# When is a Quadratic “factorisable”?

There are 3 standard ways of solving quadratic equations once they are in the form:

`ax² + bx + c = 0`

They are:

1. Factorise
2. Complete the square
3. Use the formula

I think I generally teach them in that order probably without much thought as to why. I guess the formula needs to be derived by using completing the square and factorising seems to follow on from multiplying out double brackets, which comes before all of this.  The I question that I sometimes get from students is “what’s the point in learning factorising if the two other methods always work?”.  Well, it’s quicker and you can do it on a non-calc exam is probably a standard response.

But have you tried using the formula without a calculator to solve a quadratic that you know will factorise?  Have a go.  Plug this:

`x² - 3x - 28 = 0`

into this:

…and solve without a calculator.

It’s a surprisingly satisfying experience, one that I would not want to deny my students.

You’ll need to know your square numbers because b² – 4ac will always give you a square number for quadratics that factorise. But the arithmetic is perfectly reasonable and is likely to be so for most quadratics that can be factorised.

When I did this recently, I had a great question from one of my students, as I was aching my brain trying to make up a quadratic that I knew would factorise.   “If you just picked one randomly, what are the chances that you would be able to factorise it?”

I’ve since had chance to investigate this further.  It’s a great question and there is a lesson in here, or at least an extension question to explore once the fundamentals of using the formula are secure.

I started approaching it by using the formula and focussing on b² – 4ac and what values of a, b and c would yield square numbers.  To simplify the problem, I started with a=1, so I was looking for when b² – 4c = 1,4,9,16,25, etc.

I then looked at it from the other end, i.e. starting with e.g. (x+1)(x+n) what values of a, b, and c are yielded.  Then vary further to look at (x+m)(x+n).  I just started with a few values of n and m to see if I could spot patterns.  I won’t spoil the fun by revealing those patterns, but this is very open-ended and could provide some intrigue for the right learners.