quadratic.typ

---
title: Factoring Quadratic Equations Comes Full Circle
---

A student in my math class was having trouble with factoring the following quadratic equation:

$
x^2 -5x + 4
$

She explained that no matter what numbers she tried, it was never working out.

This relates to my experience doing this kind of factoring: at the start, I would just guess numbers (they would never really go outside the range of -6 to 6, and thinking of the ways the constant term would be factored would narrow it down to only a couple options pretty quick),
then eventually, the teachers would start reusing numbers and I'd just remember the answers.

For this one I remember the factorization is:

$
(x - 4)(x - 1)
$

This always struck me as somewhat unsatisfying. The problems would be hand-picked to have nice factoring solutions, but even if we were to be presented with something mildly different, like

$
x^2 - 5.5x + 7.5
$

this approach wouldn't work. This one isn't too bad; I could multiple the whole thing by 2 and recover my whole numbers:

$
2x^2 - 11x + 15
$

then go back to my "I hope I remember the factorization" approach...

This also reminds me how my teachers would generally not us to factor anything
with an x^2 constant other than 1 (this example has a leading coefficient of 2)
since the factoring gets much more difficult.

So when the student asked this today, I brought out the equations of factoring, in the hopes of shedding some light on a general approach.

When factoring an equation of the form

$
x^2 + b x + c
$

The goal is to factor it into the form

$
(x + m)(x + n)
$

where m and n are positive or negative numbers.

Just to review, here's a simple example (and one that I remember seeing dozens of times):

$
x^2 + 5x + 4
$

can be factored as:

$
(x + 4)(x + 1)
$

So back to the equation the student was struggling with:

$
x^2 -5x + 4
$

What I did in the class was expand the polynomial:

$
(x + m)(x + n)
$

$
x^2 + m x + n x + m n
$

$
x^2 + (m + n)x + m n
$

And equate it to the original equation:

$
x^2 + (m + n)x + m n = x^2 -5x + 4
$

This produces a system of equations:

$
m + n = -5\

m times n = 4
$

At this point, the student was able to guess (-4, -1) was the solution.

But I kept going to see if I could get a closed-form solution.

First, I solved for m in the first equation:

$m = -5 -n$

Then I substituted it into the second equation:

$(-5 -n)n = 4$

$-5n -n^2 = 4$

Huh? A quadratic equation! Putting it into standard quadratic form, I get:

$n^2 + 5n + 4 = 0$

Solving _this_ quadratic equation, I get the following 2 answers: -4 and -1.

All of a sudden, the pieces started to click into place. m and n are interchangeable at this point,
and the 2 answers for m and n are -4 and -1.

Arbitrarily assigning m = -4 and n = -1, I can multiply them out to recover the original formula:

$
(x + m)(x + n)
(x - 4)(x - 1)
x^2 -5x +4
$

so I have determined that the factorization of

$x^2 -5x +4$

is:

$(x - 4)(x - 1)$

Which reminded me of one of the reasons you'd want to factor a quadratic equation in the first place: once it's factored, the solutions are whatever x will turn its expression into 0.

x = 4:

$
(4 - 4)(x - 1)\
0 times (...)\
0
$

x = 1:

$
(1 - 4)(1 - 1)\

(...) times 0 \ 

0
$

So if you can find the answers to a quadratic equation, you can factor it.

To show this, let's use the quadratic formula on the original polynomial:

$
x^2 -5x + 4 => a = 1, b = -5, c = 4.\

(-b plus.minus sqrt(b^2 - 4a c)) / (2a)\

5 plus.minus sqrt(25 - 16) / 2\

(5 + sqrt(9)) / 2,
(5 - sqrt(9)) / 2\

(5 + 3) / 2,
(5 - 3) / 2\

8 / 2,
2 / 2\

4 "and" 1.
$

Knowing these are the roots, I can immediately construct the factored form:

$
(x - 4)(x - 1)
$

Let's use this knowledge to factor a quadratic equation that has noninteger factors:

$
x^2 - 5.5x + 7.5
$

Plugging it in to the quadratic formula yields:

$
3 "and" 2.5.
$

so it can be factored thus:

$
(x - 3)(x - 2.5)
$

Checking my work by multiplying it back out:

$
(x - 3)(x - 2.5)\
x^2 -3x -2.5x + 7.5\
x^2 -5.5x + 7.5
$

Yep!