In order to factorize, it is necessary to simplify the expressions. This is necessary in order to be able to further reduce. The decomposition of a polynomial makes sense when its degree is at least two. A polynomial with the first degree is called linear.
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The article will cover all the concepts of decomposition, theoretical basis and methods for factoring the polynomial.
Theory
Theorem 1When any polynomial with degree n, having the form P n x = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0, represent as a product with a constant factor with the highest power an and n linear factors (x - xi), i = 1, 2, ..., n, then P n (x) = an (x - xn) (x - xn - 1) ... ... · (X - x 1), where x i, i = 1, 2,…, n - these are the roots of the polynomial.
The theorem is intended for roots of complex type x i, i = 1, 2,…, n and for complex coefficients a k, k = 0, 1, 2,…, n. This is the basis of any decomposition.
When coefficients of the form a k, k = 0, 1, 2, ..., n are real numbers, then complex roots that will meet in conjugate pairs. For example, the roots x 1 and x 2 referring to a polynomial of the form P n x = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 are considered to be complex conjugate, then the other roots are real, from which we obtain that the polynomial takes the form P n (x) = a n (x - x n) (x - x n - 1) ·. ... ... (X - x 3) x 2 + p x + q, where x 2 + p x + q = (x - x 1) (x - x 2).
Comment
The roots of the polynomial can be repeated. Consider the proof of the algebra theorem, a corollary of Bezout's theorem.
The main theorem of algebra
Theorem 2Any polynomial with degree n has at least one root.
Bezout's theorem
After the division of a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + has been made. ... ... + a 1 x + a 0 on (x - s), then we get the remainder, which is equal to the polynomial at the point s, then we get
P n x = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 = (x - s) Q n - 1 (x) + P n (s), where Q n - 1 (x) is a polynomial of degree n - 1.
Corollary from Bezout's theorem
When the root of the polynomial P n (x) is considered s, then P n x = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 = (x - s) Q n - 1 (x). This corollary is sufficient when used to describe a solution.
Factoring a square trinomial
A square trinomial of the form a x 2 + b x + c can be decomposed into linear factors... then we get that a x 2 + b x + c = a (x - x 1) (x - x 2), where x 1 and x 2 are roots (complex or real).
From this it is clear that the expansion itself reduces to the solution quadratic equation afterwards.
Example 1
Factorize a square trinomial.
Solution
Find the roots of the equation 4 x 2 - 5 x + 1 = 0. To do this, you need to find the value of the discriminant by the formula, then we get D = (- 5) 2 - 4 · 4 · 1 = 9. Hence we have that
x 1 = 5 - 9 2 4 = 1 4 x 2 = 5 + 9 2 4 = 1
From this we get that 4 x 2 - 5 x + 1 = 4 x - 1 4 x - 1.
The brackets must be expanded to perform the check. Then we get an expression of the form:
4 x - 1 4 x - 1 = 4 x 2 - x - 1 4 x + 1 4 = 4 x 2 - 5 x + 1
After checking, we come to the original expression. That is, we can conclude that the decomposition is performed correctly.
Example 2
Factor a square trinomial of the form 3 x 2 - 7 x - 11.
Solution
We get that it is necessary to calculate the resulting quadratic equation of the form 3 x 2 - 7 x - 11 = 0.
To find the roots, you need to determine the value of the discriminant. We get that
3 x 2 - 7 x - 11 = 0 D = (- 7) 2 - 4 3 (- 11) = 181 x 1 = 7 + D 2 3 = 7 + 181 6 x 2 = 7 - D 2 3 = 7 - 181 6
From this we get that 3 x 2 - 7 x - 11 = 3 x - 7 + 181 6 x - 7 - 181 6.
Example 3
Factor the polynomial 2 x 2 + 1.
Solution
Now you need to solve the quadratic equation 2 x 2 + 1 = 0 and find its roots. We get that
2 x 2 + 1 = 0 x 2 = - 1 2 x 1 = - 1 2 = 1 2 i x 2 = - 1 2 = - 1 2 i
These roots are called complex conjugate, which means that the decomposition itself can be represented as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.
Example 4
Decompose the square trinomial x 2 + 1 3 x + 1.
Solution
First you need to solve a quadratic equation of the form x 2 + 1 3 x + 1 = 0 and find its roots.
x 2 + 1 3 x + 1 = 0 D = 1 3 2 - 4 1 1 = - 35 9 x 1 = - 1 3 + D 2 1 = - 1 3 + 35 3 i 2 = - 1 + 35 i 6 = - 1 6 + 35 6 ix 2 = - 1 3 - D 2 1 = - 1 3 - 35 3 i 2 = - 1 - 35 i 6 = - 1 6 - 35 6 i
Having received the roots, we write
x 2 + 1 3 x + 1 = x - - 1 6 + 35 6 i x - - 1 6 - 35 6 i = = x + 1 6 - 35 6 i x + 1 6 + 35 6 i
Comment
If the value of the discriminant is negative, then the polynomials remain second-order polynomials. Hence it follows that we will not decompose them into linear factors.
Methods for factoring polynomials of degree higher than two
The decomposition assumes universal method... Most of all cases are based on a corollary from Bezout's theorem. To do this, you need to select the value of the root x 1 and lower its degree by dividing by a polynomial by 1 by dividing by (x - x 1). The resulting polynomial needs to find the root x 2, and the search process is cyclical until we get a complete decomposition.
If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic assumes the solution of equations with higher degrees and integer coefficients.
Taking the common factor out of parentheses
Consider the case when the free term is equal to zero, then the form of the polynomial becomes P n (x) = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x.
It can be seen that the root of such a polynomial will be equal to x 1 = 0, then the polynomial can be represented as the expression P n (x) = a n x n + a n - 1 x n - 1 +. ... ... + a 1 x = = x (a n x n - 1 + a n - 1 x n - 2 +... + a 1)
This method is considered as taking the common factor out of the parentheses.
Example 5
Factor the third degree polynomial 4 x 3 + 8 x 2 - x.
Solution
We see that x 1 = 0 is the root of the given polynomial, then we can take x outside the brackets of the entire expression. We get:
4 x 3 + 8 x 2 - x = x (4 x 2 + 8 x - 1)
We pass to finding the roots of the square trinomial 4 x 2 + 8 x - 1. Let's find the discriminant and roots:
D = 8 2 - 4 4 (- 1) = 80 x 1 = - 8 + D 2 4 = - 1 + 5 2 x 2 = - 8 - D 2 4 = - 1 - 5 2
Then it follows that
4 x 3 + 8 x 2 - x = x 4 x 2 + 8 x - 1 = = 4 xx - - 1 + 5 2 x - - 1 - 5 2 = = 4 xx + 1 - 5 2 x + 1 + 5 2
To begin with, let us consider the decomposition method containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0, where the coefficient at the highest power is 1.
When a polynomial has integral roots, then they are considered divisors of the free term.
Example 6
Expand the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.
Solution
Consider if there are whole roots. It is necessary to write down the divisors of the number - 18. We get that ± 1, ± 2, ± 3, ± 6, ± 9, ± 18. It follows that this polynomial has integral roots. You can check the Horner scheme. It is very convenient and allows you to quickly obtain the coefficients of the expansion of the polynomial:
It follows that x = 2 and x = - 3 are the roots of the original polynomial, which can be represented as a product of the form:
f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x 3 + 5 x 2 + 9 x + 9) = = (x - 2) (x + 3) (x 2 + 2 x + 3)
We pass to the decomposition of a square trinomial of the form x 2 + 2 x + 3.
Since the discriminant is negative, it means that there are no real roots.
Answer: f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x + 3) (x 2 + 2 x + 3)
Comment
It is allowed to use a selection of a root and division of a polynomial by a polynomial instead of the Horner scheme. We proceed to consider the expansion of a polynomial containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0, the oldest of which is equal to one.
This case takes place for rational fractional fractions.
Example 7
Factor f (x) = 2 x 3 + 19 x 2 + 41 x + 15.
Solution
It is necessary to change the variable y = 2 x, you should go to a polynomial with coefficients equal to 1 at the highest degree. You need to start by multiplying the expression by 4. We get that
4 f (x) = 2 3 x 3 + 19 2 2 x 2 + 82 2 x + 60 = = y 3 + 19 y 2 + 82 y + 60 = g (y)
When the resulting function of the form g (y) = y 3 + 19 y 2 + 82 y + 60 has integer roots, then finding them among the divisors of the free term. The entry will take the form:
± 1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, ± 12, ± 15, ± 20, ± 30, ± 60
Let us proceed to calculating the function g (y) at these points in order to obtain zero as a result. We get that
g (1) = 1 3 + 19 1 2 + 82 1 + 60 = 162 g (- 1) = (- 1) 3 + 19 (- 1) 2 + 82 (- 1) + 60 = - 4 g (2) = 2 3 + 19 2 2 + 82 2 + 60 = 308 g (- 2) = (- 2) 3 + 19 (- 2) 2 + 82 (- 2) + 60 = - 36 g (3) = 3 3 + 19 3 2 + 82 3 + 60 = 504 g (- 3) = (- 3) 3 + 19 (- 3) 2 + 82 (- 3) + 60 = - 42 g (4) = 4 3 + 19 4 2 + 82 4 + 60 = 756 g (- 4) = (- 4) 3 + 19 (- 4) 2 + 82 (- 4) + 60 = - 28 g (5) = 5 3 + 19 5 2 + 82 5 + 60 = 1070 g (- 5) = (- 5) 3 + 19 (- 5) 2 + 82 (- 5) + 60
We get that y = - 5 is the root of an equation of the form y 3 + 19 y 2 + 82 y + 60, which means that x = y 2 = - 5 2 is the root of the original function.
Example 8
It is necessary to divide with a column 2 x 3 + 19 x 2 + 41 x + 15 by x + 5 2.
Solution
Let's write and get:
2 x 3 + 19 x 2 + 41 x + 15 = x + 5 2 (2 x 2 + 14 x + 6) = = 2 x + 5 2 (x 2 + 7 x + 3)
Checking the divisors will take a lot of time, so it is more profitable to take the factorization of the resulting square trinomial of the form x 2 + 7 x + 3. Equating to zero and find the discriminant.
x 2 + 7 x + 3 = 0 D = 7 2 - 4 1 3 = 37 x 1 = - 7 + 37 2 x 2 = - 7 - 37 2 ⇒ x 2 + 7 x + 3 = x + 7 2 - 37 2 x + 7 2 + 37 2
Hence it follows that
2 x 3 + 19 x 2 + 41 x + 15 = 2 x + 5 2 x 2 + 7 x + 3 = = 2 x + 5 2 x + 7 2 - 37 2 x + 7 2 + 37 2
Artificial tricks for factoring a polynomial
Rational roots are not inherent in all polynomials. To do this, you need to use special methods for finding the multipliers. But not all polynomials can be expanded or represented as a product.
Grouping method
There are times when you can group the terms of a polynomial to find the common factor and place it outside the brackets.
Example 9
Factorize the polynomial x 4 + 4 x 3 - x 2 - 8 x - 2.
Solution
Because the coefficients are integers, then the roots, presumably, can also be integers. To check, take the values 1, - 1, 2 and - 2 in order to calculate the value of the polynomial at these points. We get that
1 4 + 4 1 3 - 1 2 - 8 1 - 2 = - 6 ≠ 0 (- 1) 4 + 4 (- 1) 3 - (- 1) 2 - 8 (- 1) - 2 = 2 ≠ 0 2 4 + 4 2 3 - 2 2 - 8 2 - 2 = 26 ≠ 0 (- 2) 4 + 4 (- 2) 3 - (- 2) 2 - 8 (- 2) - 2 = - 6 ≠ 0
From this it is clear that there are no roots, it is necessary to use a different method of decomposition and solution.
It is necessary to group:
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 4 + 4 x 3 - 2 x 2 + x 2 - 8 x - 2 = = (x 4 - 2 x 2) + (4 x 3 - 8 x) + x 2 - 2 = = x 2 (x 2 - 2) + 4 x (x 2 - 2) + x 2 - 2 = = (x 2 - 2) (x 2 + 4 x + 1)
After grouping the original polynomial, it is necessary to represent it as the product of two square trinomials... To do this, we need to do the factorization. we get that
x 2 - 2 = 0 x 2 = 2 x 1 = 2 x 2 = - 2 ⇒ x 2 - 2 = x - 2 x + 2 x 2 + 4 x + 1 = 0 D = 4 2 - 4 1 1 = 12 x 1 = - 4 - D 2 1 = - 2 - 3 x 2 = - 4 - D 2 1 = - 2 - 3 ⇒ x 2 + 4 x + 1 = x + 2 - 3 x + 2 + 3
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 2 - 2 x 2 + 4 x + 1 = = x - 2 x + 2 x + 2 - 3 x + 2 + 3
Comment
The simplicity of the grouping does not mean that it is easy enough to choose the terms. There is no definite solution, so it is necessary to use special theorems and rules.
Example 10
Factor the polynomial x 4 + 3 x 3 - x 2 - 4 x + 2.
Solution
The given polynomial has no integral roots. It is necessary to group the terms. We get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = = (x 4 + x 3) + (2 x 3 + 2 x 2) + (- 2 x 2 - 2 x) - x 2 - 2 x + 2 = = x 2 (x 2 + x) + 2 x (x 2 + x) - 2 (x 2 + x) - (x 2 + 2 x - 2) = = (x 2 + x) (x 2 + 2 x - 2) - (x 2 + 2 x - 2) = (x 2 + x - 1) (x 2 + 2 x - 2)
After factoring, we get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = x 2 + x - 1 x 2 + 2 x - 2 = = x + 1 + 3 x + 1 - 3 x + 1 2 + 5 2 x + 1 2 - 5 2
Using abbreviated multiplication formulas and Newton's binomial for factoring a polynomial
The appearance often does not always make it clear which method should be used during decomposition. After the transformations have been made, you can build a line consisting of Pascal's triangle, otherwise they are called the Newton binomial.
Example 11
Factor the polynomial x 4 + 4 x 3 + 6 x 2 + 4 x - 2.
Solution
It is necessary to convert the expression to the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3
The expression x + 1 4 indicates the sequence of sum coefficients in parentheses.
Hence, we have x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3.
After applying the difference of squares, we get
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3
Consider the expression in the second parenthesis. It is clear that there are no horses there, so you should apply the formula for the difference of squares again. We get an expression of the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3 = = x + 1 - 3 4 x + 1 + 3 4 x 2 + 2 x + 1 + 3
Example 12
Factor x 3 + 6 x 2 + 12 x + 6.
Solution
Let's do the transformation of the expression. We get that
x 3 + 6 x 2 + 12 x + 6 = x 3 + 3 2 x 2 + 3 2 2 x + 2 3 - 2 = (x + 2) 3 - 2
It is necessary to apply the formula for abbreviated multiplication of the difference of cubes. We get:
x 3 + 6 x 2 + 12 x + 6 = = (x + 2) 3 - 2 = = x + 2 - 2 3 x + 2 2 + 2 3 x + 2 + 4 3 = = x + 2 - 2 3 x 2 + x 2 + 2 3 + 4 + 2 2 3 + 4 3
A way to replace a variable when factoring a polynomial
When changing a variable, the degree is reduced and the polynomial is decomposed into factors.
Example 13
Factor a polynomial of the form x 6 + 5 x 3 + 6.
Solution
According to the condition, it is clear that it is necessary to make the replacement y = x 3. We get:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6
The roots of the resulting quadratic equation are equal to y = - 2 and y = - 3, then
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3
It is necessary to apply the formula for abbreviated multiplication of the sum of cubes. We get expressions of the form:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3 = = x + 2 3 x 2 - 2 3 x + 4 3 x + 3 3 x 2 - 3 3 x + 9 3
That is, we got the required decomposition.
The cases discussed above will help in the consideration and factorization of a polynomial in different ways.
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Factorization of polynomials is an identity transformation, as a result of which a polynomial is transformed into a product of several factors - polynomials or monomials.
There are several ways to factor polynomials.
Method 1. Taking the common factor out of the parenthesis.
This transformation is based on the distributional multiplication law: ac + bc = c (a + b). The essence of the transformation is to select the common factor in the two components under consideration and "take" it out of the brackets.
Factor the polynomial 28x 3 - 35x 4.
Solution.
1. Find the elements 28x 3 and 35x 4 common divisor... For 28 and 35 this will be 7; for x 3 and x 4 - x 3. In other words, our common factor is 7x 3.
2. Each of the elements is represented as a product of factors, one of which
7x 3: 28x 3 - 35x 4 = 7x 3 ∙ 4 - 7x 3 ∙ 5x.
3. Factor out the common factor
7x 3: 28x 3 - 35x 4 = 7x 3 ∙ 4 - 7x 3 ∙ 5x = 7x 3 (4 - 5x).
Method 2. Using abbreviated multiplication formulas. The "skill" to master this method is to notice in the expression one of the formulas for abbreviated multiplication.
Factor the polynomial x 6 - 1.
Solution.
1 TO this expression we can apply the difference of squares formula. To do this, we represent x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 - 1 = (x 3 + 1) ∙ (x 3 - 1).
2. To the resulting expression, we can apply the formula for the sum and difference of cubes:
(x 3 + 1) ∙ (x 3 - 1) = (x + 1) ∙ (x 2 - x + 1) ∙ (x - 1) ∙ (x 2 + x + 1).
So,
x 6 - 1 = (x 3) 2 - 1 = (x 3 + 1) ∙ (x 3 - 1) = (x + 1) ∙ (x 2 - x + 1) ∙ (x - 1) ∙ (x 2 + x + 1).
Method 3. Grouping. The grouping method consists in combining the components of a polynomial in such a way that it is easy to perform actions on them (addition, subtraction, removal of a common factor).
Factor the polynomial x 3 - 3x 2 + 5x - 15.
Solution.
1. Let's group the components in this way: the 1st with the 2nd, and the 3rd with the 4th
(x 3 - 3x 2) + (5x - 15).
2. In the resulting expression, put the common factors outside the brackets: x 2 in the first case and 5 - in the second.
(x 3 - 3x 2) + (5x - 15) = x 2 (x - 3) + 5 (x - 3).
3. Factor out the common factor x - 3 and get:
x 2 (x - 3) + 5 (x - 3) = (x - 3) (x 2 + 5).
So,
x 3 - 3x 2 + 5x - 15 = (x 3 - 3x 2) + (5x - 15) = x 2 (x - 3) + 5 (x - 3) = (x - 3) ∙ (x 2 + 5 ).
Let's fix the material.
Factor the polynomial a 2 - 7ab + 12b 2.
Solution.
1. Let us represent the monomial 7ab as the sum 3ab + 4ab. The expression will take the form:
a 2 - (3ab + 4ab) + 12b 2. 
Let's open the brackets and get:
a 2 - 3ab - 4ab + 12b 2.
2. Let us group the components of the polynomial as follows: 1st with 2nd and 3rd with 4th. We get:
(a 2 - 3ab) - (4ab - 12b 2).
3. Let's take the common factors out of the brackets:
(a 2 - 3ab) - (4ab - 12b 2) = a (a - 3b) - 4b (a - 3b).
4. Factor out the common factor (a - 3b):
a (a - 3b) - 4b (a - 3b) = (a - 3 b) ∙ (a - 4b).
So,
a 2 - 7ab + 12b 2 =
= a 2 - (3ab + 4ab) + 12b 2 =
= a 2 - 3ab - 4ab + 12b 2 =
= (a 2 - 3ab) - (4ab - 12b 2) =
= a (a - 3b) - 4b (a - 3b) =
= (a - 3 b) ∙ (a - 4b).
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What to do if, in the process of solving a problem from the exam or on the entrance exam in mathematics, you received a polynomial that cannot be factorized using the standard methods that you learned in school? In this article, a math tutor will tell you about one effective way, which is outside the scope of the school curriculum, but with which it will not be difficult to factor a polynomial into factors. Read this article to the end and watch the attached video tutorial. The knowledge you gain will help you on the exam.
Polynomial factorization by division
In the event that you received a polynomial greater than the second degree and were able to guess the value of the variable at which this polynomial becomes equal to zero(for example, this value is equal), be aware! This polynomial can be divided by.
For example, it is easy to see that the fourth degree polynomial vanishes at. So it can be divided without a remainder by, thus obtaining a polynomial of the third degree (less by one). That is, to represent it in the form:
where A, B, C and D- some numbers. Let's expand the brackets:
Since the coefficients at the same degrees must be the same, we get:
So we got:
Move on. It is enough to iterate over a few small integers to see that the third degree polynomial is divisible by again. This gives a polynomial of the second degree (less by one). Then let's move on to the new entry:
where E, F and G- some numbers. We open the brackets again and arrive at the following expression:
Again, from the condition of equality of the coefficients at the same degrees, we obtain:
Then we get:
That is, the original polynomial can be factorized as follows:
In principle, if desired, using the difference of squares formula, the result can also be presented in the following form:
So simple and effective method factoring polynomials. Remember it, it may come in handy for an exam or math Olympiad. Check if you have learned how to use this method. Try to solve the next problem yourself.
Factor the polynomial:
Write your answers in the comments.
Prepared by Sergey Valerievich
In this article you will find all necessary information answering the question how to factor a number into prime factors... First given general idea on the decomposition of a number into prime factors, examples of decompositions are given. The following shows the canonical form of the factorization of a number into prime factors. After that, an algorithm for decomposing arbitrary numbers into prime factors is given and examples of decomposing numbers using this algorithm are given. Also considered alternative ways that allow you to quickly decompose small integers into prime factors using divisibility criteria and multiplication tables.
Page navigation.
What does it mean to factor a number into prime factors?
First, let's figure out what prime factors are.
It is clear that since the word “factors” is present in this phrase, then there is a product of some numbers, and the qualifying word “simple” means that each factor is a prime number. For example, in a product of the form 2 · 7 · 7 · 23 there are four prime factors: 2, 7, 7 and 23.
What does it mean to factor a number into prime factors?
It means that given number must be represented as a product of prime factors, and the value of this product must be equal to the original number. As an example, consider the product of three primes 2, 3 and 5, it is equal to 30, so the factorization of 30 into prime factors is 2 · 3 · 5. Usually, the decomposition of a number into prime factors is written as an equality, in our example it will be like this: 30 = 2 · 3 · 5. We emphasize separately that prime factors in the expansion can be repeated. This is clearly illustrated by the following example: 144 = 2 · 2 · 2 · 2 · 3 · 3. But the representation of the form 45 = 3 · 15 is not a prime factorization, since the number 15 is composite.
The following question arises: "What numbers in general can be decomposed into prime factors"?
In search of an answer to it, we give the following reasoning. Prime numbers are, by definition, among those larger than ones. Given this fact and, it can be argued that the product of several prime factors is a positive integer greater than one. Therefore, the factorization takes place only for positive integers that are greater than 1.
But do all integers greater than one factor out into prime factors?
It is clear that there is no way to decompose prime integers into prime factors. This is because prime numbers have only two positive divisors - one and themselves, so they cannot be represented as a product of two or more prime numbers. If the integer z could be represented as a product of prime numbers a and b, then the notion of divisibility would allow us to conclude that z is divisible by both a and b, which is impossible due to the simplicity of the number z. However, it is believed that any prime number itself is its expansion.
What about composite numbers? Do composite numbers decompose into prime factors, and are all composite numbers subject to such a decomposition? A number of these questions are answered in the affirmative by the main theorem of arithmetic. The main theorem of arithmetic states that any integer a that is greater than 1 can be decomposed into the product of prime factors p 1, p 2, ..., pn, and the decomposition has the form a = p 1 p 2 ... the decomposition is unique, if the order of the factors is not taken into account
Canonical factorization of a number into prime factors
In the expansion of a number, prime factors can be repeated. Duplicate prime factors can be written more compactly using. Suppose that in the decomposition of a number a prime factor p 1 occurs s 1 times, a prime factor p 2 - s 2 times, and so on, p n - s n times. Then the prime factorization of the number a can be written as a = p 1 s 1 p 2 s 2… p n s n... This form of recording is the so-called canonical prime factorization.
Let us give an example of the canonical factorization of a number into prime factors. Let us know the decomposition 609 840 = 2 2 2 2 3 3 5 7 11 11, its canonical notation is 609 840 = 2 4 3 2 5 7 11 2.
The canonical factorization of a number into prime factors allows you to find all the divisors of a number and the number of divisors of a number.
Algorithm for factoring a number into prime factors
To successfully cope with the problem of factoring a number into prime factors, you need to be very familiar with the information in the article on prime and composite numbers.
The essence of the process of decomposition of an integer positive and greater than one number a is clear from the proof of the main theorem of arithmetic. The idea is to sequentially find the smallest prime divisors p 1, p 2, ..., pn of numbers a, a 1, a 2, ..., a n-1, which allows us to obtain a series of equalities 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… pn an, where an = a n-1: pn. When we get a n = 1, then the equality a = p 1 · p 2 ·… · p n will give us the required decomposition of the number a into prime factors. It should be noted here that p 1 ≤p 2 ≤p 3 ≤… ≤p n.
It remains to figure out how to find the smallest prime factors at each step, and we will have an algorithm for factoring the number into prime factors. The table of prime numbers will help us find prime factors. Let us show how to use it to obtain the smallest prime divisor of the number z.
Sequentially we take primes from the table of primes (2, 3, 5, 7, 11, and so on) and divide the given number z by them. The first prime number z is divided by one integer will be its smallest prime divisor. If the number z is prime, then its smallest prime divisor will be the number z itself. It should be recalled here that if z is not prime number, then its smallest prime divisor does not exceed the number, where is from z. Thus, if among the prime numbers not exceeding, there was not a single divisor of the number z, then we can conclude that z is a prime number (for more details, see the theory section under the heading this number is prime or composite).
As an example, we'll show you how to find the smallest prime divisor of 87. We take the number 2. Divide 87 by 2, we get 87: 2 = 43 (rest. 1) (if necessary, see the article). That is, dividing 87 by 2 results in a remainder of 1, so 2 is not a divisor of 87. We take the next prime number from the table of primes, which is 3. We divide 87 by 3, we get 87: 3 = 29. Thus, 87 is evenly divisible by 3, so 3 is the smallest prime divisor of 87.
Note that in the general case, to factor a number a into prime factors, we need a table of primes up to a number not less than. We will have to refer to this table at every step, so you need to have it at hand. For example, to factor 95 into prime factors, a table of prime numbers up to 10 will suffice (since 10 is greater than). And to decompose the number 846 653, you will already need a table of primes up to 1,000 (since 1,000 is more than).
We now have sufficient information to write prime factorization algorithm... The decomposition algorithm for the number a is as follows:
- Sequentially going through the numbers from the table of primes, we find the smallest prime divisor p 1 of the number a, after which we calculate a 1 = a: p 1. If a 1 = 1, then the number a is prime, and it is itself its prime factorization. If a 1 is not equal to 1, then we have a = p 1 · a 1 and go to the next step.
- Find the smallest prime divisor p 2 of the number a 1, for this we sequentially iterate over the numbers from the table of primes, starting with p 1, and then calculate a 2 = a 1: p 2. If a 2 = 1, then the required factorization of the number a into prime factors has the form a = p 1 · p 2. If a 2 is not equal to 1, then we have a = p 1 · p 2 · a 2 and go to the next step.
- Going through the numbers from the table of primes, starting with p 2, we find the smallest prime divisor p 3 of the number a 2, after which we calculate a 3 = a 2: p 3. If a 3 = 1, then the required factorization of the number a into prime factors has the form a = p 1 · p 2 · p 3. If a 3 is not equal to 1, then we have a = p 1 · p 2 · p 3 · a 3 and go to the next step.
- Find the smallest prime divisor p n of a n-1 by going through prime numbers, starting with p n-1, and also a n = a n-1: p n, and a n is equal to 1. This step is the last step of the algorithm, here we get the required decomposition of the number a into prime factors: a = p 1 · p 2 ·… · p n.
For clarity, all the results obtained at each step of the algorithm for decomposing a number into prime factors are presented in the form of the following table, in which, to the left of the vertical line, the numbers a, a 1, a 2, ..., an are written sequentially in a column, and to the right of the line - the corresponding least prime divisors p 1, p 2,…, pn.
It remains only to consider a few examples of the application of the obtained algorithm for the decomposition of numbers into prime factors.
Prime factorization examples
Now we will analyze in detail examples of factoring numbers into prime factors... In the decomposition, we will apply the algorithm from the previous paragraph. Let's start with simple cases, and gradually we will complicate them in order to face all the possible nuances that arise when factoring numbers into prime factors.
Example.
Divide 78 into prime factors.
Solution.
We start looking for the first smallest prime divisor p 1 of the number a = 78. To do this, we begin to sequentially iterate over the prime numbers from the table of prime numbers. We take the number 2 and divide 78 by it, we get 78: 2 = 39. The number 78 was divided by 2 without a remainder, so p 1 = 2 is the first prime factor found for 78. In this case, a 1 = a: p 1 = 78: 2 = 39. So we come to the equality a = p 1 · a 1 having the form 78 = 2 · 39. Obviously, a 1 = 39 is different from 1, so we pass to the second step of the algorithm.
Now we are looking for the smallest prime divisor p 2 of the number a 1 = 39. We start iterating over the numbers from the table of primes, starting with p 1 = 2. Divide 39 by 2, we get 39: 2 = 19 (rest. 1). Since 39 is not divisible by 2, 2 is not a divisor of it. Then we take next number from the table of prime numbers (number 3) and divide by 39, we get 39: 3 = 13. Therefore, p 2 = 3 is the smallest prime divisor of 39, while a 2 = a 1: p 2 = 39: 3 = 13. We have the equality a = p 1 p 2 a 2 in the form 78 = 2 3 13. Since a 2 = 13 is different from 1, then go to the next step of the algorithm.
Here we need to find the smallest prime divisor of the number a 2 = 13. In search of the smallest prime divisor p 3 of 13, we will sequentially iterate over the numbers from the table of primes, starting with p 2 = 3. The number 13 is not divisible by 3, since 13: 3 = 4 (rest. 1), also 13 is not divisible by 5, 7 and 11, since 13: 5 = 2 (rest. 3), 13: 7 = 1 (rest. 6) and 13:11 = 1 (rest. 2). The next prime number is 13, and 13 is divisible by it without a remainder, therefore, the smallest prime divisor p 3 of 13 is the number 13 itself, and a 3 = a 2: p 3 = 13: 13 = 1. Since a 3 = 1, then this step of the algorithm is the last, and the required decomposition of 78 into prime factors has the form 78 = 2 · 3 · 13 (a = p 1 · p 2 · p 3).
Answer:
78 = 2 3 13.
Example.
Present the number 83,006 as a product of prime factors.
Solution.
At the first step of the algorithm for decomposing a number into prime factors, we find p 1 = 2 and a 1 = a: p 1 = 83 006: 2 = 41 503, whence 83 006 = 2 · 41 503.
At the second step, we find out that 2, 3 and 5 are not prime divisors of the number a 1 = 41 503, and the number 7 is, since 41 503: 7 = 5 929. We have p 2 = 7, a 2 = a 1: p 2 = 41 503: 7 = 5 929. Thus, 83 006 = 2 7 5 929.
The smallest prime factor of a 2 = 5 929 is 7, since 5 929: 7 = 847. Thus, p 3 = 7, a 3 = a 2: p 3 = 5 929: 7 = 847, whence 83 006 = 2 7 7 847.
Then we find that the smallest prime divisor p 4 of the number a 3 = 847 is 7. Then a 4 = a 3: p 4 = 847: 7 = 121, therefore 83 006 = 2 7 7 7 7 121.
Now we find the smallest prime divisor of the number a 4 = 121, it is the number p 5 = 11 (since 121 is divisible by 11 and not divisible by 7). Then a 5 = a 4: p 5 = 121: 11 = 11, and 83 006 = 2 7 7 7 11 11.
Finally, the smallest prime factor of a 5 = 11 is p 6 = 11. Then a 6 = a 5: p 6 = 11: 11 = 1. Since a 6 = 1, then this step of the algorithm for decomposing a number into prime factors is the last, and the required decomposition has the form 83 006 = 2 · 7 · 7 · 7 · 11 · 11.
The result obtained can be written as the canonical factorization of a number into prime factors 83 006 = 2 · 7 3 · 11 2.
Answer:
83 006 = 2 7 7 7 11 11 = 2 7 3 11 2 991 is a prime number. Indeed, it does not have a single prime divisor not exceeding (can be roughly estimated as, since it is obvious that 991<40 2
), то есть, наименьшим делителем числа 991
является оно само. Тогда p 3 =991
и a 3 =a 2:p 3 =991:991=1
. Следовательно, искомое разложение числа 897 924 289
на простые множители имеет вид 897 924 289=937·967·991
.
Answer:
897 924 289 = 937 967 991.
Using divisibility criteria for prime factorization
In simple cases, you can decompose a number into prime factors without using the decomposition algorithm from the first paragraph of this article. If the numbers are not large, then for their decomposition into prime factors it is often enough to know the divisibility criteria. Here are some examples for clarification.
For example, we need to factor 10 into prime factors. From the multiplication table, we know that 2 · 5 = 10, and the numbers 2 and 5 are obviously prime, so the prime factorization of 10 is 10 = 2 · 5.
Another example. Use the multiplication table to factor 48 into prime factors. We know that six eight is forty-eight, that is, 48 = 6 · 8. However, neither 6 nor 8 are prime numbers. But we know that twice three is six, and twice four is eight, that is, 6 = 2 · 3 and 8 = 2 · 4. Then 48 = 6 8 = 2 3 2 4. It remains to remember that two times two is four, then we get the required decomposition into prime factors 48 = 2 · 3 · 2 · 2 · 2. We write this decomposition in the canonical form: 48 = 2 4 · 3.
But when decomposing the number 3 400 into prime factors, you can use the divisibility criteria. Divisibility by 10, 100 allows us to assert that 3400 is divisible by 100, while 3400 = 34100, and 100 is divisible by 10, while 100 = 1010, therefore, 3400 = 341010. And on the basis of the divisibility criterion by 2, it can be argued that each of the factors 34, 10 and 10 is divisible by 2, we get 3 400 = 34 10 10 = 2 17 2 5 2 5... All factors in the resulting decomposition are prime, so this decomposition is the desired one. It remains only to rearrange the factors so that they go in ascending order: 3400 = 2 · 2 · 2 · 5 · 5 · 17. We also write down the canonical factorization of this number into prime factors: 3 400 = 2 3 · 5 2 · 17.
When decomposing a given number into prime factors, you can use in turn both the divisibility criteria and the multiplication table. Let's represent the number 75 as a product of prime factors. Divisibility by 5 allows us to assert that 75 is divisible by 5, and we get that 75 = 5 15. And from the multiplication table we know that 15 = 3 · 5, therefore, 75 = 5 · 3 · 5. This is the required prime factorization of 75.
Bibliography.
- Vilenkin N.Ya. and other Mathematics. Grade 6: textbook for educational institutions.
- Vinogradov I.M. Fundamentals of number theory.
- Mikhelovich Sh.Kh. Number theory.
- Kulikov L.Ya. and others. Collection of problems in algebra and number theory: a textbook for students of physics and mathematics. specialties of pedagogical institutes.