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

Question

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?

Answer 1

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

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.

Answer 2

Quadratic Formula (deterministic---no guess and check about it)

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.

Answer 3

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).

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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.

Answer 4

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 $x^2 + 8x + 14$.)

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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 $f(u)$.

Answer 10

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 $s = \dfrac 1as'$

Answer 11

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.

Answer 12

1. FWIW, I'm not a jock and didn't think the example was that hard. I mean you got 5, which has integral factors of 1 and 5 (worth checking). And then 3 times 5 is suspicously similar to 14 on the middle term. Maybe minus a 1. Probably get it faster than the "ac" method.

2. What I think is harder are ones where you have to put numbers on each x...like if it were 6 and you needed a 2 and a 3...and then things just spiral.

3. The broader question (which is tricky) is do we underdo/overdo factoring exercised in general. Now, sure, there is a method (quadratic forumula) that solves all such questions and obviates ever doing factoring. That said, I wonder if general manipulational ability/games are useful in later work with trig identities and the like. It's actually a non-simple question to answer.