Decomposition of numbers into prime factors, methods and examples of decomposition. Factoring a number
What does factoring mean? This means finding numbers whose product is equal to the original number.
To understand what it means to factor, let's look at an example.
An example of factoring a number
Factor the number 8.
The number 8 can be represented as a product of 2 by 4:
Representing 8 as a product of 2 * 4 means factorization.
Note that this is not the only factorization of 8.
After all, 4 is factorized like this:
From here 8 can be represented:
8 = 2 * 2 * 2 = 2 3
Let's check our answer. Let's find what the factorization is equal to:
That is, we got the original number, the answer is correct.
Factor the number 24 into prime factors
How to decompose into prime factors number 24?
A number is called prime if it is divisible by only one and itself.
The number 8 can be represented as the product of 3 by 8:
Here the number 24 is factorized. But the assignment says “factor the number 24 into prime factors,” i.e. It is the prime factors that are needed. And in our expansion, 3 is a prime factor, and 8 is not a prime factor.
What's happened factorization? This is a way to turn an inconvenient and complex example into a simple and cute one.) A very powerful technique! It is found at every step in both elementary and higher mathematics.
Such transformations in mathematical language are called identical transformations of expressions. For those who are not in the know, take a look at the link. There is very little, simple and useful.) The meaning of any identity transformation is a recording of the expression in another form while maintaining its essence.
Meaning factorization extremely simple and clear. Right from the name itself. You may forget (or not know) what a multiplier is, but you can figure out that this word comes from the word “multiply”?) Factoring means: represent an expression in the form of multiplying something by something. May mathematics and the Russian language forgive me...) That's all.
For example, you need to expand the number 12. You can safely write:
So we presented the number 12 as a multiplication of 3 by 4. Please note that the numbers on the right (3 and 4) are completely different than on the left (1 and 2). But we understand perfectly well that 12 and 3 4 same. The essence of the number 12 from transformation hasn't changed.
Is it possible to decompose 12 differently? Easily!
12=3·4=2·6=3·2·2=0.5·24=........
The decomposition options are endless.
Factoring numbers is a useful thing. It helps a lot, for example, when working with roots. But factoring algebraic expressions is not only useful, it is necessary! Just for example:
Simplify:
Those who do not know how to factor an expression rest on the sidelines. Those who know how - simplify and get:
The effect is amazing, right?) By the way, the solution is quite simple. You'll see for yourself below. Or, for example, this task:
Solve the equation:
x 5 - x 4 = 0
It is decided in the mind, by the way. Using factorization. We will solve this example below. Answer: x 1 = 0; x 2 = 1.
Or, the same thing, but for the older ones):
Solve the equation:
In these examples I showed main purpose factorization: simplifying fractional expressions and solving some types of equations. Here's a rule of thumb to remember:
If we have a scary fractional expression in front of us, we can try factoring the numerator and denominator. Very often the fraction is reduced and simplified.
If we have an equation in front of us, where on the right there is zero, and on the left - I don’t understand what, we can try to factorize the left side. Sometimes it helps).
Basic methods of factorization.
Here they are, the most popular methods:
4. Expansion of a quadratic trinomial.
These methods must be remembered. Exactly in that order. Complex examples are checked for all possible ways decomposition. And it’s better to check in order so as not to get confused... So let’s start in order.)
1. Taking the common factor out of brackets.
A simple and reliable way. Nothing bad comes from him! It happens either well or not at all.) That’s why he comes first. Let's figure it out.
Everyone knows (I believe!) the rule:
a(b+c) = ab+ac
Or, more general view:
a(b+c+d+.....) = ab+ac+ad+....
All equalities work both from left to right and vice versa, from right to left. You can write:
ab+ac = a(b+c)
ab+ac+ad+.... = a(b+c+d+.....)
That's the whole point of taking the common factor out of brackets.
On the left side A - common multiplier for all terms. Multiplied by everything that exists). On the right is the most A is already located outside the brackets.
Practical use Let's look at the method using examples. At first the option is simple, even primitive.) But on this option I will note ( green) Very important points for any factorization.
Factorize:
ah+9x
Which general does the multiplier appear in both terms? X, of course! We will put it out of brackets. Let's do this. We immediately write X outside the brackets:
ax+9x=x(
And in parentheses we write the result of division each term on this very X. In order:
That's all. Of course, there is no need to describe it in such detail, this is done in the mind. But it is advisable to understand what’s what). We record in memory:
We write the common factor outside the brackets. In parentheses we write the results of dividing all terms by this common factor. In order.
So we have expanded the expression ah+9x by multipliers. Turned it into multiplying x by (a+9). I note that in the original expression there was also a multiplication, even two: a·x and 9·x. But it was not factorized! Because in addition to multiplication, this expression also contained addition, the “+” sign! And in expression x(a+9) There is nothing but multiplication!
How so!? - I hear the indignant voice of the people - And in brackets!?)
Yes, there is addition inside the parentheses. But the trick is that while the brackets are not opened, we consider them like one letter. And we do all the actions with brackets entirely, as with one letter. In this sense, in the expression x(a+9) There is nothing except multiplication. This is the whole point of factorization.
By the way, is it possible to somehow check whether we did everything correctly? Easily! It’s enough to multiply back what you put out (x) by brackets and see if it worked original expression? If it works, everything is great!)
x(a+9)=ax+9x
Happened.)
There are no problems in this primitive example. But if there are several terms, and even with different signs... In short, every third student messes up). Therefore:
If necessary, check the factorization by inverse multiplication.
Factorize:
3ax+9x
We are looking for a common factor. Well, everything is clear with X, it can be taken out. Is there more general factor? Yes! This is a three. You can write the expression like this:
3ax+3 3x
Here it is immediately clear that the common factor will be 3x. Here we take it out:
3ax+3 3x=3x(a+3)
Spread out.
What happens if you take it out only x? Nothing special:
3ax+9x=x(3a+9)
This will also be a factorization. But in this fascinating process, it is customary to lay out everything to the limit while there is an opportunity. Here in brackets there is an opportunity to put out a three. It will turn out:
3ax+9x=x(3a+9)=3x(a+3)
The same thing, only with one extra action.) Remember:
When taking the common factor out of brackets, we try to take out maximum common factor.
Shall we continue the fun?)
Factor the expression:
3akh+9х-8а-24
What will we take away? Three, X? Nope... You can't. I remind you that you can only take out general multiplier that is in all terms of the expression. That's why he general. There is no such multiplier here... What, you don’t have to expand it!? Well, yes, we were so happy... Meet:
2. Grouping.
Actually, it’s hard to name the group in an independent way factorization. It's more of a way to get out complex example.) We need to group the terms so that everything works out. This can only be shown by example. So, we have the expression:
3akh+9х-8а-24
It can be seen that there are some common letters and numbers. But... General there is no multiplier to be in all terms. Let's not lose heart and break the expression into pieces. Grouping. So that each piece has a common factor, there is something to take away. How do we break it? Yes, we just put parentheses.
Let me remind you that parentheses can be placed anywhere and however you want. Just the essence of the example hasn't changed. For example, you can do this:
3akh+9х-8а-24=(3ах+9х)-(8а+24)
Please pay attention to the second brackets! They are preceded by a minus sign, and 8a And 24 turned positive! If, to check, we open the brackets back, the signs will change, and we get original expression. Those. the essence of the expression from the brackets has not changed.
But if you just inserted parentheses without taking into account the change of sign, for example, like this:
3akh+9х-8а-24=(3ax+9x) -(8a-24 )
it would be a mistake. On the right - already other expression. Open the brackets and everything will become visible. You don’t have to decide further, yes...)
But let's return to factorization. Let's look at the first brackets (3ax+9x) and we think, is there anything we can take out? Well, we solved this example above, we can take it 3x:
(3ax+9x)=3x(a+3)
Let's study the second brackets, we can add an eight there:
(8a+24)=8(a+3)
Our entire expression will be:
(3ax+9x)-(8a+24)=3x(a+3)-8(a+3)
Factored? No. The result of decomposition should be only multiplication but with us the minus sign spoils everything. But... Both terms have a common factor! This (a+3). It was not for nothing that I said that the entire brackets are, as it were, one letter. This means that these brackets can be taken out of brackets. Yes, that's exactly what it sounds like.)
We do as described above. We write the common factor (a+3), in the second brackets we write the results of dividing the terms by (a+3):
3x(a+3)-8(a+3)=(a+3)(3x-8)
All! There is nothing on the right except multiplication! This means that factorization has been completed successfully!) Here it is:
3ax+9x-8a-24=(a+3)(3x-8)
Let us briefly repeat the essence of the group.
If the expression does not general multiplier for everyone terms, we break the expression into brackets so that inside the brackets the common factor was. We take it out and see what happens. If you are lucky and there are absolutely identical expressions left in the brackets, we move these brackets out of brackets.
I will add that grouping is a creative process). It doesn't always work out the first time. It's OK. Sometimes you have to swap terms and consider different variants groups until a successful one is found. The main thing here is not to lose heart!)
Examples.
Now, having enriched yourself with knowledge, you can solve tricky examples.) At the beginning of the lesson there were three of these...
Simplify:
In essence, we have already solved this example. Unbeknownst to ourselves.) I remind you: if we are given a terrible fraction, we try to factor the numerator and denominator. Other simplification options simply no.
Well, the denominator here is not expanded, but the numerator... We have already expanded the numerator during the lesson! Like this:
3ax+9x-8a-24=(a+3)(3x-8)
We write the result of the expansion into the numerator of the fraction:
According to the rule of reducing fractions (the main property of a fraction), we can divide (at the same time!) the numerator and denominator by the same number, or expression. Fraction from this does not change. So we divide the numerator and denominator by the expression (3x-8). And here and there we will get ones. The final result of the simplification:
I would like to especially emphasize: reducing a fraction is possible if and only if in the numerator and denominator, in addition to multiplying expressions there is nothing. That is why the transformation of the sum (difference) into multiplication so important for simplification. Of course, if the expressions different, then nothing will be reduced. It will happen. But factorization gives a chance. This chance without decomposition is simply not there.
Example with equation:
Solve the equation:
x 5 - x 4 = 0
We take out the common factor x 4 out of brackets. We get:
x 4 (x-1)=0
We realize that the product of factors is equal to zero then and only then, when any of them is zero. If in doubt, find me a couple of non-zero numbers that, when multiplied, will give zero.) So we write, first the first factor:
With such an equality, the second factor does not concern us. Anyone can be, but in the end it will still be zero. What number to the fourth power does zero give? Only zero! And no other... Therefore:
We figured out the first factor and found one root. Let's look at the second factor. Now we don’t care about the first factor anymore.):
Here we found a solution: x 1 = 0; x 2 = 1. Any of these roots fits our equation.
Very important note. Please note that we solved the equation piece by piece! Each factor was equal to zero, regardless of other factors. By the way, if in such an equation there are not two factors, like ours, but three, five, as many as you like, we will solve similar. Piece by piece. For example:
(x-1)(x+5)(x-3)(x+2)=0
Anyone who opens the brackets and multiplies everything will be stuck on this equation forever.) A correct student will immediately see that there is nothing on the left except multiplication, and zero on the right. And he will begin (in his mind!) to equate all brackets in order to zero. And he will receive (in 10 seconds!) the right decision: x 1 = 1; x 2 = -5; x 3 = 3; x 4 = -2.
Cool, right?) Such an elegant solution is possible if the left side of the equation factorized. Got the hint?)
Well, one last example, for the older ones):
Solve the equation:
It’s somewhat similar to the previous one, don’t you think?) Of course. It's time to remember that in seventh grade algebra, sines, logarithms, and anything else can be hidden under the letters! Factoring works throughout mathematics.
We take out the common factor lg 4 x out of brackets. We get:
log 4 x=0
This is one root. Let's look at the second factor.
Here is the final answer: x 1 = 1; x 2 = 10.
I hope you've realized the power of factoring in simplifying fractions and solving equations.)
In this lesson we learned about common factoring and grouping. It remains to deal with the formulas for abbreviated multiplication and the quadratic trinomial.
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Any composite number can be factorized into prime factors. There can be several methods of decomposition. Either method produces the same result.
How to factor a number into prime factors in the most convenient way? Let's look at how best to do this using specific examples.
Examples. 1) Factor the number 1400 into prime factors.
1400 is divisible by 2. 2 is a prime number; there is no need to factor it. We get 700. Divide it by 2. We get 350. We also divide 350 by 2. The resulting number 175 can be divided by 5. The result is 35 - divide by 5 again. Total - 7. It can only be divided by 7. We get 1, division over.
The same number can be factorized differently:
1400 is conveniently divided by 10. 10 is not prime number, so it needs to be factorized into simple factors: 10=2∙5. The result is 140. We divide it again by 10=2∙5. We get 14. If 14 is divided by 14, then it should also be decomposed into a product of prime factors: 14=2∙7.
Thus, we again came to the same decomposition as in the first case, but faster.
Conclusion: when decomposing a number, it is not necessary to divide it only into prime factors. We divide by what is more convenient, for example, by 10. You just need to remember to decompose the compound divisors into simple factors.
2) Factor the number 1620 into prime factors.
The most convenient way to divide the number 1620 is by 10. Since 10 is not a prime number, we represent it as a product of prime factors: 10=2∙5. We got 162. It is convenient to divide it by 2. The result is 81. The number 81 can be divided by 3, but by 9 it is more convenient. Since 9 is not a prime number, we expand it as 9=3∙3. We get 9. We also divide it by 9 and expand it into the product of prime factors.
Each natural number, besides one, has two or more divisors. For example, the number 7 is divisible without a remainder only by 1 and 7, that is, it has two divisors. And the number 8 has divisors 1, 2, 4, 8, that is, as many as 4 divisors at once.
What is the difference between prime and composite numbers?
Numbers that have more than two divisors are called composite numbers. Numbers that have only two divisors: one and the number itself are called prime numbers.
The number 1 has only one division, namely the number itself. One is neither a prime nor a composite number.
- For example, the number 7 is prime and the number 8 is composite.
First 10 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. The number 2 is the only even prime number, all other prime numbers are odd.
The number 78 is composite, since in addition to 1 and itself, it is also divisible by 2. When divided by 2, we get 39. That is, 78 = 2*39. In such cases, they say that the number was factored into factors of 2 and 39.
Any composite number can be decomposed into two factors, each of which is greater than 1. This trick will not work with a prime number. So it goes.
Factoring a number into prime factors
As noted above, any composite number can be decomposed into two factors. Let's take, for example, the number 210. This number can be decomposed into two factors 21 and 10. But the numbers 21 and 10 are also composite, let's decompose them into two factors. We get 10 = 2*5, 21=3*7. And as a result, the number 210 was decomposed into 4 factors: 2,3,5,7. These numbers are already prime and cannot be expanded. That is, we factored the number 210 into prime factors.
When factoring composite numbers into prime factors, they are usually written in ascending order.
It should be remembered that any composite number can be decomposed into prime factors and in a unique way, up to permutation.
- Usually, when decomposing a number into prime factors, divisibility criteria are used.
Let's factor the number 378 into prime factors
We will write down the numbers, separating them with a vertical line. The number 378 is divisible by 2, since it ends in 8. When divided, we get the number 189. The sum of the digits of the number 189 is divisible by 3, which means the number 189 itself is divisible by 3. The result is 63.
The number 63 is also divisible by 3, according to divisibility. We get 21, the number 21 can again be divided by 3, we get 7. Seven is divided only by itself, we get one. This completes the division. To the right after the line are the prime factors into which the number 378 is decomposed.
378|2
189|3
63|3
21|3
This article gives answers to the question of factoring a number on a sheet. Let's look at the general idea of decomposition with examples. Let us analyze the canonical form of the expansion and its algorithm. All alternative methods will be considered using divisibility signs and multiplication tables.
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What does it mean to factor a number into prime factors?
Let's look at the concept of prime factors. It is known that every prime factor is a prime number. In a product of the form 2 · 7 · 7 · 23 we have that we have 4 prime factors in the form 2, 7, 7, 23.
Factorization involves its representation in the form of products of primes. If we need to decompose the number 30, then we get 2, 3, 5. The entry will take the form 30 = 2 · 3 · 5. It is possible that the multipliers may be repeated. A number like 144 has 144 = 2 2 2 2 3 3.
Not all numbers are prone to decay. Numbers that are greater than 1 and are integers can be factorized. Prime numbers, when factored, are only divisible by 1 and themselves, so it is impossible to represent these numbers as a product.
When z refers to integers, it is represented as a product of a and b, where z is divided by a and b. Composite numbers are factored using the fundamental theorem of arithmetic. If the number is greater than 1, then its factorization p 1, p 2, ..., p n takes the form a = p 1 , p 2 , … , p n . The decomposition is assumed to be in a single variant.
Canonical factorization of a number into prime factors
During expansion, factors can be repeated. They are written compactly using degrees. If, when decomposing the number a, we have a factor p 1, which occurs s 1 times and so on p n – s n times. Thus the expansion will take the form a=p 1 s 1 · a = p 1 s 1 · p 2 s 2 · … · p n s n. This entry is called the canonical factorization of a number into prime factors.
When expanding the number 609840, we get that 609 840 = 2 2 2 2 3 3 5 7 11 11, its canonical form will be 609 840 = 2 4 3 2 5 7 11 2. Using canonical expansion, you can find all the divisors of a number and their number.
To correctly factorize, you need to have an understanding of prime and composite numbers. The point is to obtain a sequential number of divisors of the form p 1, p 2, ..., p n numbers a , a 1 , a 2 , … , a n - 1, this makes it possible to get a = p 1 a 1, where a 1 = a: p 1 , a = p 1 · a 1 = p 1 · p 2 · a 2 , where a 2 = a 1: p 2 , … , a = p 1 · p 2 · … · p n · a n , where a n = a n - 1: p n. Upon receipt a n = 1, then the equality a = p 1 · p 2 · … · p n we obtain the required decomposition of the number a into prime factors. notice, that p 1 ≤ p 2 ≤ p 3 ≤ … ≤ p n.
To find least common factors, you need to use a table of prime numbers. This is done using the example of finding the smallest prime divisor of the number z. When taking prime numbers 2, 3, 5, 11 and so on, and dividing the number z by them. Since z is not a prime number, it should be taken into account that the smallest prime divisor will not be greater than z. It can be seen that there are no divisors of z, then it is clear that z is a prime number.
Example 1
Let's look at the example of the number 87. When it is divided by 2, we have that 87: 2 = 43 with a remainder of 1. It follows that 2 cannot be a divisor; division must be done entirely. When divided by 3, we get that 87: 3 = 29. Hence the conclusion is that 3 is the smallest prime divisor of the number 87.
When factoring into prime factors, you must use a table of prime numbers, where a. When factoring 95, you should use about 10 primes, and when factoring 846653, about 1000.
Let's consider the decomposition algorithm into prime factors:
- finding the smallest factor of divisor p 1 of a number a by the formula a 1 = a: p 1, when a 1 = 1, then a is a prime number and is included in the factorization, when not equal to 1, then a = p 1 · a 1 and follow to the point below;
- finding the prime divisor p 2 of a number a 1 by sequentially enumerating prime numbers using a 2 = a 1: p 2 , when a 2 = 1 , then the expansion will take the form a = p 1 p 2 , when a 2 = 1, then a = p 1 p 2 a 2 , and we move on to the next step;
- searching through prime numbers and finding a prime divisor p 3 numbers a 2 according to the formula a 3 = a 2: p 3 when a 3 = 1 , then we get that a = p 1 p 2 p 3 , when not equal to 1, then a = p 1 p 2 p 3 a 3 and move on to the next step;
- the prime divisor is found p n numbers a n - 1 by enumerating prime numbers with pn - 1, and a n = a n - 1: p n, where a n = 1, the step is final, as a result we get that a = p 1 · p 2 · … · p n .
The result of the algorithm is written in the form of a table with the decomposed factors with a vertical bar sequentially in a column. Consider the figure below.
The resulting algorithm can be applied by decomposing numbers into prime factors.
When factoring into prime factors, the basic algorithm should be followed.
Example 2
Factor the number 78 into prime factors.
Solution
In order to find the smallest prime divisor, you need to go through all the prime numbers in 78. That is 78: 2 = 39. Division without a remainder means this is the first simple divisor, which we denote as p 1. We get that a 1 = a: p 1 = 78: 2 = 39. We arrived at an equality of the form a = p 1 · a 1 , where 78 = 2 39. Then a 1 = 39, that is, we should move on to the next step.
Let's focus on finding the prime divisor p2 numbers a 1 = 39. You should go through the prime numbers, that is, 39: 2 = 19 (remaining 1). Since division with a remainder, 2 is not a divisor. When choosing the number 3, we get that 39: 3 = 13. This means that p 2 = 3 is the smallest prime divisor of 39 by a 2 = a 1: p 2 = 39: 3 = 13. We obtain an equality of the form a = p 1 p 2 a 2 in the form 78 = 2 3 13. We have that a 2 = 13 is not equal to 1, then we should move on.
The smallest prime divisor of the number a 2 = 13 is found by searching through numbers, starting with 3. We get that 13: 3 = 4 (remaining 1). From this we can see that 13 is not divisible by 5, 7, 11, because 13: 5 = 2 (rest. 3), 13: 7 = 1 (rest. 6) and 13: 11 = 1 (rest. 2). It can be seen that 13 is a prime number. According to the formula it looks like this: a 3 = a 2: p 3 = 13: 13 = 1. We found that a 3 = 1, which means the completion of the algorithm. Now the factors are written as 78 = 2 · 3 · 13 (a = p 1 · p 2 · p 3) .
Answer: 78 = 2 3 13.
Example 3
Factor the number 83,006 into prime factors.
Solution
The first step involves factoring p 1 = 2 And a 1 = a: p 1 = 83,006: 2 = 41,503, where 83,006 = 2 · 41,503.
The second step assumes that 2, 3 and 5 are not prime divisors for the number a 1 = 41,503, but 7 is a prime divisor, because 41,503: 7 = 5,929. We get that p 2 = 7, a 2 = a 1: p 2 = 41,503: 7 = 5,929. Obviously, 83,006 = 2 7 5 929.
Finding the smallest prime divisor of p 4 to the number a 3 = 847 is 7. It can be seen that a 4 = a 3: p 4 = 847: 7 = 121, so 83 006 = 2 7 7 7 121.
To find the prime divisor of the number a 4 = 121, we use the number 11, that is, p 5 = 11. Then we get an expression of the form a 5 = a 4: p 5 = 121: 11 = 11, and 83,006 = 2 7 7 7 11 11.
For number a 5 = 11 number p 6 = 11 is the smallest prime divisor. Hence a 6 = a 5: p 6 = 11: 11 = 1. Then a 6 = 1. This indicates the completion of the algorithm. The factors will be written as 83 006 = 2 · 7 · 7 · 7 · 11 · 11.
The canonical notation of the answer will take the form 83 006 = 2 · 7 3 · 11 2.
Answer: 83 006 = 2 7 7 7 11 11 = 2 7 3 11 2.
Example 4
Factor the number 897,924,289.
Solution
To find the first prime factor, search through the prime numbers, starting with 2. The end of the search occurs at the number 937. Then p 1 = 937, a 1 = a: p 1 = 897 924 289: 937 = 958 297 and 897 924 289 = 937 958 297.
The second step of the algorithm is to iterate over smaller prime numbers. That is, we start with the number 937. The number 967 can be considered prime because it is a prime divisor of the number a 1 = 958,297. From here we get that p 2 = 967, then a 2 = a 1: p 1 = 958 297: 967 = 991 and 897 924 289 = 937 967 991.
The third step says that 991 is a prime number, since it does not have a single prime factor that does not exceed 991. The approximate value of the radical expression is 991< 40 2 . Иначе запишем как 991 < 40 2 . This shows that p 3 = 991 and a 3 = a 2: p 3 = 991: 991 = 1. We find that the decomposition of the number 897 924 289 into prime factors is obtained as 897 924 289 = 937 967 991.
Answer: 897 924 289 = 937 967 991.
Using divisibility tests for prime factorization
To factor a number into prime factors, you need to follow an algorithm. When there are small numbers, it is permissible to use the multiplication table and divisibility signs. Let's look at this with examples.
Example 5
If it is necessary to factorize 10, then the table shows: 2 · 5 = 10. The resulting numbers 2 and 5 are prime numbers, so they are prime factors for the number 10.
Example 6
If it is necessary to decompose the number 48, then the table shows: 48 = 6 8. But 6 and 8 are not prime factors, since they can also be expanded as 6 = 2 3 and 8 = 2 4. Then the complete expansion from here is obtained as 48 = 6 8 = 2 3 2 4. The canonical notation will take the form 48 = 2 4 · 3.
Example 7
When decomposing the number 3400, you can use the signs of divisibility. IN in this case The criteria for divisibility by 10 and 100 are relevant. From here we get that 3,400 = 34 · 100, where 100 can be divided by 10, that is, written as 100 = 10 · 10, which means that 3,400 = 34 · 10 · 10. Based on the divisibility test, we find that 3 400 = 34 10 10 = 2 17 2 5 2 5. All factors are prime. The canonical expansion takes the form 3 400 = 2 3 5 2 17.
When we find prime factors, we need to use divisibility tests and multiplication tables. If you imagine the number 75 as a product of factors, then you need to take into account the rule of divisibility by 5. We get that 75 = 5 15, and 15 = 3 5. That is, the desired expansion is an example of the form of the product 75 = 5 · 3 · 5.
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