C 37 factorization of polynomials. Factorization

In order to factorize, it is necessary to simplify the expressions. This is necessary so that it can be further reduced. The expansion of a polynomial makes sense when its degree is not lower than 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 of factoring a polynomial.

Theory

Theorem 1

When 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, are represented as a product with a constant factor with the highest degree a n and n linear factors (x - x i), i = 1, 2, ..., n, then P n (x) = a n (x - x n) (x - x n - 1) · . . . · (x - x 1) , where x i, i = 1, 2, …, n 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 occur in conjugate pairs. For example, roots x 1 and x 2 related 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 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 a polynomial can be repeated. Let's consider the proof of the algebra theorem, a consequence of Bezout's theorem.

Fundamental theorem of algebra

Theorem 2

Any polynomial with degree n has at least one root.

Bezout's theorem

After dividing a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 on (x - s), then we get the remainder, which is equal to the polynomial at 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 with degree n - 1.

Corollary to Bezout's theorem

When the root of the polynomial P n (x) is considered to be 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 the solution.

Factoring a quadratic trinomial

A square trinomial of the form a x 2 + b x + c can be factorized 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).

This shows that the expansion itself reduces to solving the quadratic equation subsequently.

Example 1

Factor the quadratic trinomial.

Solution

It is necessary to 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 using the formula, then we get D = (- 5) 2 - 4 · 4 · 1 = 9. From here 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.

To perform the check, you need to open the parentheses. 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 arrive at the original expression. That is, we can conclude that the decomposition was performed correctly.

Example 2

Factor the quadratic trinomial of the form 3 x 2 - 7 x - 11 .

Solution

We find 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 we 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 the expansion itself can be depicted as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.

Example 4

Decompose the quadratic 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 · i x 2 = - 1 3 - D 2 · 1 = - 1 3 - 35 3 · i 2 = - 1 - 35 · i 6 = - 1 6 - 35 6 · i

Having obtained 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 discriminant value is negative, then the polynomials will remain second-order polynomials. It follows from this that we will not expand them into linear factors.

Methods for factoring a polynomial of degree higher than two

When decomposing, a universal method is assumed. Most of all cases are based on a corollary of Bezout's theorem. To do this, you need to select the value of the root x 1 and reduce 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 obtain a complete expansion.

If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic posits a solution to the equations with higher degrees and integer coefficients.

Taking the common factor out of brackets

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 to be taking the common factor out of brackets.

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 remove x from the brackets of the entire expression. We get:

4 x 3 + 8 x 2 - x = x (4 x 2 + 8 x - 1)

Let's move on 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 x x - - 1 + 5 2 x - - 1 - 5 2 = = 4 x x + 1 - 5 2 x + 1 + 5 2

To begin with, let us take into consideration a 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 of the highest degree is 1.

When a polynomial has integer roots, then they are considered divisors of the free term.

Example 6

Decompose the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.

Solution

Let's consider whether there are complete 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 integer roots. You can check using Horner's scheme. It is very convenient and allows you to quickly obtain the expansion coefficients of a 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 proceed to the expansion of a quadratic trinomial of the form x 2 + 2 x + 3.

Since the discriminant is negative, it means real roots No.

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 root selection and division of a polynomial by a polynomial instead of Horner's scheme. Let's move on to considering 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 highest of which is equal to one.

This case occurs for rational fractions.

Example 7

Factorize f (x) = 2 x 3 + 19 x 2 + 41 x + 15 .

Solution

It is necessary to replace the variable y = 2 x, you should move on 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 their location is among the divisors of the free term. The entry will look like:

±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60

Let's move on to calculating the function g (y) at these points in order to get 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 find that y = - 5 is the root of an equation of the form y 3 + 19 y 2 + 82 y + 60, which means 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 it down 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 factorize the resulting quadratic trinomial of the form x 2 + 7 x + 3. By equating to zero we 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

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 techniques for factoring a polynomial

Rational roots are not inherent in all polynomials. To do this you need to use in special ways to find factors. But not all polynomials can be expanded or represented as a product.

Grouping method

There are cases when you can group the terms of a polynomial to find a common factor and put it out of brackets.

Example 9

Factor the polynomial x 4 + 4 x 3 - x 2 - 8 x - 2.

Solution

Because the coefficients are integers, then the roots can presumably 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

This shows that there are no roots; it is necessary to use another method of expansion 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, you need to represent it as the product of two square trinomials. To do this we need to factorize. 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 grouping does not mean that choosing terms is easy enough. There is no specific solution method, 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 integer roots. The terms should be grouped. 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 factorization 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 to factor a polynomial

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 Newton’s 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 sequence of coefficients of the sum in parentheses is indicated by the expression x + 1 4 .

This means 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 that is in the second bracket. It is clear that there are no knights there, so we should apply the difference of squares formula 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

Factorize x 3 + 6 x 2 + 12 x + 6 .

Solution

Let's start transforming 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 method for replacing a variable when factoring a polynomial

When replacing a variable, the degree is reduced and the polynomial is factored.

Example 13

Factor the 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 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 obtained the desired decomposition.

The cases discussed above will help in considering and factoring a polynomial in different ways.

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Let's look at specific examples, how to factor a polynomial.

We will expand the polynomials in accordance with .

Factor polynomials:

Let's check if there is a common factor. yes, it is equal to 7cd. Let's take it out of brackets:

The expression in parentheses consists of two terms. There is no longer a common factor, the expression is not a formula for the sum of cubes, which means the decomposition is complete.

Let's check if there is a common factor. No. The polynomial consists of three terms, so we check to see if there is a formula for a complete square. Two terms are the squares of the expressions: 25x²=(5x)², 9y²=(3y)², the third term is equal to the double product of these expressions: 2∙5x∙3y=30xy. This means that this polynomial is a perfect square. Since the double product has a minus sign, it is:

We check whether it is possible to take the common factor out of brackets. There is a common factor, it is equal to a. Let's take it out of brackets:

There are two terms in brackets. We check if there is a formula for difference of squares or difference of cubes. a² is the square of a, 1=1². This means that the expression in brackets can be written using the difference of squares formula:

There is a common factor, it is equal to 5. Let’s take it out of brackets:

in brackets there are three terms. We check whether the expression is a perfect square. Two terms are squares: 16=4² and a² - the square of a, the third term is equal to the double product of 4 and a: 2∙4∙a=8a. Therefore, it is a perfect square. Since all terms have a “+” sign, the expression in parentheses is the perfect square of the sum:

We take the general multiplier -2x out of brackets:

In parentheses is the sum of two terms. We check whether this expression is a sum of cubes. 64=4³, x³- cube x. This means that the binomial can be expanded using the formula:

There is a common multiplier. But, since the polynomial consists of 4 terms, we will first, and only then, take the common factor out of brackets. Let’s group the first term with the fourth, and the second with the third:

From the first brackets we take out the common factor 4a, from the second - 8b:

There is no common multiplier yet. To get it, we take out the “-“ from the second brackets, and each sign in the brackets changes to the opposite:

Now let’s take the common factor (1-3a) out of brackets:

In the second brackets there is a common factor 4 (this is the same factor that we did not put out of brackets at the beginning of the example):

Since the polynomial consists of four terms, we perform grouping. Let’s group the first term with the second, the third with the fourth:

In the first brackets there is no common factor, but there is a formula for the difference of squares, in the second brackets the common factor is -5:

A common multiplier has appeared (4m-3n). Let's take it out of the equation.

Factoring a large number is not an easy task. Most people have trouble figuring out four or five digit numbers. To make the process easier, write the number above the two columns.

• Let's factorize the number 6552.

Divide given number the smallest prime divisor (other than 1) that divides a given number without a remainder. Write this divisor in the left column, and write the result of the division in the right column. As noted above, even numbers are easy to factor, since their smallest prime factor will always be the number 2 (odd numbers have smallest prime factors are different).

• In our example, 6552 is an even number, so 2 is its smallest prime factor. 6552 ÷ 2 = 3276. Write 2 in the left column and 3276 in the right column.

Next, divide the number in the right column by the smallest prime factor (other than 1) that divides the number without a remainder. Write this divisor in the left column, and in the right column write the result of the division (continue this process until there are no 1 left in the right column).

• In our example: 3276 ÷ 2 = 1638. Write 2 in the left column, and 1638 in the right column. Next: 1638 ÷ 2 = 819. Write 2 in the left column, and 819 in the right column.

You got an odd number; For such numbers, finding the smallest prime divisor is more difficult. If you get an odd number, try dividing it by the smallest prime odd numbers: 3, 5, 7, 11.

• In our example, you received an odd number 819. Divide it by 3: 819 ÷ 3 = 273. Write 3 in the left column and 273 in the right column.
• When looking for factors, try all the prime numbers up to the square root of the largest factor you find. If no divisor divides the number by a whole, then you most likely have a prime number and can stop calculating.

Continue the process of dividing numbers by prime factors until you are left with a 1 in the right column (if you get a prime number in the right column, divide it by itself to get a 1).

• Let's continue the calculations in our example:
  • Divide by 3: 273 ÷ 3 = 91. There is no remainder. Write down 3 in the left column and 91 in the right column.
  • Divide by 3. 91 is divisible by 3 with a remainder, so divide by 5. 91 is divisible by 5 with a remainder, so divide by 7: 91 ÷ 7 = 13. No remainder. Write down 7 in the left column and 13 in the right column.
  • Divide by 7. 13 is divisible by 7 with a remainder, so divide by 11. 13 is divisible by 11 with a remainder, so divide by 13: 13 ÷ 13 = 1. There is no remainder. Write 13 in the left column and 1 in the right column. Your calculations are complete.

The left column shows the prime factors of the original number. In other words, when you multiply all the numbers in the left column, you will get the number written above the columns. If the same factor appears more than once in the list of factors, use exponents to indicate it. In our example, 2 appears 4 times in the list of multipliers; write these factors as 2 4 rather than 2*2*2*2.

• In our example, 6552 = 2 3 × 3 2 × 7 × 13. You factored 6552 into prime factors (the order of the factors in this notation does not matter).

Very often, the numerator and denominator of a fraction are algebraic expressions that must first be factored, and then, having found identical ones among them, divide both the numerator and denominator by them, that is, reduce the fraction. An entire chapter of the 7th grade algebra textbook is devoted to the task of factoring a polynomial. Factorization can be done 3 ways, as well as a combination of these methods.

1. Application of abbreviated multiplication formulas

As is known, to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial and add the resulting products. There are at least 7 (seven) frequently occurring cases of multiplying polynomials that are included in the concept. For example,

Table 1. Factorization in the 1st way

2. Taking the common factor out of brackets

This method is based on the application of the distributive multiplication law. For example,

We divide each term of the original expression by the factor that we take out, and we get an expression in parentheses (that is, the result of dividing what was by what we take out remains in parentheses). First of all you need determine the multiplier correctly, which must be taken out of the bracket.

The common factor can also be a polynomial in brackets:

When performing the “factorize” task, you need to be especially careful with the signs when putting the total factor out of brackets. To change the sign of each term in a parenthesis (b - a), let’s take the common factor out of brackets -1 , and each term in the bracket will be divided by -1: (b - a) = - (a - b) .

If the expression in brackets is squared (or to any even power), then numbers inside brackets can be swapped completely freely, since the minuses taken out of brackets will still turn into a plus when multiplied: (b - a) 2 = (a - b) 2, (b - a) 4 = (a - b) 4 and so on…

3. Grouping method

Sometimes not all terms in an expression have a common factor, but only some. Then you can try group terms in brackets so that some factor can be taken out of each one. Grouping method- this is a double removal of common factors from brackets.

4. Using several methods at once

Sometimes you need to apply not one, but several methods of factoring a polynomial at once.

This is a summary of the topic "Factorization". Select next steps:

• Go to next summary:

Expanding polynomials to obtain a product can sometimes seem confusing. But it's not that difficult if you understand the process step by step. The article describes in detail how to factor a quadratic trinomial.

Many people do not understand how to factor a square trinomial, and why this is done. At first it may seem like a futile exercise. But in mathematics nothing is done for nothing. The transformation is necessary to simplify the expression and ease of calculation.

A polynomial of the form – ax²+bx+c, called a quadratic trinomial. The term "a" must be negative or positive. In practice, this expression is called a quadratic equation. Therefore, sometimes they say it differently: how to expand a quadratic equation.

Interesting! A polynomial is called a square because of its largest degree, the square. And a trinomial - because of the 3 components.

Some other types of polynomials:

• linear binomial (6x+8);
• cubic quadrinomial (x³+4x²-2x+9).

Factoring a quadratic trinomial

First, the expression is equal to zero, then you need to find the values ​​of the roots x1 and x2. There may be no roots, there may be one or two roots. The presence of roots is determined by the discriminant. You need to know its formula by heart: D=b²-4ac.

If the result D is negative, there are no roots. If positive, there are two roots. If the result is zero, the root is one. The roots are also calculated using the formula.

If, when calculating the discriminant, the result is zero, you can use any of the formulas. In practice, the formula is simply shortened: -b / 2a.

Formulas for different meanings discriminants differ.

If D is positive:

If D is zero:

Online calculators

On the Internet there is online calculator. It can be used to perform factorization. Some resources provide the opportunity to view the solution step by step. Such services help to better understand the topic, but you need to try to understand it well.

Useful video: Factoring a quadratic trinomial

Examples

We invite you to view simple examples, how to factor a quadratic equation.

Example 1

This clearly shows that the result is two x's because D is positive. They need to be substituted into the formula. If the roots turn out to be negative, the sign in the formula changes to the opposite.

We know the formula for factoring a quadratic trinomial: a(x-x1)(x-x2). We put the values ​​in brackets: (x+3)(x+2/3). There is no number before a term in a power. This means that there is one there, it goes down.

Example 2

This example clearly shows how to solve an equation that has one root.

We substitute the resulting value:

Example 3

Given: 5x²+3x+7

First, let's calculate the discriminant, as in previous cases.

D=9-4*5*7=9-140= -131.

The discriminant is negative, which means there are no roots.

After receiving the result, you should open the brackets and check the result. The original trinomial should appear.

Alternative solution

Some people were never able to make friends with the discriminator. There is another way to factorize a quadratic trinomial. For convenience, the method is shown with an example.

Given: x²+3x-10

We know that we should get 2 brackets: (_)(_). When the expression looks like this: x²+bx+c, at the beginning of each bracket we put x: (x_)(x_). The remaining two numbers are the product that gives “c”, i.e. in this case -10. The only way to find out what numbers these are is by selection. The substituted numbers must correspond to the remaining term.

For example, multiplying the following numbers gives -10:

• -1, 10;
• -10, 1;
• -5, 2;
• -2, 5.

1. (x-1)(x+10) = x2+10x-x-10 = x2+9x-10. No.
2. (x-10)(x+1) = x2+x-10x-10 = x2-9x-10. No.
3. (x-5)(x+2) = x2+2x-5x-10 = x2-3x-10. No.
4. (x-2)(x+5) = x2+5x-2x-10 = x2+3x-10. Fits.

This means that the transformation of the expression x2+3x-10 looks like this: (x-2)(x+5).

Important! You should be careful not to confuse the signs.

Expansion of a complex trinomial

If “a” is greater than one, difficulties begin. But everything is not as difficult as it seems.

To factorize, you first need to see if anything can be factored out.

For example, given the expression: 3x²+9x-30. Here the number 3 is taken out of brackets:

3(x²+3x-10). The result is the already well-known trinomial. The answer looks like this: 3(x-2)(x+5)

How to decompose if the term that is in the square is negative? IN in this case The number -1 is taken out of brackets. For example: -x²-10x-8. The expression will then look like this:

The scheme differs little from the previous one. There are just a few new things. Let's say the expression is given: 2x²+7x+3. The answer is also written in 2 brackets that need to be filled in (_)(_). In the 2nd bracket is written x, and in the 1st what is left. It looks like this: (2x_)(x_). Otherwise, the previous scheme is repeated.

The number 3 is given by the numbers:

• -1, -3;
• -3, -1;
• 3, 1;
• 1, 3.

We solve equations by substituting these numbers. The last option is suitable. This means that the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).

Other cases

It is not always possible to convert an expression. With the second method, solving the equation is not required. But the possibility of transforming terms into a product is checked only through the discriminant.

It's worth practicing to decide quadratic equations so that there are no difficulties when using formulas.

Useful video: factoring a trinomial

Conclusion

You can use it in any way. But it’s better to practice both until they become automatic. Also, learning how to solve quadratic equations well and factor polynomials is necessary for those who are planning to connect their lives with mathematics. All the following mathematical topics are built on this.