Factoring Quadratics Misery

One of the topics covered in an Algebra sequence is about factoring quadratics.

Typically, a student is shown an example like $$x^{2} + 3x + 2$$ and is asked to factor it. What mysterious methods must exist for factoring quadratics? Here, the student is taught that they have to “find two numbers that multiply to 2 and add to three” in order to factor the quadratic. Thus, we have \(a = 2\) and \(b = 1\) since \(a + b = 3\) and \(ab = 2\). Therefore, we have $$x^{2} + 3x + 2 = (x + 2)(x+1)$$ and huzzah! This is a standard in all Algebra courses and many books have several sections devoted to this.

But I despise this method. It is too narrow of a technique and provides little additional information about the quadratic other than its zeros. Though, even with respect to the zeros of the quadratic, students are often not shown how factoring a quadratic tells one what the zeros are. The exercise of “find two numbers that multiply to something and add to something else” only is sensible when the coefficient of the linear term and the constant term are relatively tame (small integers).

I asked my students how they would factor $$x^{2} + 4x + \frac{6}{7}$$ and I posed it in the only way they had been taught how — what two numbers multiply to 4 and add to six-sevenths?

Of course, no one knew. No one even knew how to begin. Heck, I don’t know by intuition nor even through sophisticated guessing (of course, I can factor this).

What compounds the woes of Algebra students on the topic of factoring quadratics is that they are shown the following:

  • Factor via the “find two numbers that multiply to something and add to something else”
  • Memorize the quadratic formula $$x_{1,2} = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$$ to solve \(ax^{2} + bx + c = 0\)
  • “Complete the square” method for graphing quadratics: that is, write rewrite the quadratic as \(a(x – p)^{2} + q\) which tells us that it is a parabola with vertex at \((p,q)\) (or some equivalent variant of this).
  • Memorize that \(x^{2} – a^{2}\) factors to \((x – a)(x + a)\)
  • Memorize that \(x^{2} + a^{2}\) “can’t be factored” (because they haven’t been taught about complex numbers yet)
  • Memorize that the vertex of a quadratic is at \(x = -\frac{b}{2a}\)


It’s almost as if there were too much time to cover all the material in an Algebra II course, so that this stuff was invented to make things purposefully confusing and lengthen the course. The bullet points above can easily take several one hours college lecture periods and in high school, several months.

What on Earth for? What’s wacky is that completing the square actually takes care of root finding, graphing, and factoring. Additionally, it doesn’t matter if the problem is to factor \(x^{2} + 3x + 2\) or to factor \(x^{2} + 4x + \frac{6}{7}\) — the method does not change.

So let’s begin and since most who read this teach mathematics, I’ll skip the simple examples and just start in the general case. Then I will give a few examples. When teaching, my approach is the reverse. I give a bevy of examples first, then generalize. Thus, the progression in example seen here is not how I teach it, but rather is a justification of why I omit all the senseless of the above bullet points. If you’d like to see an example first then jump to here.

Our goal is to write, for \(a \neq 0, ax^{2} + bx + c\) as \(a(x + r)(x + s)\). Thus, we will start as follows.

ax^{2} + bx + c & = a\Big(x^{2} + \frac{b}{a}x\Big) + c\\
& = a\Big(\Big(x + \frac{b}{2a}\Big)^{2} – \frac{b^{2}}{4a^{2}}\Big) + c\\
& = a\Big(x + \frac{b}{2a}\Big)^{2} – \frac{b^{2}}{4a} + c

Notice that at this point, we’ve completed the square! And for those of us who have a familiarity with the quadratic formula, we can see some familiar terms. We can also notice that the vertex of the quadratic is at \(x = -\frac{b}{2a}\) since it is at that value of \(x\) the square term disappears leaving us with \(-\frac{b^{2}}{4a} + c\) as the \(y\) coordinate of the vertex. And from the sign of \(a\) we can determine if the quadratic (parabola) “opens up” or “opens down”. So that handles most of the graphing stuff (we still have to find zeros).

Now setting
a\Big(x + \frac{b}{2a}\Big)^{2} – \frac{b^{2}}{4a} + c &= 0\\
\implies a\Big(x + \frac{b}{2a}\Big)^{2} &= \frac{b^{2}}{4a}-c\\
\implies \Big(x + \frac{b}{2a}\Big)^{2} & = \frac{b^{2}-4ac}{4a^{2}}\\
\implies x & = \frac{-b}{2a} \pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}\\
\implies x & = \frac{-b}{2a} \pm \frac{\sqrt{b^{2}-4ac}}{2|a|}

and we can drop the absolute value since the \(\pm\) will remain regardless of the sign of \(a\) and thus we have the quadratic formula

$$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$

So, the point here is that starting with the \(ax^{2} + bx + c\) and going down the path of completing the square and finding zeros, we obtain a form for graphing a parabola (and not shown here, but with slight rearrangement we can obtain the standard form so that we can identify the focus and directrix of a parabola), we automatically and always derive the quadratic formula, and now that we have the zeros, we also have the factorization of the quadratic. Namely, if we let $$p = \frac{b}{2a}$$ and $$q = \frac{\sqrt{b^{2}-4ac}}{2a}$$ then, $$ax^{2} + bx + c = a(x + p – q)(x + p + q)$$

Of course, the problem with teaching completing the square as a general method for arriving at many of the other useful properties of quadratics is that students are terrible with fractions. Thus, to get through Algebra, students are asked (required) to memorize facts about quadratics / parabolas. And why are students terrible with fractions? Because they were asked (required) to memorize a bunch of steps and with a bunch of gimmicky rules. See my presentation at #NJPAECET2.

Like interest rates, math misery compounds. Memorizing math is like having high interest rate debt. Eventually, it (facts to keep memorized (debt)) just becomes unmanageable and everything comes crashing down.

An Example

Factor $$x^{2} + 3x + 2$$
So, let’s complete the square
x^{2} + 3x + 2 & = \Big(x + \frac{3}{2}\Big)^{2} – \frac{1}{4}\\
\Big(x + \frac{3}{2}\Big)^{2} – \frac{1}{4} & = 0\mbox{ setting the right hand side to zero}\\
\implies \Big(x + \frac{3}{2}\Big)^{2} &= \frac{1}{4}\\
\implies x + \frac{3}{2} & = \pm\frac{1}{2}\\
\implies x & = \{-2,-1\}\\
\mbox{thus, } x^{2} + 3x + 2 &= (x + 2)(x+1) \mbox{ as expected}

Now, we can tie in some geometry with this question:

Find the side lengths of a rectangle with area equal to two square units and with perimter equal to six units.

To solve this, we would just have to factor the quadratic $$x^{2} + 3x + 2$$ Why? Because in order to factor this quadratic, we have to find \(p\) and \(q\) such that $$(x + p)(x + q) = x^{2} + 3x + 2$$ And expanding the left hand side we obtain $$(x + p)(x + q) = x^{2} + (p + q)x + pq$$

Thus, \(pq = 2\) (the area) and \(p + q = 3\) (half the perimeter). What’s nice though is that rather than picking problems that require some clever numerical guessing, we can choose any area and perimeter, so long as \(b^{2} – 4ac >= 0\). What happens if \(b^{2} – 4ac = 0\)? We have a perfect square both from the standpoint of an exponent and via geometry. And that can help to give a more firm connection with the power of two and why we may want to call it a square power.

In some sense we can view \(x^{2} + bx + c\) as a “generating function” for a rectangle with property that the area is equal to \(c\) and that the perimeter is equal to \(2b\). If the generating function yields \(p\) and \(q\) positive in the factorization of \((x + p)(x + q)\), then such a rectangle exists.

Anyhow, my point here is that if we are going to spend \(n\) lecture periods discussing quadratics, it’s better spent understanding how / why completing the square works and then using that as a jumping point for exploring all the other neat things about quadratics / parabolas, rather than memorizing formulas and (arbitrary) facts. All of those things that are memorized can be easily derived if we pay the upfront cost of a little bit of mathematical labor.

I know that this can seem controversial, especially because all the other material that I say is not worth teaching is actually part of a standard Algebra (II) course and that textbooks and online tutorials devote time and space to it. But I think the external resources are in a strange feedback loop. Math, especially at this level doesn’t really change and if a resource (eg, textbook) omits standard topics, then it looks like an omission. Thus making the resource less desirable and less likely to be adopted by a given educational institution. And so, the standard material finds its way into educational resources because the material was standard. Strange feedback loop.

Feel free to disagree in the comments.

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5 thoughts on “Factoring Quadratics Misery

  1. Bill Wood

    I find it interesting how all students seem to have the quadratic formula memorized, but few understand that it gives the zeroes of a quadratic function.

  2. Brad Luthman

    I do enjoy the side mental benefits of teaching a student to factor. I find that after getting them good at factoring, they become way more proficient at exponent rules, multiplying polynomials, combining like terms,etc. I understand the limited nature of traditional factoring but still enjoy the “puzzle solving” type skills that it brings out in a student’s brain. I will however enjoy using your post here to frame my thinking on diving into the study of graphing parabola’s with my students.

    1. Manan Shah Post author

      Absolutely, Brad! Factoring is good from a puzzle standpoint and using it to that end is good. The integer-based factoring could even be taught as a special case of a general mechanic, and I would have no objections to that.

  3. Pingback: Fractions Are Part Of The Whole Algebra Thing … | Math Misery?

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