My first post in this series was about using vectors for enlargements. Today's post is about a way to save time when factorising a non-monic quadratic expression (ie a quadratic where the coefficient of x

^{2}is greater than 1) by inspection.

Back in September I had a post published on La Salle's blog about methods for factorising non-monics. I showed that by looking at old textbooks we can see how methods have changed over time.

I explained that I prefer to factorise by inspection, but most teachers these days factorise by grouping (ie 'splitting the middle term'). In response to that post I had lots of teachers either (a) try to convince me that grouping is better than inspection (b) try to convince me that another method is better than inspection. Often teachers argue that their method is great because their students really liked it in the lesson, but I'm more interested in the extent to which methods 'stick' in the long term. When I used to teach factorising by grouping my students were always happy with it at the time, but a few months later they would struggle to remember the full procedure.

Factorising by inspection is intuitive and logical, so there's no procedure to memorise. I appreciate that most teachers still prefer the grouping method, and I do not intend to try to convince anyone to change their mind. But for those of you who like factorising by inspection, here's a tip from Susan Russo (@Dsrussosusan):

Let's say you have to factorise 6x

Factorising by inspection is super-quick once you get the hang of it, but here both 6 and 12 have multiple factors so this one might take a bit longer than others.

If you list all the possibilities and check each one, there are twelve cases to check.

(6x + 1)(x + 12)

(6x + 12)(x + 1)

(6x + 2)(x + 6)

(6x + 6)(x + 2)

(6x + 3)(x + 4)

(6x + 4)(x + 3)

(2x + 1)(3x + 12)

(2x + 12)(3x + 1)

(2x + 2)(3x + 6)

(2x + 6)(3x + 2)

(2x + 3)(3x + 4)

(2x + 4)(3x + 3)

Yawn!

An expert would probably work out the correct combination fairly quickly without writing down all the options. For a novice it's a pain that there are twelve options to think about here. At first it seems like it might take a while to select the correct combination.

Susan pointed out something which should be totally obvious but hadn't occurred to me before. We can teach our students to refine their guesses in order to make this method more efficient. Here's the key:

Let's look at that list again and immediately disregard any option where there's a common factor in one or both brackets.

(6x + 1)(x + 12)

~~(6x + 12)(x + 1)~~

~~(6x + 2)(x + 6)~~

~~(6x + 6)(x + 2)~~

~~(6x + 3)(x + 4)~~

~~(6x + 4)(x + 3)~~

~~(2x + 1)(3x + 12)~~

~~(2x + 12)(3x + 1)~~

~~(2x + 2)(3x + 6)~~

~~(2x + 6)(3x + 2)~~

(2x + 3)(3x + 4)

~~(2x + 4)(3x + 3)~~

It turns out there are actually only two cases to check by inspection. Students fluent in expanding brackets should be able to do it in seconds. You can immediately see that the first option will give a large coefficient of x, so we check (2x + 3)(3x + 4) and find that it works.

For some reason I've never shared this time-saving tip with my students. I'm very grateful to Susan Russo for bringing it to my attention. Let's try it again with one more example: factorise 12x

(12x + 15)(x - 1)

(12x + 1)(x - 15)

(12x + 5)(x - 3)

(12x + 3)(x - 5)

(6x + 15)(2x - 1)

(6x + 1)(2x - 15)

(6x + 5)(2x - 3)

(6x + 3)(2x - 5)

(3x + 15)(4x - 1)

(3x + 1)(4x - 15)

(3x + 5)(4x - 3)

(3x + 3)(4x - 5)

(12x - 15)(x + 1)

(12x - 1)(x + 15)

(12x - 5)(x + 3)

(12x - 3)(x + 5)

(6x - 15)(2x + 1)

(6x - 1)(2x + 15)

(6x - 5)(2x + 3)

(6x - 3)(2x + 5)

(3x - 15)(4x + 1)

(3x - 1)(4x + 15)

(3x - 5)(4x + 3)

(3x - 3)(4x + 5)

Removing those with a common factor in one or both brackets gives us this:

~~(12x + 15)(x - 1)~~

(12x + 1)(x - 15)

(12x + 5)(x - 3)

~~(12x + 3)(x - 5)~~

~~(6x + 15)(2x - 1)~~

(6x + 1)(2x - 15)

(6x + 5)(2x - 3)

~~(6x + 3)(2x - 5)~~

~~(3x + 15)(4x - 1)~~

(3x + 1)(4x - 15)

(3x + 5)(4x - 3)

~~(3x + 3)(4x - 5)~~

~~(12x - 15)(x - 1)~~

(12x - 1)(x + 15)

(12x - 5)(x + 3)

~~(12x - 3)(x + 5)~~

~~(6x - 15)(2x + 1)~~

(6x - 1)(2x + 15)

(6x - 5)(2x + 3)

~~(6x - 3)(2x + 5)~~

~~(3x - 15)(4x + 1)~~

(3x - 1)(4x + 15)

(3x - 5)(4x + 3)

~~(3x - 3)(4x + 5)~~

So this time we eliminated half the possibilities. It's not as time-saving as in the first example, but still helpful. My next step would be to try the less extreme numbers (ie not those involving a 12x or a 15) so that gives me only four options to test initially. For an experienced factoriser it's fairly quick to see that (3x + 5)(4x - 3) works.

Of course, in reality we never really list out all the options and then decide what to eliminate. What most people actually do when faced with 12x

This isn't a big game-changer and it doesn't help with every quadratic, but I like things that save us time. When solving a long, complicated problem at A level, it's good to able to factorise quadratics efficiently.

If you hadn't realised that you can quickly eliminate options in this way then I hope this was helpful.

If you want to have a play with this, there are lots of non-monic expressions to factorise here.

I explained that I prefer to factorise by inspection, but most teachers these days factorise by grouping (ie 'splitting the middle term'). In response to that post I had lots of teachers either (a) try to convince me that grouping is better than inspection (b) try to convince me that another method is better than inspection. Often teachers argue that their method is great because their students really liked it in the lesson, but I'm more interested in the extent to which methods 'stick' in the long term. When I used to teach factorising by grouping my students were always happy with it at the time, but a few months later they would struggle to remember the full procedure.

Factorising by inspection is intuitive and logical, so there's no procedure to memorise. I appreciate that most teachers still prefer the grouping method, and I do not intend to try to convince anyone to change their mind. But for those of you who like factorising by inspection, here's a tip from Susan Russo (@Dsrussosusan):

Let's say you have to factorise 6x

^{2}+ 17x + 12.Factorising by inspection is super-quick once you get the hang of it, but here both 6 and 12 have multiple factors so this one might take a bit longer than others.

If you list all the possibilities and check each one, there are twelve cases to check.

(6x + 1)(x + 12)

(6x + 12)(x + 1)

(6x + 2)(x + 6)

(6x + 6)(x + 2)

(6x + 3)(x + 4)

(6x + 4)(x + 3)

(2x + 1)(3x + 12)

(2x + 12)(3x + 1)

(2x + 2)(3x + 6)

(2x + 6)(3x + 2)

(2x + 3)(3x + 4)

(2x + 4)(3x + 3)

Yawn!

An expert would probably work out the correct combination fairly quickly without writing down all the options. For a novice it's a pain that there are twelve options to think about here. At first it seems like it might take a while to select the correct combination.

Susan pointed out something which should be totally obvious but hadn't occurred to me before. We can teach our students to refine their guesses in order to make this method more efficient. Here's the key:

**each bracketed expression shouldn't contain any common factors**. For example if you have a 2x then you can't put an even number in with it.Let's look at that list again and immediately disregard any option where there's a common factor in one or both brackets.

(6x + 1)(x + 12)

(2x + 3)(3x + 4)

It turns out there are actually only two cases to check by inspection. Students fluent in expanding brackets should be able to do it in seconds. You can immediately see that the first option will give a large coefficient of x, so we check (2x + 3)(3x + 4) and find that it works.

For some reason I've never shared this time-saving tip with my students. I'm very grateful to Susan Russo for bringing it to my attention. Let's try it again with one more example: factorise 12x

^{2}+ 11x - 15. Here's the massive list of 24 options to consider:(12x + 15)(x - 1)

(12x + 1)(x - 15)

(12x + 5)(x - 3)

(12x + 3)(x - 5)

(6x + 15)(2x - 1)

(6x + 1)(2x - 15)

(6x + 5)(2x - 3)

(6x + 3)(2x - 5)

(3x + 15)(4x - 1)

(3x + 1)(4x - 15)

(3x + 5)(4x - 3)

(3x + 3)(4x - 5)

(12x - 15)(x + 1)

(12x - 1)(x + 15)

(12x - 5)(x + 3)

(12x - 3)(x + 5)

(6x - 15)(2x + 1)

(6x - 1)(2x + 15)

(6x - 5)(2x + 3)

(6x - 3)(2x + 5)

(3x - 15)(4x + 1)

(3x - 1)(4x + 15)

(3x - 5)(4x + 3)

(3x - 3)(4x + 5)

Removing those with a common factor in one or both brackets gives us this:

(12x + 1)(x - 15)

(12x + 5)(x - 3)

(6x + 1)(2x - 15)

(6x + 5)(2x - 3)

(3x + 1)(4x - 15)

(3x + 5)(4x - 3)

(12x - 1)(x + 15)

(12x - 5)(x + 3)

(6x - 1)(2x + 15)

(6x - 5)(2x + 3)

(3x - 1)(4x + 15)

(3x - 5)(4x + 3)

So this time we eliminated half the possibilities. It's not as time-saving as in the first example, but still helpful. My next step would be to try the less extreme numbers (ie not those involving a 12x or a 15) so that gives me only four options to test initially. For an experienced factoriser it's fairly quick to see that (3x + 5)(4x - 3) works.

Of course, in reality we never really list out all the options and then decide what to eliminate. What most people actually do when faced with 12x

^{2}+ 11x - 15 is write down (3x )(4x ) and then (often in their head rather than on paper) try some numbers that multiply to give -15. So it's helpful to remember that there's no point putting and 3 or a 15 in the first bracket.This isn't a big game-changer and it doesn't help with every quadratic, but I like things that save us time. When solving a long, complicated problem at A level, it's good to able to factorise quadratics efficiently.

If you hadn't realised that you can quickly eliminate options in this way then I hope this was helpful.

If you want to have a play with this, there are lots of non-monic expressions to factorise here.

I just realized another strategy today: If you're trying to simplify an expression like: (x^4 + x^2 + 1)/(x^2+ x + 1) Rather than trying to factor the denominator just give polynomial division a try.

ReplyDeleteNice post Jo, thanks. Its got ke thinking about ho2 i t2ach and how i would do it.

ReplyDeleteregarding stickability of the grouping method - i think it sticks much better if you use it for monic quadratics too. The issue with forgetting comes when you teach them to factorise by inspection for monic then bust an entirely different method for non-monic. It alsohelps is you teach full distributive method for expanding, rather than foil or crab claw or smiley face type methods.

Totally agree that the grouping method will stick better if you use the same method for monics. Nice post about that here . Problem is, I find that using grouping for monics is unnecessarily inefficient - I don't want to teach a multi-step procedure for something that can just be written down.

DeleteInteresting about the full distributive method for expanding. I watched it taught that way recently ie (x + 3)(x + 4) = x(x + 4) + 3(x + 4). I'm not sure about that one. Again, I worry about using unnecessary steps when it can just be done as 'each by each' (no smiley faces!), but perhaps there are advantages to doing it that way. Not sure.

After many years of teaching different methods of factorising trinomials, I find this method below (which I call Slide and Divide and Bottoms up) the one that sticks best. However, it does seem like magic Maths. When I tried to explain why this method works, I completely lost the kids, so now I don't even bother.

ReplyDeleteSLIDE AND DIVIDE AND BOTTOMS UP METHOD

Example: Factorise 12x²-5x-2

STEPS:

On the right hand side of the page (rough work section), multiply the coefficient of x² by the constant i.e. we slide the 12 across to join the -2 and multiply them.

x²-5x-24

Factorise this trinomial like you've been doing, by inspection

(x-8)(x+3)

Now divide the constants in each bracket by the coefficient of x² i.e. 12

(x-8/12)(x+3/12)

=(x-2/3)(x+1/4) Simplify the fractions

Bring the denominators up to be the coefficients of the x’s in brackets i.e. bottoms up!

(3x-2)(4x+1) This is now your final answer and can

be written on the neat side of your

page.

I've seen that a few times before and I'm not a fan tbh. I'm surprised you find it sticks in the long term - how have you assessed that? There are a number of steps to remember (and those steps have no logical connection) which would make it hard to remember. As you say, it seems like magic!

DeleteI'd argue that if you can't explain a technique well then its better to use an alternative that is comprehensible. And providing no explanation reinforces a view of math as an arbitrary set of "incantations" to be memorized rather than a logical system with structure. That actually causes a bit of damage.

ReplyDeleteI agree that it's not a perfect method (Is there a perfect method?) but I disagree that no explanation is a problem and that any damage is caused. As long as they can see the relationship between the question and the answer; understand what factorisation means i.e. breaking an expression/quantity into a product of two or more quantities and be able to factorise trinomials in different situations, then they do not need to know why it works.We do this all the time in Maths. We don't always show them why something works before teaching them how to do it.I think that if we did that all the time, that is when damage is caused.

ReplyDeleteI agree with you on that - insisting on full conceptual understanding before teaching procedures isn't always helpful. I'm more concerned with the fact that students are less likely to remember a string of steps with no logical connection.

DeleteIs there an argument against finding the roots of the function first and factorising second? I go straight for the discriminant because it tells you immediately whether or not it is possible to factorise it, and once you know that, it's only a hop and a skip to the factors...

ReplyDeleteIt just seems a rather convoluted method for something that can be done in seconds.

DeleteI'd thought of doing it that way too - once you have the first factor it's easy to factorise by inspection (even if it's a quartic or greater) - but it seems like overkill for just quadratics. I do show that it also works for quadratics when teaching factor theorem & factorising for cubics though.

DeleteEven with 12x2 + 11x - 15? As someone who enjoyed mental maths, I personally always liked inspection because of the Countdown-esque buzz of chasing down the numbers, but there are still plenty of A* grade pupils who struggle to plough through all the possibilities and dislike the trial and error element of inspection.

ReplyDeleteFor less number-crunchy students who take time to work through the options, it gives them a linear route to the answer and reinforces the quadratic formula and the connection between roots and factors. Plus if it's in a calculator paper, they get the roots (and hence the answer) pretty much instantly.

I always present classes with a range of options for factorising non-monics. When the Classwiz with its equation solver gets involved, they find this approach so easy it feels like cheating.

I tend not to persist with inspection with kids who aren't up for it - do you find that most kids, even less able ones, settle into it?

I'm thinking of this at GCSE where my students do not have the Classwiz. You're right, if they have a Classwiz they can just use equation solver and this post isn't relevant to them.

DeleteI feel what you say about inspection rings true, Jo. One thing I find really helps when I teach inspection as a method is to start by giving the students a ton of examples with a and c both prime numbers. They'll see the benefit of inspection straight away!

ReplyDelete