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 x2 is greater than 1) by inspection.
Back in September I had a post published a blog post for La Salle 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 6x2 + 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)
(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 12x2 + 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 + 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 12x2 + 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 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 6x2 + 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 12x2 + 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 12x2 + 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.