2 + 8x + 14$.)

• 4th paragraph -2dx, right? – JTP - Apologise to Monica Dec 16 '14 at 4:01
• @JoeTaxpayer: Thanks, fixed. – Daniel Hast Dec 16 '14 at 4:09
• -1, I do not believe this answers the question. This answer seems to say that to factor $3x^2 - 14x - 5$, you tell your students to first complete the square, writing it as $3(x - \frac73)^2 - \frac{64}{3}$? Then you have them set it to zero, solve for the two roots, and then write $3(x - r_1)(x - r_2)$? I can't believe you really advocate this as a primary strategy for factoring quadratic polynomials. – Chris Cunningham Dec 17 '14 at 18:10
• @ChrisCunningham: The quadratic formula gives $\frac{14 \pm \sqrt{14^2 - 4 \cdot 3 \cdot (-5)}}{6}$, which is still pretty messy. Anyway, there's no need to go through those extra steps: after writing $3x^2 - 14x - 5 = 3(x^2 - \frac{14}{3} x - \frac{5}{3}) = 3[(x - \frac{7}{3})^2 - \frac{49 + 15}{9}] = 3[(x - \frac{7}{3})^2 - (\frac{8}{3})^2]$, one can directly factor using the identity $s^2 - t^2 = (s - t)(s + t)$, yielding $3 (x - \frac{7}{3} - \frac{8}{3}) (x - \frac{7}{3} + \frac{8}{3})$. This also finds the axis of symmetry at $x = \frac{7}{3}$ at the same time. – Daniel Hast Dec 17 '14 at 18:42
• @DanielHast So the answer is "yes," then -- you advocate that the first step in factoring a quadratic polynomial should be "complete the square!" I'll remove my -1 since my complaint now is only a strong disagreement. – Chris Cunningham Dec 17 '14 at 19:23

Assuming the problem comes from the happy world of textbook problems... a typically successful method is to assume integer factorizations: $$(3x+a)(x+b) = 0$$ where $ab=-5$. In the world free of those complicated fractions, we have just $a= \pm 1$ and $b= \mp 5$ to choose. So, our options are: $$(3x+1)(x-5) \qquad \& \qquad (3x-5)(x+1)$$ $$(3x-1)(x+5) \qquad \& \qquad (3x+5)(x-1)$$ and multiplication reveals $(3x+1)(x-5)=0$ is the winner. In summary, guided guess and check.

• surely there exist users who can improve my woefully unalgorithmic answer :) – James S. Cook Dec 15 '14 at 18:10
• A "meta" thing that can be inserted in your method is to make things even happier by pointing out that the leading and constant coefficients are not full of factors--like 12 or 60. – user52817 Dec 15 '14 at 23:09
• @JamesS.Cook just wondering, has anyone ever required an integer coefficient factorisation of a quadratic except in the happy world of textbook problems or early renaissance maths competitions? – Richard Dec 16 '14 at 0:15
• @Richard there exist these creatures who only love numbers. Such people might require only these, although, in a moment of imperfection, I suppose rational coefficients might be allowed. I personally am open to all manner of factorizations, real, complex, hyperbolic, tricyclic, n-cyclic, whatever. – James S. Cook Dec 17 '14 at 2:50

The more little tricks and techniques we teach our students, the more they see math as an arcane toolbox of things to remember until the next exam and forget thereafter.

Instead of teaching an N-step process for each problem type, whenever possible we should try to find a memorable, generalizable, useful concept that unites all similar problems. In this case, we should use the same idea to factor any difficult quadratic, including things like $x^2 - 7x + 1$ or $x^2 + 9$ or OP's $3x^2−14x−5$:

You should only ever try to factor a quadratic as long as you find it interesting to keep trying. As soon as you are sick of factoring, just use the quadratic formula to find the roots. It's okay if you give up immediately when $a \neq 0$.

If the roots of a quadratic are $r_1$ and $r_2$, then the quadratic factors as $y = a(x - r_1)(x - r_2)$.

I don't see any reason to learn factoring tricks or skills beyond this one. This method has several advantages:

• It reinforces the fundamental intuition of algebra: that the factors of a polynomial line up with its zero-places.
• It works in many more situations and even generalizes to higher degree polynomials.
• It is a fundamentally geometric viewpoint, so you can draw pictures of it to illustrate.
• I agree that it's important to teach generalizable, useful concepts, but I think it's absolutely essential to teach completing the square, preferably before the quadratic formula. After all, completing the square explains where the quadratic formula comes from and why it works. Without that reasoning backing it up, the quadratic formula is just another meaningless formula to memorize. Once students understand how to complete the square, I don't see any problem with using the quadratic formula as a shortcut for the outcome of completing the square, provided it's clear where it comes from. – Daniel Hast Dec 16 '14 at 6:40
• The quadratic formula itself is a trick and can be completely ignored. A method for factoring a quadratic is to complete the square which is useful in cases beyond solving quadratic equations, such as showing that this $$1 - 2 x + x^2 + y^2 = 1$$ is a circle centered at (1,0). – azdahak Dec 16 '14 at 19:59
• @ChrisCunningham: The quadratic formula is just completing the square in disguise. How do you propose to teach the quadratic formula, if not as a direct consequence of completing the square? – Daniel Hast Dec 17 '14 at 18:18
• @ChrisCunningham: Fair enough — as long as you present the quadratic formula as a consequence of completing the square, I don't see a huge difference between the two methods. I also wouldn't argue that the quadratic formula should be abolished, as it's a very direct way of finding roots, and multiple presentations of the same ideas are definitely useful. – Daniel Hast Dec 18 '14 at 0:03
• Many of us (me for example) may be in the situation of teaching basic algebra courses where factoring is part of the curriculum, but the quadratic formula is not. If we think about the quadratic formula just for this issue alone, then we also have add completing the square to prove it, which itself involves factoring a binomial square (circular pedagogy!), and risk (or guarantee) overflowing our students' memory stack. In most curricula the position of factoring polynomials in integers is meant as a low-level warm-up to more generalized solutions in later semesters. – Daniel R. Collins Jun 19 '17 at 14:04

Edit 2:

Andrew Sanfratello, in a new answer to this question, seems to have discovered that the method I describe below is called the "Berry method"(for reasons unknown). My explanation shows that it is actually a refinement of the substitution method.

Comparison of the ac and berry methods:

Edit:

I just taught this today. The curriculum requires that they solve these problems, but all the examples have at least one of a or c as prime. This doubles the problem space compared to a monic quadratic, but it is far from combinatorial explosion.

Completing the square/quadratic formula are really over the top, "ac" method or substitution introduce another unnecessary layer and sometimes a very large "ac" to factor. In reality they had no problem at all simply extending the guessing method used for monic quadratics.

Original:

While I would tend to teach completing the square/quadratic formula, I have an updated alternative to the "ac method" which is a cut down version of the substitution method also presented as an answer.

I was surprised how similar the "ac method" is to the substitution method.

To factor:$ax^2 + bx + c$

As in the "ac method" find the factors $d,e$ of $ac$ that add to $b$. As in the "ac method" this step does not need to be written down. $$u^2 + bu + ac = (u+d)(u+e)$$

Different from the "ac method", simply plug the found factors $d,e$ into the following equation $$ax^2 + bx + c = (ax + d)(ax + e)\div a$$

$a,d$ and/or $a,e$ will have common factors that will cancel with the $\div a$ if needed to give the simplest integer factorisation if possible. If you simply want to find zeros, the $\div a$ can be ignored.

This method works by $u=ax$ and multiplying the whole equation by $a$, but the student can actually safely ignore the u step. The only step that needs writing is the final step. I think the full u substitution is beautiful for understanding, but tedious for lots of problem solving.

This method is of limited usefulness as it only works with integer factorisation. I am interested in what people think about this and whether there are any improvements to be made.

I have (just recently) become aware of a method for factoring non-monic quadratics called the "Berry Method". At first glance it seems like a variation of the ac method discussed in a different answer, though perhaps more attainable for weaker students. I will continue to research this method and report back, but has anyone had experience using this method? Included here are some early Google results.

• This appears to be identical to the method I discovered a few months ago, inspired by the substitution method but simplified so the substitution is hidden. Nice to know it has a name! I also discovered that it is very similar to the ac method, but I prefer the way the berry/substitution method simplifies 2 factors as the final step rather than how the ac method splits the "x" term into 2 parts before being factored. – Richard Feb 24 '15 at 10:21
• As my answer above explains, I trialled it with adults as it seems much nicer than the ac method, but it was unnecessary for the curriculum that I needed to teach, so I stuck with "less is more" and let them use the normal guessing method which was more direct and intuitive. – Richard Feb 24 '15 at 10:31
• For many cases, the "ac" multiplication used in the ac and berry methods actually produced very large numbers (sometimes over 100), which made factoring more cumbersome than the original, which negated any gains it may have had over the basic guessing method. – Richard Feb 24 '15 at 10:40

I have never understood the infatuation of teachers with "factoring quadratics".

1. As a concept, factoring on the ring of integers is way down on the list of what one should spend time on. Plus, most students deemed good at it stall on $x^{2}-2$ and $x^{2}+x+1$.
2. A concept much higher on the list is that of a change of variable. But then, to find the zeros of a quadratic function $q$, just "localize", that is make the change of variable $x=x_{0}+u$ and kill the coefficient of $u$ to find $x_{0-\text{slope}}$. Then, $q_{x_{0-\text{slope}}}(u)$ will have no $u$ term and $q_{x_{0-\text{slope}}}(u)=0$ can be solved.
3. The big advantage is that this takes its place in the much "bigger picture" of the investigation of elementary functions. E.g. the inflection of a cubic is found the same way. And it also shows the limitations of the approach: since $=$ has only two sides, we need to be able to kill enough terms so that there remain only two terms in $f(u)$.
• I'm unfamiliar with the approach you describe in 2. Can you link to an example or add an example? – Richard Mar 4 '15 at 11:30
• @Richard For the sake of comparison, here is "the" canonical example: Given $x \xrightarrow {\;f\;} f(x)=x^{2}-5x+6$, $f({x_{0}+u}) = (x_{0}+u)^{2} -5(x_{0}+u)+6=\left[x_{0}^{2}-5x_{0}+6\right]+\left[2x_{0}-5\right]u+u^{2}$ So, $x_{0-\text{slope}}=+\frac{5}{2}$ and $f({\frac{5}{2}+u}) = (\frac{5}{2}+u)^{2} -5(\frac{5}{2}+u)+6=-\frac{1}{4}+u^{2}$ So, $u_{0-\text{height}}=\pm\frac{1}{2}$ and $x_{0-\text{height}}=\frac{5}{2}\pm\frac{1}{2}$ – schremmer Mar 5 '15 at 15:17
• It looks to me like you are using some form of calculus, as $\frac{dy}{du}=2x_0 - 5 + 2u$ then you set $u=0$and $\frac{dy}{du}=0$ and solve for $x_0$. It seems much more complicated than the standard calculus approach. You then find $y_{vertex}$ by finding the $c$ coefficient of $f(u)$ produced by $f(x_{vertex}+u)$, rather than simply finding the numerical value $f(x_{vertex})$. It is pretty, and is probably very interesting theoretically, but it is an awful lot of work, most of which does not affect the result. I tried factorising a cubic in this method, but it was too hard by hand. – Richard Mar 5 '15 at 20:49
• I think I understand your notation ... $x_{0-slope}$ means $x$ when the slope is 0? And $x_{0-height}$ is $x$ when $f(x)=0$ . I got confused as it overlaid your use of $x_0$ . Or is $x_0$ a short form of $x_{0-slope}$? – Richard Mar 5 '15 at 21:31
• @Richard Re. Calculus. The other way round: you can define the $n^{\text{th}}$ derivative from the polynomial expansion. Re. $x_{0\text{-slope}}$ instead of $x_{\text{vertex}}$: I am not doing geometry (hence no $y$). Also for uniform notation. Re. Conflict with $x_{0}$: you are right---although none of my students, who often nitpick my book to death, ever complained about that.) Re. Cubic. I wouldn't even dream of trying to factor a cubic: what for? But killing the coefficient of $h^{2}$ in a cubic gives ... $x_{0\text{-concavity}}$. For more context, see [freemathtexts.org/] – schremmer Mar 6 '15 at 15:09

This is yet another variant of the ac method.

$$3x^2 -14x - 5 = 0$$

Solve the "companion" quadratic ($ax^2+bx+c=0 \to x^2+bx+ac=0)$

$$x^2 -14x - 15 = 0$$

$$(x-15)(x+1) = 0$$

$$x \in \{15, -1\}$$

Roots of original polynomial are above roots divided by $a$.

$$\text{Roots are \left\{\dfrac{15}{3}, -\dfrac{1}{3}\right\} = \left\{5, -\dfrac{1}{3}\right\}}$$

Reconstruct factors

$$(x-5)(3x+1)$$

PROOF

$ax^2 + bx + c = 0$ has roots $r=\dfrac{-b + \sqrt{b^2-4ac}}{2a}$ and $s=\dfrac{-b - \sqrt{b^2-4ac}}{2a}$.

$x^2 + bx + ac = 0$ has roots $r'=\dfrac{-b + \sqrt{b^2-4ac}}{2}$ and $s'=\dfrac{-b - \sqrt{b^2-4ac}}{2}$.

Note that $r = \dfrac 1ar'$ and $s = \dfrac 1as'$

• Can you edit to add what "companion" does? It looks like a variant of AC method, no? – JTP - Apologise to Monica Jun 19 '17 at 14:22
• @JTP-ApologisetoMonica - Yes. It seems it is called the "swing method". – Steven Gregory Feb 18 at 19:22

A very general method for any polinomial is the rational root test: Any rational zero of $$a x^2 + b x + c$$ must be of the form $$p / q$$, where $$p$$ is a factor of $$c$$ and $$q$$ a factor of $$a$$ (don't forget signs!). This affords a factor $$q x - p$$, for a quadratic the other must be $$(a / q) x - (c / p)$$. As a bonus point, the (simple) proof of the test and the resulting factorization in this case exercise some non-trivial manipulations.