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.

The formulas for different discriminant values ​​are different.

If D is positive:

If D is zero:

Online calculators

There is an online calculator on the Internet. 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 suggest looking at simple examples of 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 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 is worth practicing solving quadratic equations so that when using the formulas there are no difficulties.

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.

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Factoring polynomials is an identity transformation, as a result of which a polynomial is transformed into the product of several factors - polynomials or monomials.

There are several ways to factor polynomials.

Method 1. Taking the common factor out of brackets.

This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to isolate the common factor in the two components under consideration and “take” it out of brackets.

Let us factor the polynomial 28x 3 – 35x 4.

Solution.

1. Find a common divisor for the elements 28x3 and 35x4. For 28 and 35 it will be 7; for x 3 and x 4 – x 3. In other words, our common factor is 7x 3.

2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x.

3. We take the common factor out of brackets
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x = 7x 3 (4 – 5x).

Method 2. Using abbreviated multiplication formulas. The “mastery” of using this method is to notice one of the abbreviated multiplication formulas in the expression.

Let us factor the polynomial x 6 – 1.

Solution.

1. We can apply the difference of squares formula to this expression. To do this, imagine 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. We can apply the formula for the sum and difference of cubes to the resulting expression:
(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 is to combine the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, subtraction of a common factor).

Let's factor the polynomial x 3 – 3x 2 + 5x – 15.

Solution.

1. Let's group the components in this way: 1st with 2nd, and 3rd with 4th
(x 3 – 3x 2) + (5x – 15).

2. In the resulting expression, we take the common factors out of 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. We take the common factor x – 3 out of brackets 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 secure 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's group the components of the polynomial in this way: 1st with 2nd and 3rd with 4th. We get:
(a 2 – 3ab) – (4ab – 12b 2).

3. Let’s take the common factors out of brackets:
(a 2 – 3ab) – (4ab – 12b 2) = a(a – 3b) – 4b(a – 3b).

4. Let’s take the common factor (a – 3b) out of brackets:
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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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.

The article will cover all the concepts of decomposition, theoretical foundations 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 the complex roots 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 involves solving equations with higher powers 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 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 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 by 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 special methods 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 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 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

Factorize 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 .

So 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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In general, this task requires a creative approach, since there is no universal method for solving it. But let's try to give a few tips.

In the overwhelming majority of cases, the factorization of a polynomial is based on a corollary of Bezout’s theorem, that is, the root is found or selected and the degree of the polynomial is reduced by one by dividing by . The root of the resulting polynomial is sought and the process is repeated until complete expansion.

If the root cannot be found, then specific expansion methods are used: from grouping to introducing additional mutually exclusive terms.

Further presentation is based on the skills of solving equations of higher degrees with integer coefficients.

Bracketing out the common factor.

Let's start with the simplest case, when the free term is equal to zero, that is, the polynomial has the form .

Obviously, the root of such a polynomial is , that is, we can represent the polynomial in the form .

This method is nothing more than putting the common factor out of brackets.

Example.

Factor a third degree polynomial.

Solution.

Obviously, what is the root of the polynomial, that is X can be taken out of brackets:

Let's find the roots of the quadratic trinomial

Thus,

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Factoring a polynomial with rational roots.

First, let's consider a method for expanding a polynomial with integer coefficients of the form , the coefficient of the highest degree is equal to one.

In this case, if a polynomial has integer roots, then they are divisors of the free term.

Example.

Solution.

Let's check if there are intact roots. To do this, write down the divisors of the number -18 : . That is, if a polynomial has integer roots, then they are among the written numbers. Let's check these numbers sequentially using Horner's scheme. Its convenience also lies in the fact that in the end we obtain the expansion coefficients of the polynomial:

That is, x=2 And x=-3 are the roots of the original polynomial and we can represent it as a product:

It remains to expand the quadratic trinomial.

The discriminant of this trinomial is negative, therefore it has no real roots.

Answer:

Comment:

Instead of Horner's scheme, one could use the selection of the root and subsequent division of the polynomial by a polynomial.

Now consider the expansion of a polynomial with integer coefficients of the form , and the coefficient of the highest degree is not equal to one.

In this case, the polynomial can have fractionally rational roots.

Example.

Factor the expression.

Solution.

By performing a variable change y=2x, let's move on to a polynomial with a coefficient equal to one at the highest degree. To do this, first multiply the expression by 4 .

If the resulting function has integer roots, then they are among the divisors of the free term. Let's write them down:

Let us sequentially calculate the values ​​of the function g(y) at these points until zero is reached.

That is, y=-5 is the root , therefore, is the root of the original function. Let's divide the polynomial by a column (corner) into a binomial.

Thus,

It is not advisable to continue checking the remaining divisors, since it is easier to factorize the resulting quadratic trinomial

Hence,

Unknown polynomials. The theorem about the distribution of polynomials in additions of unknowns. Canonical layout of a polynomial.

A polynomial is an expression consisting of the sum of monomials. The latter are the product of a constant (number) and the root (or roots) of the expression to the power of k. In this case, we speak of a polynomial of degree k. The expansion of a polynomial involves a transformation of the expression in which the terms are replaced by factors. Let's consider the main ways to carry out this kind of transformation.

Method of expanding a polynomial by isolating a common factor

This method is based on the laws of the distribution law. So, mn + mk = m * (n + k).

• Example: expand 7y 2 + 2uy and 2m 3 – 12m 2 + 4lm.

7y 2 + 2uy = y * (7y + 2u),

2m 3 – 12m 2 + 4lm = 2m(m 2 – 6m + 2l).

However, the factor that is necessarily present in each polynomial may not always be found, so this method is not universal.

Polynomial expansion method based on abbreviated multiplication formulas

Abbreviated multiplication formulas are valid for polynomials of any degree. In general, the transformation expression looks like this:

u k – l k = (u – l)(u k-1 + u k-2 * l + u k-3 *l 2 + … u * l k-2 + l k-1), where k is a representative of natural numbers .

The formulas most often used in practice are for polynomials of the second and third orders:

u 2 – l 2 = (u – l)(u + l),

u 3 – l 3 = (u – l)(u 2 + ul + l 2),

u 3 + l 3 = (u + l)(u 2 – ul + l 2).

• Example: expand 25p 2 – 144b 2 and 64m 3 – 8l 3.

25p 2 – 144b 2 = (5p – 12b)(5p + 12b),

64m 3 – 8l 3 = (4m) 3 – (2l) 3 = (4m – 2l)((4m) 2 + 4m * 2l + (2l) 2) = (4m – 2l)(16m 2 + 8ml + 4l 2 ).

Polynomial expansion method - grouping terms of an expression

This method in some way has something in common with the technique of deriving the common factor, but has some differences. In particular, before isolating a common factor, the monomials should be grouped. The grouping is based on the rules of combinational and commutative laws.

All monomials presented in the expression are divided into groups, in each of which a common value is given such that the second factor will be the same in all groups. In general, this decomposition method can be represented as the expression:

pl + ks + kl + ps = (pl + ps) + (ks + kl) ⇒ pl + ks + kl + ps = p(l + s) + k(l + s),

pl + ks + kl + ps = (p + k)(l + s).

• Example: spread out 14mn + 16ln – 49m – 56l.

14mn + 16ln – 49m – 56l = (14mn – 49m) + (16ln – 56l) = 7m * (2n – 7) + 8l * (2n – 7) = (7m + 8l)(2n – 7).

Polynomial expansion method - forming a perfect square

This method is one of the most effective in the expansion of a polynomial. At the initial stage, it is necessary to determine monomials that can be “collapsed” into the square of the difference or sum. To do this, use one of the relations:

(p – b) 2 = p 2 – 2pb + b 2 ,

• Example: expand the expression u 4 + 4u 2 – 1.

Among its monomials, we select the terms that form a complete square: u 4 + 4u 2 – 1 = u 4 + 2 * 2u 2 + 4 – 4 – 1 =

= (u 4 + 2 * 2u 2 + 4) – 4 – 1 = (u 4 + 2 * 2u 2 + 4) – 5.

Complete the transformation using the abbreviated multiplication rules: (u 2 + 2) 2 – 5 = (u 2 + 2 – √5)(u 2 + 2 + √5).

That. u 4 + 4u 2 – 1 = (u 2 + 2 – √5)(u 2 + 2 + √5).