How to factor a quadratic trinomial into linear factors. Examples of factoring polynomials. Bracketing out the common factor

8 examples of factorization of polynomials are given. They include examples of solving quadratic and biquadratic equations, examples of reciprocal polynomials, and examples of finding integer roots of third- and fourth-degree polynomials.

Content

See also: Methods for factoring polynomials
Roots of a quadratic equation
Solving cubic equations

1. Examples with solving a quadratic equation

Example 1.1

x 4 + x 3 - 6 x 2.

We take out x 2 outside of brackets:
.
2 + x - 6 = 0:
.
Roots of the equation:
, .

.

Example 1.2

Factor the third degree polynomial:
x 3 + 6 x 2 + 9 x.

Let's take x out of brackets:
.
Solving the quadratic equation x 2 + 6 x + 9 = 0:
Its discriminant: .
Since the discriminant is zero, the roots of the equation are multiples: ;
.

From here we obtain the factorization of the polynomial:
.

Example 1.3

Factor the fifth degree polynomial:
x 5 - 2 x 4 + 10 x 3.

We take out x 3 outside of brackets:
.
Solving the quadratic equation x 2 - 2 x + 10 = 0.
Its discriminant: .
Since the discriminant is less than zero, the roots of the equation are complex: ;
, .

The factorization of the polynomial has the form:
.

If we are interested in factorization with real coefficients, then:
.

Examples of factoring polynomials using formulas

Examples with biquadratic polynomials

Example 2.1

Factor the biquadratic polynomial:
x 4 + x 2 - 20.

Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).

;
.

Example 2.2

Factor the polynomial that reduces to a biquadratic one:
x 8 + x 4 + 1.

Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b):

;

;
.

Example 2.3 with recurrent polynomial

Factor the reciprocal polynomial:
.

A reciprocal polynomial has odd degree. Therefore it has root x = - 1 . Divide the polynomial by x - (-1) = x + 1. As a result we get:
.
Let's make a substitution:
, ;
;


;
.

Examples of factoring polynomials with integer roots

Example 3.1

Factor the polynomial:
.

Let's assume that the equation

6
-6, -3, -2, -1, 1, 2, 3, 6 .
(-6) 3 - 6·(-6) 2 + 11·(-6) - 6 = -504;
(-3) 3 - 6·(-3) 2 + 11·(-3) - 6 = -120;
(-2) 3 - 6·(-2) 2 + 11·(-2) - 6 = -60;
(-1) 3 - 6·(-1) 2 + 11·(-1) - 6 = -24;
1 3 - 6 1 2 + 11 1 - 6 = 0;
2 3 - 6 2 2 + 11 2 - 6 = 0;
3 3 - 6 3 2 + 11 3 - 6 = 0;
6 3 - 6 6 2 + 11 6 - 6 = 60.

So, we found three roots:
x 1 = 1 , x 2 = 2 , x 3 = 3 .
Since the original polynomial is of the third degree, it has no more than three roots. Since we found three roots, they are simple. Then
.

Example 3.2

Factor the polynomial:
.

Let's assume that the equation

has at least one whole root. Then it is a divisor of the number 2 (member without x). That is, the whole root can be one of the numbers:
-2, -1, 1, 2 .
We substitute these values ​​one by one:
(-2) 4 + 2·(-2) 3 + 3·(-2) 3 + 4·(-2) + 2 = 6 ;
(-1) 4 + 2·(-1) 3 + 3·(-1) 3 + 4·(-1) + 2 = 0 ;
1 4 + 2 1 3 + 3 1 3 + 4 1 + 2 = 12;
2 4 + 2 2 3 + 3 2 3 + 4 2 + 2 = 54.

So, we found one root:
x 1 = -1 .
Divide the polynomial by x - x 1 = x - (-1) = x + 1:


Then,
.

Now we need to solve the third degree equation:
.
If we assume that this equation has an integer root, then it is a divisor of the number 2 (member without x). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Let's substitute x = -1 :
.

So, we have found another root x 2 = -1 . It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:
.

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 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).

From this it is clear that the expansion itself reduces to the solution 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

Expand 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 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 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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In this lesson we will learn how to factor quadratic trinomials into linear factors. To do this, we need to remember Vieta’s theorem and its converse. This skill will help us quickly and conveniently expand quadratic trinomials into linear factors, and will also simplify the reduction of fractions consisting of expressions.

So let's go back to the quadratic equation, where .

What we have on the left side is called a quadratic trinomial.

The theorem is true: If are the roots of a quadratic trinomial, then the identity holds

Where is the leading coefficient, are the roots of the equation.

So, we have a quadratic equation - a quadratic trinomial, where the roots of the quadratic equation are also called the roots of the quadratic trinomial. Therefore, if we have the roots of a square trinomial, then this trinomial is decomposed into linear factors.

Proof:

The proof of this fact is carried out using Vieta’s theorem, which we discussed in previous lessons.

Let's remember what Vieta's theorem tells us:

If are the roots of a quadratic trinomial for which , then .

The following statement follows from this theorem:

We see that, according to Vieta’s theorem, i.e., by substituting these values ​​into the formula above, we obtain the following expression

Q.E.D.

Recall that we proved the theorem that if are the roots of a square trinomial, then the expansion is valid.

Now let's remember an example of a quadratic equation, to which we selected roots using Vieta's theorem. From this fact we can obtain the following equality thanks to the proven theorem:

Now let's check the correctness of this fact by simply opening the brackets:

We see that we factored correctly, and any trinomial, if it has roots, can be factorized according to this theorem into linear factors according to the formula

However, let's check whether such factorization is possible for any equation:

Take, for example, the equation . First, let's check the discriminant sign

And we remember that in order to fulfill the theorem we learned, D must be greater than 0, so in this case, factorization according to the theorem we learned is impossible.

Therefore, we formulate a new theorem: if a square trinomial has no roots, then it cannot be decomposed into linear factors.

So, we have looked at Vieta's theorem, the possibility of decomposing a quadratic trinomial into linear factors, and now we will solve several problems.

Task No. 1

In this group we will actually solve the problem inverse to the one posed. We had an equation, and we found its roots by factoring it. Here we will do the opposite. Let's say we have the roots of a quadratic equation

The inverse problem is this: write a quadratic equation using its roots.

There are 2 ways to solve this problem.

Since are the roots of the equation, then is a quadratic equation whose roots are given numbers. Now let's open the brackets and check:

This was the first way in which we created a quadratic equation with given roots, which does not have any other roots, since any quadratic equation has at most two roots.

This method involves using converse theorem Vieta.

If are the roots of the equation, then they satisfy the condition that .

For the reduced quadratic equation , , i.e. in this case, and .

Thus, we have created a quadratic equation that has the given roots.

Task No. 2

It is necessary to reduce the fraction.

We have a trinomial in the numerator and a trinomial in the denominator, and the trinomials can either be factorized or not. If both the numerator and the denominator are factored, then among them there may be equal factors that can be reduced.

First of all, you need to factor the numerator.

First, you need to check whether this equation can be factorized, let’s find the discriminant. Since , the sign depends on the product (must be less than 0), in this example, i.e. given equation has roots.

To solve, we use Vieta’s theorem:

In this case, since we are dealing with roots, it will be quite difficult to simply select the roots. But we see that the coefficients are balanced, that is, if we assume that , and substitute this value into the equation, we get the following system: , i.e. 5-5=0. Thus, we have selected one of the roots of this quadratic equation.

We will look for the second root by substituting what is already known into the system of equations, for example, , i.e. .

Thus, we have found both roots of the quadratic equation and can substitute their values ​​into the original equation to factor it:

Let's remember the original problem, we needed to reduce the fraction .

Let's try to solve the problem by substituting .

It is necessary not to forget that in this case the denominator cannot be equal to 0, i.e. , .

If these conditions are met, then we have reduced the original fraction to the form .

Problem No. 3 (task with a parameter)

At what values ​​of the parameter is the sum of the roots of the quadratic equation

If the roots given equation exist, then , question: when.

It is a square, and it consists of three terms (). So it turns out - a square trinomial.

Examples Not square trinomials:

\(x^3-3x^2-5x+6\) - cubic quadrinomial
\(2x+1\) - linear binomial

Square root of the trinomial:

Example:
The trinomial \(x^2-2x+1\) has a root \(1\), because \(1^2-2 1+1=0\)
The trinomial \(x^2+2x-3\) has roots \(1\) and \(-3\), because \(1^2+2-3=0\) and \((-3)^ 2-6-3=9-9=0\)

For example: if you need to find roots for the quadratic trinomial \(x^2-2x+1\), we equate it to zero and solve the equation \(x^2-2x+1=0\).

\(D=4-4\cdot1=0\)
\(x=\frac(2-0)(2)=\frac(2)(2)=1\)

Ready. The root is \(1\).

Decomposition of a quadratic trinomial into:

The square trinomial \(ax^2+bx+c\) can be expanded as \(a(x-x_1)(x-x_2)\) if the equations \(ax^2+bx+c=0\) are greater than zero \ (x_1\) and \(x_2\) are roots of the same equation).

For example, consider the trinomial \(3x^2+13x-10\).
The quadratic equation \(3x^2+13x-10=0\) has a discriminant equal to 289 (greater than zero) and roots equal to \(-5\) and \(\frac(2)(3)\). Therefore \(3x^2+13x-10=3(x+5)(x-\frac(2)(3))\). It is easy to verify the correctness of this statement - if we , then we will obtain the original trinomial.

The square trinomial \(ax^2+bx+c\) can be represented as \(a(x-x_1)^2\) if the discriminant of the equation \(ax^2+bx+c=0\) is zero.

For example, consider the trinomial \(x^2+6x+9\).
The quadratic equation \(x^2+6x+9=0\) has a discriminant equal to \(0\) and a unique root equal to \(-3\). This means \(x^2+6x+9=(x+3)^2\) (here the coefficient is \(a=1\), so it is not written before the bracket - there is no need). Please note that the same conversion can be done by .

The square trinomial \(ax^2+bx+c\) is not factorized if the discriminant of the equation \(ax^2+bx+c=0\) is less than zero.

For example, the trinomials \(x^2+x+4\) and \(-5x^2+2x-1\) have a discriminant less than zero. Therefore, it is impossible to factor them.

Example . Factor \(2x^2-11x+12\).
Solution :
Let's find the roots of the quadratic equation \(2x^2-11x+12=0\)

\(D=11^2-4 \cdot 2 \cdot 12=121-96=25>0\)
\(x_1=\frac(11-5)(4)=1.5;\) \(x_2=\frac(11+5)(4)=4.\)

So, \(2x^2-11x+12=2(x-1.5)(x-4)\)
Answer : \(2(x-1.5)(x-4)\)

The resulting answer may be written differently: \((2x-3)(x-4)\).

Example . (Assignment from the OGE) The square trinomial is factored \(5x^2+33x+40=5(x++ 5)(x-a)\). Find \(a\).
Solution:
\(5x^2+33x+40=0\)
\(D=33^2-4 \cdot 5 \cdot 40=1089-800=289=17^2\)
\(x_1=\frac(-33-17)(10)=-5\)
\(x_2=\frac(-33+17)(10)=-1.6\)
\(5x^2+33x+40=5(x+5)(x+1.6)\)
Answer : \(-1,6\)