Dear Uncle Colin,

A while back, you shared an easy way to factorise nasty quadratics. Why does it work?

Dutifully Indulging Students’ Curiousity & Reasoning In Maths. I’m Not A Nasty Teacher!

Hi, DISCRIMINANT, and thanks for your message!

Let’s start by recapping the method in the post, go through an example, and then I’ll run through the logic behind it. (I say ‘the logic behind it’ - the logic, at least for me, came long after the method.)

### The method

• Start with your quadratic, $ax^2 + bx + c = 0$
• Find two numbers, $p$ and $q$ such that:
• $pq = ac$
• $p + q = b$
• Write down $\left(ax^2 + px\right) + \left(qx + c\right)$ (which is equivalent to the original quadratic)
• Fully factorise the two brackets
• … and recombine any common factors.

### An example

Let’s go for something horrific: $18x^2 + 3x - 10$.

I need to find $p$ and $q$ such that $pq = -180$ and $p + q = 3$. What factors of 180 are relatively close to each other? -10 and +18 are too far apart, but +15 and -12 will work!

Now I write down $\left(18x^2 + 15x\right) + \left(-12x - 10\right)$ and try to factorise the two brackets.

I get $3x\left(6x + 5\right) - 2\left(6x +5\right)$, and I have a common factor of $6x+5$ between the terms…

… so I can make it $(3x-2)(6x+5)$, which I could check by expanding. It’s right!

### Why does it work?

This is going to be a little bit algebra-heavy, but you did ask. Here’s the logic:

• Suppose the quadratic $ax^2 + bx + c = 0$ factorises as $(Px+Q)(Rx+S)$.
• This expands (in baby steps) to $Px(Rx+S) + Q(Rx+S)$
• … and to $PR x^2 + (PS + QR)x + QS$
• So we have $PR = a$, $QS =c$, and $(PS+QR)=b$
• The sum of $PS$ and $QR$ is $b$
• The product of $PS$ and $QR$ is $ac$
• … so if I call these two numbers $p$ and $q$ above, I can work backwards through the steps to get the factorisation I need!

I hope that clears it up!

- Uncle Colin

* Edited 2020-07-15 to remove a spurious square. Thanks, @chrishazell72!