Shortly after we start multiplying, we’re taught to *factor* numbers. So often it feels like a pointless rite of passage that we’re forced to endure. We learn it devoid of all the contexts it will be useful in the future. We learn algorithms and strategies that feel prescriptive and scripted but really are anything but. Which is confusing. You want me to do it this way, but I get to start wherever I want? It feels like a contradiction and like we must be misunderstanding something.

*Factoring* is the act of *unmultiplying*: finding the numbers (or in algebra, the expressions) that *could be multiplied together* to get the number we’re asked to factor. If we’re simply asked to *factor* a number, any collection of numbers (or expressions) that, when multiplied, make the number is a legitimate answer. If we want a more specific factoring, we have to look for an adjective or other information, such as *prime factors* or *list of factor pairs*.

## Step 0: Know your basic multiplication facts

Factoring is just one of many math topics where any weaknesses in multiplication facts will slow you down. I have worked with many, many people who have struggled with math, and the vast majority of them would have had a better experience if they had known the multiplication facts enough to start becoming automatic. Practice factoring can be one piece of helping them become automatic, but it should only be one small piece. Knowing the facts separately and fluently will make first factoring, and then many, many, many other math topics much more accessible. If you need more research-backed ideas how to increase multiplication fluency, check out my post How to Learn Multiplication Facts: A Roundup.

## Step 1: Know what “factor” means

**noun:** A *factor* of a certain Number is the same or different number that evenly divides the given Number.

**verb:** To *factor* a certain Number means to find pairs or groups of numbers that multiply together to make the given Number.

You might notice that I had to resort to capitalizing that word “Number” vs “number” in the above definitions in order to hold any hope of knowing which was which. Perhaps some official math terminology will help.

In math, a multiplication problem has the following parts: *factor* $\times$ *factor* $=$ *product*.

The *product *is the result of multiplying — the “answer” of a multiplication problem, if you will. The *factors* are the numbers that are *being* multiplied together. So, using these terms in our definitions:

**noun:** A *factor* is a number that evenly divides the given product.

**verb:** To *factor* is to find pairs or groups of factors (see noun above) that equal the product when multiplied together.

When we are asked to *factor a number*, we are asked to find numbers that *could* and *would* multiply together to make that original number.

With this understanding, is it becoming obvious why it’s important to know your multiplication facts well? We’re about to use them, in some sense “backwards”. It’s hard to turn any knowledge backward and inside out if we don’t know it really, really well.

## Step 2: Find one factor

I’ll talk more specifically about different variations and directions below. But for all of them, let’s start by finding just one factor of the given number. Even if it’s a factor we don’t use in our final answer (it’s rare not to), it tells us something about the number. There are two main ways to find this first factor. Pick the one that works for you and for the number you’re factoring:

- If the number is the result of a known multiplication fact, such as 12 ($3 \times 4=12$ or $2 \times 6=12$ or whichever relevant fact you remember first), then bonus! You’ve just found a factor
*pair*, not just a single factor, and have completed multiple steps at once! In the example of 12, you could use either factor pair, whichever you think of first or whichever is your favorite:- 3 and 4 are
*factors of*12 - 2 and 6 are
*factors of*12

- 3 and 4 are
- If the number is
*not*part of the multiplication tables you know well, then you can perform*divisibility tests*to find out what numbers are factors.

You can download and print this for free in either color or black-and-white

### Divisibility Tests

- It is common strategy to start small and methodically build up to bigger numbers until you find one that divides the given number evenly, and it can also be efficient to
*test only*(the ones on the following list), with one exception that I’ll list first because it’s the easiest:**prime**numbers*Is it divisible by 10?*Does it end in 0? If so, it’s divisible by 10 and 10 is a factor.*Is it divisible by 2?*There are three questions below that all mean the same thing. Pick whichever of the following is your favorite.- Is the number even? If yes, then it is divisible by 2 and 2 is a factor.
- Can the number be halved evenly (without fractions in the result)? If yes, then it is divisible by 2 and 2 is a factor.
- Does the number end in 0, 2, 4, 6, or 8? If yes, then it is divisible by 2 and 2 is a factor.

*Is it divisible by 3?*Add up the digits of the given number. If the sum of those digits is divisible by 3, then so is the original number! In this case, 3 is a factor.*Example:*157 is*not*divisible by 3 because $1+5+7=13$ and 13 is not on the 3s times table. 3 isa factor of 157.**not***Example:*156*is*divisible by 3 because $1+5+6=12$ and 12*is*on the 3s times table ($3\times 4=12$). 3a factor of 156.**is***Example:*123,456,789, well, let’s find out: $1+2+3+4+5+6+7+8+9=45$. $45$ is too big for most memorized 3s times tables, so let’s figure out whether $45$ is divisible by 3 using the strategy again: $4+5=9$. $9$divisible by 3 ($3\times 3=9$), so $45$ is divisible by 3, so 123,456,789 is also divisible by 3 and 3**is**123,456,789.**is a factor of**

*Is it divisible by*5? Does the number end in $5$ or $0$? If so, it is divisible by 5 and 5 is a factor.*Is it divisible by 7?** One of the harder strategies to remember, and perhaps the one that makes the least sense. Take the last digit and double it. Then consider a “new” number that is all the digits*except*the last digit. Subtract your doubled number from that “new” number. If the result is on the 7s table, then the original number is divisible by 7 and 7 is a factor of the original number.*Example:*157 is*not*divisible by 7 because:- $7 \times 2=14$
- $15-14=1$
- 1 is
*not*divisible by 7

- 12,345 well… let’s find out
- Take the last digit and double: $5 \times 2=10$
- “New” number without last digit minus doubled: $1234-10=1224$
- Is 1224 divisible by 7? Too big to know from 7s table, so let’s find out…
- Take the last digit and double: $4 \times 2=8$
- “New” number without last digit minus doubled: $122-8=114$
- Is 114 divisible by 7? Too big to know from 7s table, so let’s find out…
- Double last digit: $4 \times 2=8$
- Subtract from “rest”: $11-8=3$
- 3 is
divisible by 7, so 114 is**not**divisible by 7*not*

- So 1224 is
divisible by 7**not**

- So 12,345 is
divisible by 7. So 7 is**not**a factor of 12,345.**not**

- You can find a lot more examples by doing a web search on “divisibility by 7.”

*Is it divisible by 11?*Here we need to consider*every other digit*. It can be summarized as “a number is divisible by 11**if the difference between the sum of its digits in odd places and the sum of the digits in even places is either 0 or a multiple of 11**.” But there are a lot of math-specific words in there, and so I think it’s better understood with an example.*Example*: Is 11 a factor of 132,465,978?- Sum of digits in “odd” places of
**1**3**2**,4**6**5,**9**7**8**: $8+9+6+2+1=26$ - Sum of digits in “even” places of 1
**3**2,**4**6**5**,9**7**8: $3+4+5+7=24$ - Difference of these sums: $26-24=2$. 2 is
divisible by 11.**not** - So 11 is
a factor of 132,465,978.**not**

- Sum of digits in “odd” places of
*Example:*Is 11 a factor of 4,070?- Sum of digits in “odd” places of 4,
**0**7**0**: $0+0=0$ - Sum of digits in “even” places of
**4**,0**7**0: $7+4=11$ - Difference of these sums: $11-0=11$. 11
on the 11s table, $11\times 1 = 11$.**is** - So 11
a factor of 4,070**is**

- Sum of digits in “odd” places of 4,

There are similar divisibility rules for larger numbers, but they get more and more complicated (you can see this already starting to happen with the 7s and 11s). Most textbooks and teachers find the 11s rule a convenient place to stop teaching the divisibility rules. To test larger numbers, use your favorite division method to find out if it divides without remainders. (An advantage of using a division method to find out is that you end up with a *factor pair*, not just a single factor!)

*This rule is relatively new to the common population. When I was a kid, we just had to divide and see if it came out with or without remainders!

## Step 3: Know the specifics of the question being asked

Now that you know one factor (or maybe two), it’s time to read the question again to make sure you know what your answer should look like. Are you being asked to “find a factor pair,” “list the factor pairs,” “is ___ a factor of ____,” or “find the prime factors of”? (There may be other variations I’m not thinking of right now. I’ll add them as they come up.) The next step depends on the specific question being asked.

## Step 4: Find the paired factor

So in Step 2, you found one factor. If you used a method that found a *factor pair* (such as knowing it as a multiplication fact or using direct division), then congratulations! You’ve already completed this step and can move onto the next one.

If, however, you used a divisibility rule to find the first factor, you’ll need to do some division to find the other, paired factor. Divide your original number by the factor you found. The result will be the paired factor.

*Example*: Find a factor pair of 4,070.

*Solution*: We determined that 11 * is* a factor of 4,070. So now we need the paired factor. Divide 4,070 by 11:

[image here]

Since $4070 \div 11=370$, * a factor pair of 4,070 is 11 and 370*.

If the question only asked you to find * a* factored pair, then you can skip to step 7!

## Step 5: Repeat until all factors are found

The question might ask you to “find all factor pairs” or to “list all the factors” or something similar. In that case, we will need to repeat the above steps until we find all of them. If you approach this methodically and organized, it can be a lot easier to make sure you’ve found them all.

We’ll use the rules from Step 2 above, but we’ll also add some rules for *composite* (not prime numbers — namely, the ones we skipped). The most common way to do this is to go through the divisibility rules from smallest to largest factors. Let’s go through and example step-by-step. We’ll factor 1,100.

*Is it divisible by 2?*1,100 is even, it can be halved, and it ends in 0. So, yes, it’s divisible by 2. If we divide it in half to find the paired factor using your favorite division algorithm. I like to do this in my head by thinking of $1100=1000+100$ and halving each part. Half of 1000 is 500. Half of 100 is 50. So half of 1,100 is $500+50=550$.**One factor pair is 2 and 550.***Is it divisible by 3?*I add the digits: $1+1+0+0=2$. The result is small enough but ison the 3s times table, so 1,100 is**not**divisible by 3.**not***Is it divisible by 4?*If it’s*not*divisible by 2, then it’s*not*divisible by 4. But our example*is*divisible by 2, so it*might be*divisible by 4. Find the paired factor that goes with the 2. Is that paired factor also divisible by 2? Then this number is divisible by 4. The paired factor to 2 is 550. 550 is still divisible by 2, so let’s do that division: $550 \div2 = (500+50) \div 2 = 250+25=275$.**One factor pair is 4 and 275.***Is it divisible by 5?*1,100 ends in a 0, so yes, it’s divisible by 5. Divide it out your favorite way. I can do something similar to what I did for the 2s, or I can use long division or my other favorite algorithm. $1100 \div 5=(1000+100) \div 5 = 200+20=220$.**One factor pair is 5 and 220.***Is it divisible by 6?*Since $6=2\times 3$, the number has to be divisible by*both*2 and 3 to be divisible by 6. 1,100 is*not*divisible by 3, so it’s also*not*divisible by 6.*Is it divisible by 7?*This was the weird rule, and you might actually prefer just to divide it directly. But here let’s practice the rule. The last digit of 1,100 is 0. Double that: $0\times 2 =0$. Subtract the result from thedigits as if they were a number by themselves: $110-0=110$. Is it on the 7s table? Well, it’s bigger than what most of us memorize to, so let’s repeat. The last digit of 110 is 0. Double that: $0\times 2 = 0$. Subtract the results from the**other**digits: $11-0=11$. 11 is**other**on the 7s table, so 1,100 is**not**divisible by 7.**not***Is it divisible by 8?*One way we can think of multiplying by 8 is as a “double-double-double” since $8=2\times 2\times 2$. So to divide by 8, we could think of it as “halve-halve-halve”. Look at your 4 factor pair. If we can halve that factor paired with 4 one more time, then the number is divisible by 8. We found that $1100=4\times 275$. But 275 is odd and not divisible by 2. So 1,100 isdivisible by 8.**not***Is it divisible by 9?*Since $9=3\times 3$, if a number is*not*divisible by 3, then it is also*not*divisible by 9. (If it*is*divisible by 3, then you have to compare the paired factor and find out if it’s*also*divisible by 3, similar to the test for 4 except using the 3-rule instead of the 2-rule.) Since 1,100 was*not*divisible by 3, then it also isn’t divisible by 9.*Is it divisible by 10?*1,100 ends in a zero, so it’s divisible by 10! $1100=10*110$, so**a factor pair is 10 and 110.***Is it divisible by 11?*Remember the 11 rule had us adding together*every other digit*and subtracting the odds-placed digits and evens-placed digits. In 1,100, this looks like $1+0=1$ and $1+0=1$, then subtracting $1-1=0$. The result is 0 (or a multiple of 11), so 11 is a factor. We divide in order to find out what the paired factor is. Use your favorite division algorithm.think of the short-hand version of how many of us casually pronounce that number: “eleven hundred.” That casual pronunciation is not wrong, it’s just casual (which is why some teachers will count it as wrong). Anyway, that language makes it clear that 1,100 is 11 hundreds, or $1100=11\times 100$. So**Or****one factor pair is 11 and 100**.

Now here’s where things get a bit tedious for these bigger numbers. We’ve run out of divisibility rules. But there are potentially a lot of numbers left to check. Let’s think about the list of the next 21 numbers: 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32. Can we rule any out?

- 12: $12=3\times 4$. But 1,100 is
*not*divisible by 3, so it also won’t be divisible by 12. In fact, every multiple of three will be ruled out for this reason. - We
need to check 12, 15, 18, 21, 24, 27, or 30 because they’re multiples of 3.**don’t**

~~12~~, 13, 14, ~~15~~, 16, 17, ~~18~~, 19, 20, ~~21~~, 22, 23, ~~24~~, 25, 26, ~~27~~, 28, 29, ~~30~~, 31, 32

- 1,100 was also
divisible by 7, which knocks out all the numbers on the 7s table as well. In this case, eliminating 14 and 28 (21 was already eliminated).**not**

~~12~~, 13, ~~14~~, ~~15~~, 16, 17, ~~18~~, 19, 20, ~~21~~, 22, 23, ~~24~~, 25, 26, ~~27~~, ~~28~~, 29, ~~30~~, 31, 32

- Similar for 8: 1,100 was
divisible by 8, and 16 and 32 are on the 8s table, so they are also eliminated.**not**

~~12~~, 13, ~~14~~, ~~15~~, ~~16~~, 17, ~~18~~, 19, 20, ~~21~~, 22, 23, ~~24~~, 25, 26, ~~27~~, ~~28~~, 29, ~~30~~, 31, ~~32~~

We’ve now cut our list in half. We do need to test these, which will involve some division. But at least it’s not a lot of division problems.

*Divisible by 13?*$1100\div 13 = 84 R 8$, so 13 does*not*divide 1,100 evenly, and thus isa factor.**not***Divisible by 17?*$1100\div 17=64 R12$, so 17 does*not*divide 1,100 evenly, and thus isa factor.**not***Divisible by 19?*$1100\div 19=57 R17$, so 19 does*not*divide 1,100 evenly, and thus isa factor.**not***Divisible by 20?*$1100\div 20=55$ with no remainder, so**20 and 55 are a factor pair.***Divisible by 22?*$1100\div 22=50$ with no remainder, so**22 and 50 are a factor pair.***Divisible by 23?*$1100\div 23=47 R19$, so 23 does*not*divide 1,100 evenly, and thus isa factor.**not***Divisible by 25?*$1100\div 25=44$ with no remainder, so**25 and 44 are a factor pair.***Divisible by 26?*$1100\div 26=42 R8$, so 26 does*not*divide 1,100 evenly, and thus isa factor.**not***Divisible by 29?*$1100\div 29=37 R27$, so 29 does*not*divide 1,100 evenly, and thus isa factor.**not***Divisible by 31?*$1100\div 31=35 R15$, so 31 does*not*divide 1,100 evenly, and thus isa factor.**not**

**Important note: this process can be shorter with a good understanding of prime factorization! For example, if we found the prime factorization of 1,100 ($1100=2\times 2\times 5\times 5\times 11$), we would know right away that no other prime numbers can be factors, thus eliminating 13, 17, 19, 23, 29, and 31 right away (as well as 26 from its being $2\times 13$). But that’s a topic for a different blog post.**

## Step 6: Make sure you’ve found them all

Did you go through the list methodically in order, without skipping any number tests? I know it’s tedious and boring, but it’s the best way to make sure you don’t miss any.

Assuming you did, how do you know when you can stop? Doing every number between 1 and 1,100 seems unreasonable.

Well, have you noticed that the factors in each factor pair get closer and closer together? At some point, you end up with the first number starting to be bigger than the second number. When this happens, you’ll be repeating the same factor pairs over but in reverse order. That tells you that you’ve finished.

For our example, consider 33. Since 33 is on the 3s table, we know it won’t be a factor, but what does happen when we divide $1100 \div 33$?

$1100 \div 33=33 R11$

Ignoring the remainder for just a moment, do you see how this is the turning point for the factors to flip roles? The tested number is approximately the same size as the result of the division. This tells us we’re done.

## Step 7: State your answer clearly, visibly, and directly.

In our example of 1,100…

The * factor pairs* of 1,100 are 2 and 550, 4 and 275, 5 and 220, 10 and 110, 11 and 100, 20 and 55, 22 and 50, 25 and 44

The * factors* of 1,100 are 2, 4, 5, 10, 11, 20, 22, 25, 44, 50, 55, 100, 110, 220, 275, and 550.

You’ll notice that I almost always list this, or something similar, as the final step. I can not underemphasize the importance of this. Individual teachers may have specific ways they ask you to do this, but this is the point each time. There are multiple reasons for stating your answer clearly, visibly, and directly:

- It helps you know that you
*answered the actual question that was asked*and not some other question that distracted you along the way. - It sometimes helps you do a quick
*check for appropriateness*. If you end up writing an illogical sentence (such as one person buying 1,021 watermelons), you might notice the absurdity and rethink the problem before it is graded. - It helps your teacher spot your answer. They have a
*lot*of work to review, grade, as well as prepare. So anything to help them be more efficient is a good thing. - It helps you stay organized even if you weren’t organized in the middle.
- It helps you move the process into long-term memory as writing up the solution requires a small amount of reviewing what you did and what the goals were.
- And perhaps most importantly: It builds good habits of effective communication. Skills that are useful in
areas of life, not just math class.**all**

**You’ve Got This!**