Factorising quadratics by focusing on the sum first

This blog post is about my way of helping students factorise quadratic expressions by inspection, which is the opposite of how most people do it.

Factorising

When you multply out two monic linear factors to make a quadratic, the same thing always happens:

You end up with the sum of the two constant terms as the coefficient of $x$, and the product of the two constant terms as the final constant term.

Therefore, if you want to do this process backwards – that is, to factorise a quadratic expression – then you need to think of two numbers that add to give the coefficient of $x$ and multiply to give the constant term.

For example, to factorise $x^2+5x+6$, you need to think of two numbers that add to give 5 and multiply to give 6.

There are at least two ways you could go about doing this systematically.

One way is to think of pairs of numbers that multiply to give 6, and then test them to see if they also add to give 5. So, you’d think of 1×6 and test 1+6=7, which isn’t right. And you’d think of 2×3 and test 2+3=5, which is right. So your factorisation is $(x+2)(x+3)$.

Another way is to think of pairs of numbers that add to give 5, and test them to see if they also multiply to give 6. So, you’d think of 1+4 and test 1×4=4, which isn’t right. And you’r think of 2+3, and test 2×3=6, which is right. So your factorisation is $(x+2)(x+3)$.

Every maths teacher I’ve ever met tells students to list the product first and check the sum. I think that it’s much better to tell students to do the sum first and check the product.

Examples

Let me do several examples to compare the sum first approach with the product first approach.

Example 1: $x^2+13x+40$

Product first

I need two numbers that multiply to give 40, which could be 1×40, 2×20, 4×10, 5×8 and I think that’s it. The matching sums are 41, 22, 14, 13. So the numbers I need are 5 and 8 and the factorisation is $(x+5)(x+8)$.

Sum first

I need two numbers that add to give 13. I’ll start at 10+3, and the product is 10×3=30, which is too low. Now I’ll try 9+4, and the product is 9×4=36, which is higher but still too low. Now I’ll try 8+5, and the product is 8×5=40, which is just right. So the numbers I need are 8 and 5, and the factorisation is $(x+8)(x+5).$

Example 2: $x^2+20x+91$.

Product first

I need two numbers that multiply to give 91. 1×91 obviously, and the matching sum is 1+91=92. So I need something else. What else? 2? Doesn’t go. 3? Doesn’t go. 5? Doesn’t go. 7? Oh yes that does work because 91=70+21, which is 10 and 3 sevens, so 91=7×13. The matching sum is 7+13=20, so that works. The factorisation is $(x+7)(x+13)$.

Sum first

I need two numbers that add to give 20. My first thought is 10+10, and the matching product is 10×10=100, which is too high. Now 11+9=20, and 11×9=99, which is still too high, but lower. Next, 12+8=20 and 12×8=80+16=96, which is still too high, but lower. Next 13+7=20, and 13×7=70+21=91, which is just right. The factorisation is $(x+13)(x+7)$.

Example 3: $x^2+30x+144$.

Product first

I need two numbers that multiply to give 144. What goes into 144? It’s 12×12, so 1, 2, 3, 4, 6, 12 will all work. Have I missed anything? Oh 9, taking a 3 from each 12. Anything over 12 will go with one of the small numbers. Right, so what have we got?
1×144, but 1+144 is way too big.
2×72, but 2+72 is too big.
3×48, but 3+48 is too big.
4×36, but 4+36 is too big.
6×24, and 6+24 is just right.
So the factorisation is $(x+6)(x+24)$.

Sum first

I need two numbers that add to 30. How about starting with 10 and 20?
10+20=30, 10×20=200, too big.
11+19=30, 11×19=110+99=209, that’s worse. I should be going the other way.
9+21=30, 9×21=189, still too big, but the right direction.
8+22=30, 8×22=160+16=176, closer.
7+23=30, 7×23=140+21=161, closer.
6+24=30, 6×24=120+24=144, just right.
So the factorisation is $(x+6)(x+24)$.

Reasons for a sum first approach

The above examples point to the many reasons why I think focusing on the sum first is better than focusing on the product first. I’m going to list them, but they overlap quite a bit, so be prepared for me to repeat myself a lot in the explanations below each reason.

Reason 1

Doing the product first requires you to know or figure out what numbers divide into another. Doing the sum first doesn’t require any special knowledge about factors.

Look at Example 2. We had to figure out that 91 had 7 as a factor at all before we could get to the answer. With the sum, it just fell out along the way.

You may argue that guessing factors is a really important skill, and I don’t disagree, but honestly students don’t have much practice at that when they start factorising quadratics, and it’s a huge barrier to success. Focusing on the sum first allows them early success without the need for this skill. And you know what, they do a lot of multiplications along the way and might even notice what numbers tend to be multiples of what other numbers.

Also look at Example 3. The number 144 has a lot of factors, and you kind of need to find all of them to be able to have things to try to see if they come out to the right sum. Most worked examples for students dont even list all the options for factors, but just zero in on the magically right one, picking from an unspoken list in the teacher’s head. With the sums first approach, it doesn’t matter if you missed a factor.

And look, all the work you’ve done in the past to get good at seeing factors isn’t wasted! If one of the sums is 7+23 so you test 7×23 going for 144, you can actually say to yourself that 144 isn’t a multiple of 7 and just skip that one. I actually think developing this instinct for ways to shortcut the process can be quite an exciting idea to students.

Reason 2

With sums first, you can get started right away.

When you do product first, you have to think of some factors to begin with, and it’s very rare that 1×something is going to work, so there’s this job to do before you can even get started. When you do sums first, it’s not hard to think of a sum that works and you can just get on with it.

And there’s no wrong place to start either. You can just do a couple and you’ll know then if you’re going in the right direction. (See what happened with the one with 144.) So there’s no need to worry about your first inspiration – you can just get going.

Reason 3

With sums first, you feel like you’re getting somewhere.

When you investigate the sum first, you systematically change them by 1 each time and the product changes along with it, getting closer and closer to the right answer. There is a real feeling of progress, like the work is paying off. And to reinforce the previous Reason, this feeling happens right at the start, rather than having to wait for finding factors first.

I will concede that you can be systematic with the product first approach too, as you saw in my example with the 144. But to many students, the examples they see seem random, or worse, go straight to the right answer with no trial and improvement. If you do want to do product first, then I recommend being more systematic about it so that students can feel like they’re getting somewhere, rather than waiting for the lightning strike of the right one.

Note that the feeling of getting somewhere has another advantage: if you’re a long way away from the right result, it makes you feel safer to skip some steps to get there quicker. This way lies developing instincts for when some combinations of numbers are unlikely to work.

Reason 4

With sums first, there’s cool things you can help students to notice.

I personally think the experience of running through several possible sums and testing the products is some excellent fuel for helping students notice cool things, which are totally lost on a products-first approach.

For example, in the example with $20x$, the highest possible product happened when the sum was 10+10=20. That is, when it was two of the same number. This is very cool and that way lies completing the square. Also the further apart the numbers were, the further away the next product is from this one. Indeed, the differences were two apart.

And I’ve already mentioned students noticing that a certain sum would require 144 to be a multiple of 7 and skipping it, that sort of thing makes the skill of noticing factors feel like a cheat code they’ve discovered, rather than a burden upon them. That sort of noticing is empowering for lots of students.

More examples

You’ve probably noticed that all the examples I’ve shown so far have had all positive coefficients, and they’re all monic (the coefficient of $x^2$ is 1). Well it’s time for some examples to deal with that. First I’ll deal with the negative coefficients, then later I’ll deal with non-monic quadratics. Mostly I’ll just do them straight using the sum first approach as if I didn’t know the answers yet, rather than compare them to a product first approach.

Example 4: $x^2-13x+40$

We have to think of two numbers that add to -13 and multiply to 40.

Positive numbers won’t add to a negative number, so I need two negative numbers, which will indeed multiply to a postive number.

(-10)+(-3)=-13, (-10)×(-3)=30, too low.
(-11)+(-2)=-13, (-11)×(-2)=22, even lower. Need to go the other way.
(-9)+(-4)=-13, (-9)×(-4)=36, higher.
(-8)+(-5)=-13, (-8)×(-5)=40, correct!

So the factorisation is $(x-8)(x-5)$.

Example 5: $x^2+3x-40$

We have to think of two numbers that add to 3 and multiply to -40.

If you think about the product first, there’s twice as many options as there were before, because while 4×10=40, both (-4)×10 and 4×(-10) are -40 and you have to decide which one. If you only think about the product long enough to realise you need one positive and one negative, then you can start your search with sums that add to 3 like this:
4+(-1)=3, 4×(-1)=-4, not low enough.
5+(-2)=3, 5×(-2)=-10, lower, so I’m going the right way.
6+(-3)=3, 6×(-3)=-18, lower.
7+(-4)=3, 7×(-4)=-28, getting there.
8+(-5)=3, 8×(-5)=-40, and we’re there.

I probably could have skipped a couple since there was a long way to go, but it was so pleasant watching it get closer.
Anyway, the factorisation is $(x+8)(x-5)$.

Example 6: $x^2-11x-26$

We have to think of numbers that add to -11 and multiply to give -26.

I’ll need a positive and a negative number to get a negative product, so let me start with -12+1.

-12+1=-11, (-12)×1=-12, not low enough
-13+2=-11, (-13)×2=-26, correct!

So the factorisation is $(x-13)(x+2)$.

What if I had decided to start with something less obvious, like -20+9?

-20+9=-11, (-20)×9=-180, way too low.
-21+10=-11, (-21)×10=-210, even lower, so I need to go the other way.
I’ll skip some since I was so far away.
-15+4=-11, (-15)×4=-60, getting closer.
-14+3=-11, (-14)×3=-42, getting closer.
-13+2=-13, (-13)×2=-26, just right!

Example 7: $x^2+11x+26$

We have to think of numbers that add to 11 and multiply to give 26.

Two positive numbers will work, so I’ll start with 10+1.

10+1=11, 10×1=10, too low.
9+2=11, 9×2=18, higher so I’m going the right way.
8+3=11, 8×3=24, closer.
7+4=11, 7×4=28, too big.

So there’s definitely a factorisation that will work with roots somewhere between 7 and 8 and between 3 and 4, but there’s not one with integers.

Example 8: $x^2+8x+20$

We have to think of numbers that add to 8 and multiply to give 20.

I’ll start with 1+7=8, 1×7=7, too low.
2+6=8, 2×6=12, too low, but closer, so I’m going the right way.
3+5=8, 3×5=15, too low.
4+4=8, 4×4=16, too low.
But there’s nowhere else to go from here. I’ll never get to 20.
So this one doesn’t factorise at all.

Interlude

It’s time to stop for a short break. I’m hoping that this set of examples has convinced you that this approach has some merit for helping students understand how quadratic equations work, and indeed making the process a bit more playful.

I also sneakily wanted to cover some objections people have brought up, such as how you could be sure it doesn’t factorise if there’s infinitely many choices for numbers that add to the x-coefficient.

I just have one more thing to deal with, which is what do do with a non-monic quadratic. I’m just going to do one example in two ways.

Two more examples

Example 9a: $6x^2+x-12$

There is this method called by many “the ac method” which allows you to factorise a non-monic polynomial. I didn’t learn it at school, so I don’t think of it first, but it’s always something people bring up when I talk about factorising quadratics.

The way it works is you multiply the constant term and the leading coefficient, and then think of two numbers that add to the x-coefficient and multiply to give this new answer. (If your quadratic was $ax^2+bx+c$, that means making the sum $b$ and the product $ac$, hence the name of “ac method”.) Then you split the x-term into two parts with these numbers as the coefficents and continue from there.

(For a proof, consider the product of two linear factors:

Notice how the numbers and add to give the x-coefficient and multiply to the same answer as the x²-coefficient times the constant term.)

Anyway, this still requires you to find two numbers with a specific sum and a specific product, so you can still do sum first.

The quadratic is $6x^2+x-12$. So I need numbers that multiply to give 6×(-12)=-72 and add to give 1. I’ll need a positive and a negative.

2+(-1)=1, 2×(-1)=-2, which is way too high.
3+(-2)=1, 3×(-2)=-6, which is lower, so I’m going the right way, but I have a long way to go. I’ll skip some.
6+(-5)=1, 6×(-5)=-30, which still has a long way. I’ll skip some more.
10+(-9)=1, 10×(-9)=-90, which is too far, but quite close.
9+(-8)=1, 9×(-8)=-72, which is just right

So I need to split the $x$ into $9x$ and $-8x$.

(You could argue that if I went product first, I might have realised immediately that 8 and 9 would be right, but I can guarantee you that a heap of students would not realise that. This way, they’ll get there in the end.)

So,

If you wanted this in fully factorised form so that it shows the roots, you’d have to pull out a 3 from one factor and a 2 from the other to get

(It’s worth noting that for many people, this “splitting the middle term and then factorising twice” thing the way that you’re supposed to do all quadratic factorisations, including the monic ones, which I can see the appeal of if I’m honest. But I’m not rewriting my entire set of examples now.)

Example 9b: $6x^2+x-12$

There is a far more prosaic approach than the ac method, which is just to do what I’ve been doing all along but with fractions. Let me show you:

Now I’ll factorise the monic quadratric in the brackets there. I need two numbers that add to give 1/6 and multiply to give -2. They’ll have to be a positive and a negative.

2/6+(-1/6)=1/6, 2/6×(-1/6)=1/3×(-1/6)=-1/18, which is not low enough.
3/6+(-2/6)=1/6, 3/6×(-2/6)=1/2×(-1/3)=-1/6, which is lower but not low enough, and I’ve got quite a long way to go.
7/6+(-6/6)=1/6, 7/6×(-6/6)=7/6×(-1)=-7/6, so much closer.
8/6+(-7/6)=1/6, 8/6×(-7/7)=… yeah that won’t work out right.
9/6+(-8/6)=1/6, 9/6×(-8/6)=3/2×(-4/3)=-2 yay!

So the factorisation is $6\left(x+\frac32\right)\left(x-\frac43\right)$.

I have to say I prefer this one to the other one in a lot of ways. But yes a big fly in the ointment is the fraction arithmetic. But honestly this seems to me to be quite low stakes, and it certainly gives a lot of practice! You have to decide how you want to play it.

Oh, and why did I choose to count in sixths? Well it turns out that in a monic quadratic with rational coefficients, if there are any rational solutions, they’ll be able to be written with the common denominator of the coefficients. (But that’s another story and shall be told at another time.)

Conclusion

So, I’ve given a lot of examples to show how the reasoning works when you factorise quadratic expressions by first focusing on the sum and checking the product, rather than the other way around as is more traditional. And I’ve tried to describe why I think it has a lot of advantages. I hope you give it some consideration when you next help students with their factorising.