Factoring Quadratic Equations Comes Full Circle

2025-11-07

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

x^{2} - 5 x + 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 factoring the constant term would narrow it down to only a couple options pretty quick). 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)

But 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} - 5.5 x + 7.5

this approach wouldn’t work.

So when the student asked this today, I tried to look for a general procedure.

When factoring a quadratic 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} + 5 x + 6

This can be factored as:

(x + 2) (x + 3)

Back to the equation the student was struggling with:

x^{2} - 5 x + 4

To show the student how to solve this, I expanded the general factored representation:

(x + m) (x + n)

x^{2} + m x + n x + m n

x^{2} + (m + n) x + m n

And equated it to the original equation:

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

This produced a system of equations:

m + n = - 5

m \times n = 4

The problem can now be stated “we’re looking for 2 numbers that add up to -5 and multiply to 4.” The student was able to guess (-4, -1) was the solution, and moved on.

But I kept going to see if I could get some insight into solving the problem generally.

First, I solved for $m$ in the first equation:

m = - 5 - n

Then I substituted it into the second equation:

(- 5 - n) n = 4

- 5 n - n^{2} = 4

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

n^{2} + 5 n + 4 = 0

Solving this quadratic equation, I got 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 solving the new quadratic equation gave me their 2 possible values: -4 and -1.

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

(x + m) (x + n)

(x - 4) (x - 1)

x^{2} - 5 x + 4

so I have determined that the factorization of

x^{2} - 5 x + 4

is:

(x - 4) (x - 1)

This 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.

For x = 4:

(4 - 4) (x - 1)

0 \times (\dots)

0

For x = 1:

(1 - 4) (1 - 1)

(\dots) \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} - 5 x + 4 \Rightarrow a = 1, b = - 5, c = 4 .

\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

\frac{5 \pm \sqrt{25 - 16}}{2}

\frac{5 + \sqrt{9}}{2}, \frac{5 - \sqrt{9}}{2}

\frac{5 + 3}{2}, \frac{5 - 3}{2}

\frac{8}{2}, \frac{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} - 5.5 x + 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} - 3 x - 2.5 x + 7.5

x^{2} - 5.5 x + 7.5

Yep!