The factorization theorem for a square trinomial. Factorization of square trinomials: examples and formulas

Expanding polynomials to get a product sometimes seems confusing. But it is not so difficult if you understand the process step by step. The article details how to factorize a square trinomial.

Many do not understand how to factorize a square trinomial, and why this is done. At first it may seem that this is a useless exercise. But in mathematics, nothing is done just like that. The transformation is necessary to simplify the expression and the convenience of calculation.

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

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

Some other kinds of polynomials:

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

Factorization of a square 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. Its formula must be known by heart: D=b²-4ac.

If the result of 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 by the formula.

If the calculation of the discriminant results in zero, you can apply any of the formulas. In practice, the formula is simply abbreviated: -b / 2a.

Formulas for different values discriminant are different.

If D is positive:

If D is zero:

Online calculators

The Internet has online calculator. It can be used to factorize. Some resources provide the opportunity to see the solution step by step. Such services help to better understand the topic, but you need to try to understand well.

Useful video: Factoring a square trinomial

Examples

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

Example 1

Here it is clearly shown that the result will be two x, because D is positive. They need to be substituted into the formula. If the roots are negative, the sign in the formula is reversed.

We know the decomposition formula square trinomial multipliers: a(x-x1)(x-x2). We put the values ​​in brackets: (x+3)(x+2/3). There is no number before the term in the exponent. This means that there is a unit, it is lowered.

Example 2

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

Substitute the resulting value:

Example 3

Given: 5x²+3x+7

First, we calculate the discriminant, as in the previous cases.

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

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

After receiving the result, it is worth opening the brackets and checking the result. The original trinomial should appear.

Alternative solution

Some people have never been able to make friends with the discriminant. There is another way to factorize a square trinomial. For convenience, the method is shown in an example.

Given: x²+3x-10

We know that we should end up with 2 brackets: (_)(_). When the expression looks like this: x² + bx + c, we put x at the beginning of each bracket: (x_) (x_). The remaining two numbers are the product that gives "c", i.e. -10 in this case. To find out what these numbers are, you can only use the selection method. Substituted numbers must match 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.

So, 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.

Decomposition of a complex trinomial

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

In order to factorize, one must first see if it is possible to factor something 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 known trinomial. The answer looks like this: 3(x-2)(x+5)

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

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

The number 3 gives the numbers:

• -1, -3;
• -3, -1;
• 3, 1;
• 1, 3.

We solve equations by substituting the given numbers. The last option fits. So the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).

Other cases

It is not always possible to transform an expression. In the second method, the solution of the equation is not required. But the possibility of converting terms into a product is checked only through the discriminant.

It is worth practicing solving quadratic equations so that there are no difficulties when using formulas.

Useful video: factorization of a trinomial

Conclusion

You can use it in any way. But it is better to work both to automatism. Also, those who are going to connect their lives with mathematics need to learn how to solve quadratic equations well and decompose polynomials into factors. All the following mathematical topics are built on this.

The factorization of square trinomials refers to school assignments that everyone will face sooner or later. How to do it? What is the formula for factoring a square trinomial? Let's go through it step by step with examples.

General formula

The factorization of square trinomials is carried out by solving a quadratic equation. This is a simple problem that can be solved by several methods - by finding the discriminant, using the Vieta theorem, there exists and graphic way solutions. The first two methods are studied in high school.

The general formula looks like this:lx 2 +kx+n=l(x-x 1)(x-x 2) (1)

Task execution algorithm

In order to factorize square trinomials, you need to know Wit's theorem, have a program for solving at hand, be able to find a solution graphically or look for the roots of an equation of the second degree through the discriminant formula. If a square trinomial is given and it must be factored, the algorithm of actions is as follows:

1) Equate the original expression to zero to get the equation.

2) Give similar terms (if necessary).

3) Find the roots of any known way. The graphical method is best used if it is known in advance that the roots are integers and small numbers. It must be remembered that the number of roots is equal to the maximum degree of the equation, that is, the quadratic equation has two roots.

4) Substitute value X into expression (1).

5) Write down the factorization of square trinomials.

Examples

Practice allows you to finally understand how this task is performed. Examples illustrate the factorization of a square trinomial:

you need to expand the expression:

Let's use our algorithm:

1) x 2 -17x+32=0

2) similar terms are reduced

3) according to the Vieta formula, it is difficult to find the roots for this example, therefore it is better to use the expression for the discriminant:

D=289-128=161=(12.69) 2

4) Substitute the roots we found in the main formula for expansion:

(x-2.155) * (x-14.845)

5) Then the answer will be:

x 2 -17x + 32 \u003d (x-2.155) (x-14.845)

Let's check whether the solutions found by the discriminant correspond to the Vieta formulas:

14,845 . 2,155=32

For these roots, Vieta's theorem is applied, they were found correctly, which means that the factorization we obtained is also correct.

Similarly, we expand 12x 2 + 7x-6.

x 1 \u003d -7 + (337) 1/2

x 2 \u003d -7- (337) 1/2

In the previous case, the solutions were non-integer, but real numbers, which are easy to find with a calculator in front of you. Now consider more complex example, in which the roots will be complex: factorize x 2 + 4x + 9. According to the Vieta formula, the roots cannot be found, and the discriminant is negative. The roots will be on the complex plane.

D=-20

Based on this, we get the roots we are interested in -4 + 2i * 5 1/2 and -4-2i * 5 1/2 because (-20) 1/2 = 2i*5 1/2 .

We obtain the desired expansion by substituting the roots into the general formula.

Another example: you need to factorize the expression 23x 2 -14x + 7.

We have the equation 23x 2 -14x+7 =0

D=-448

So the roots are 14+21,166i and 14-21,166i. The answer will be:

23x 2 -14x+7 =23(x- 14-21,166i )*(X- 14+21.166i ).

Let us give an example that can be solved without the help of the discriminant.

Let it be necessary to decompose the quadratic equation x 2 -32x + 255. Obviously, it can also be solved by the discriminant, but it is faster in this case to find the roots.

x 1 =15

x2=17

Means x 2 -32x + 255 =(x-15)(x-17).

The world is immersed in a huge number of numbers. Any calculations occur with their help.

People learn numbers in order not to fall for deception in later life. It is necessary to devote a huge amount of time to be educated and calculate your own budget.

Mathematics is an exact science that plays a big role in life. At school, children learn numbers, and then, actions on them.

Actions on numbers are completely different: multiplication, expansion, addition, and others. In addition to simple formulas, more complex actions are also used in the study of mathematics. There are a huge number of formulas by which any values ​​\u200b\u200bare known.

At school, as soon as algebra appears, simplification formulas are added to the life of a student. There are equations when there are two unknown numbers, but find in a simple way will not work. A trinomial is a compound of three monomials, with the help of simple method subtractions and additions. The trinomial is solved using the Vieta theorem and the discriminant.

The formula for factoring a square trinomial into factors

There are two correct and simple solutions example:

• discriminant;
• Vieta's theorem.

A square trinomial has an unknown squared, as well as a number without a square. The first option for solving the problem uses the Vieta formula. It's a simple formula if the digits that come before unknown will be the minimum value.

For other equations, where the number is in front of the unknown, the equation must be solved through the discriminant. It's over difficult decision, but the discriminant is used much more often than Vieta's theorem.

Initially, to find all the variables of the equation, it is necessary to raise the example to 0. The solution of the example can be checked and find out whether the numbers are adjusted correctly.

Discriminant

1. It is necessary to equate the equation to 0.

2. Each number before x will be called numbers a, b, c. Since there is no number before the first square x, it equates to 1.

3. Now the solution of the equation begins through the discriminant:

4. Now we have found the discriminant and find two x. The difference is that in one case b will be preceded by a plus, and in the other by a minus:

5. By solving two numbers, it turned out -2 and -1. Substitute under the original equation:

6. In this example, it turned out two correct options. If both solutions are correct, then each of them is true.

More complex equations are also solved through the discriminant. But if the value of the discriminant itself is less than 0, then the example is wrong. The discriminant in the search is always under the root, and a negative value cannot be in the root.

Vieta's theorem

It is used to solve easy problems, where the first x is not preceded by a number, that is, a=1. If the option matches, then the calculation is carried out through the Vieta theorem.

To solve any trinomial it is necessary to raise the equation to 0. The first steps for the discriminant and the Vieta theorem are the same.

2. Now there are differences between the two methods. Vieta's theorem uses not only "dry" calculation, but also logic and intuition. Each number has its own letter a, b, c. The theorem uses the sum and product of two numbers.

Remember! The number b is always added with the opposite sign, and the number c remains unchanged!

Substituting data values ​​in the example , we get:

3. Using the logic method, we substitute the most suitable numbers. Consider all possible solutions:

1. The numbers are 1 and 2. When added, we get 3, but if we multiply, we don’t get 4. Not suitable.
2. Value 2 and -2. When multiplied, it will be -4, but when added, it turns out 0. Not suitable.
3. Numbers 4 and -1. Since the multiplication contains a negative value, it means that one of the numbers will be with a minus. Suitable for addition and multiplication. Correct option.

4. It remains only to check, laying out the numbers, and see if the chosen option is correct.

5. Thanks to an online check, we found out that -1 does not match the condition of the example, which means it is the wrong solution.

When adding negative value in the example, you need to put the number in brackets.

In mathematics there will always be simple tasks and complex. Science itself includes a variety of problems, theorems and formulas. If you understand and correctly apply knowledge, then any difficulties with calculations will be trifling.

Mathematics does not need constant memorization. You need to learn to understand the solution and learn a few formulas. Gradually, according to logical conclusions, it is possible to solve similar problems, equations. Such a science may seem very difficult at first glance, but if one plunges into the world of numbers and tasks, then the view will change dramatically in better side.

Technical specialties always remain the most sought after in the world. Now, in the world modern technologies Mathematics has become an indispensable attribute of any field. You must always remember about useful properties mathematics.

Decomposition of a trinomial with brackets

In addition to solving in the usual ways, there is another one - decomposition into brackets. Used with Vieta's formula.

1. Equate the equation to 0.

ax 2 + bx+ c= 0

2. The roots of the equation remain the same, but instead of zero, they now use bracket expansion formulas.

ax 2 + bx + c = a (x-x 1) (x-x 2)

2 x 2 – 4 x – 6 = 2 (x + 1) (x – 3)

4. Solution x=-1, x=3

Factorization of a square trinomial can be useful when solving inequalities from problem C3 or problem with parameter C5. Also, many B13 word problems will be solved much faster if you know Vieta's theorem.

This theorem, of course, can be considered from the standpoint of the 8th grade, in which it is first passed. But our task is to prepare well for the exam and learn how to solve exam tasks as efficiently as possible. Therefore, in this lesson, the approach is slightly different from the school one.

The formula for the roots of the equation according to Vieta's theorem know (or at least have seen) many:

$$x_1+x_2 = -\frac(b)(a), \quad x_1 x_2 = \frac(c)(a),$$

where `a, b` and `c` are the coefficients of the square trinomial `ax^2+bx+c`.

To learn how to use the theorem easily, let's understand where it comes from (it will be really easier to remember this way).

Let us have the equation `ax^2+ bx+ c = 0`. For further convenience, we divide it by `a` and get `x^2+\frac(b)(a) x + \frac(c)(a) = 0`. Such an equation is called a reduced quadratic equation.

Important lesson points: any square polynomial that has roots can be decomposed into brackets. Suppose ours can be represented as `x^2+\frac(b)(a) x + \frac(c)(a) = (x + k)(x+l)`, where `k` and ` l` - some constants.

Let's see how the brackets open:

$$(x + k)(x+l) = x^2 + kx+ lx+kl = x^2 +(k+l)x+kl.$$

Thus, `k+l = \frac(b)(a), kl = \frac(c)(a)`.

This is slightly different from the classical interpretation Vieta's theorems- in it we are looking for the roots of the equation. I propose to look for terms for bracket expansions- so you don't need to remember about the minus from the formula (meaning `x_1+x_2 = -\frac(b)(a)`). It is enough to choose two such numbers, the sum of which is equal to the average coefficient, and the product is equal to the free term.

If we need a solution to the equation, then it is obvious: the roots `x=-k` or `x=-l` (since in these cases one of the brackets will be zero, which means that the whole expression will be equal to zero).

For example, I will show the algorithm, how to decompose a square polynomial into brackets.

Example one. Algorithm for Factoring a Square Trinomial

The path we have is the square trinomial `x^2+5x+4`.

It is reduced (coefficient of `x^2` equal to one). He has roots. (To be sure, you can estimate the discriminant and make sure that it is greater than zero.)

Next steps (they need to be learned by doing all training tasks):

1. Make the following notation: $$x^2+5x+4=(x \ldots)(x \ldots).$$ Leave free space instead of dots, we will add appropriate numbers and signs there.
2. View all possible options, how you can decompose the number `4` into the product of two numbers. We get pairs of "candidates" for the roots of the equation: `2, 2` and `1, 4`.
3. Estimate from which pair you can get the average coefficient. Obviously it's `1, 4`.
4. Write $$x^2+5x+4=(x \quad 4)(x \quad 1)$$.
5. The next step is to place signs in front of the inserted numbers.

   How to understand and remember forever what signs should be in front of the numbers in brackets? Try to expand them (brackets). The coefficient before `x` to the first power will be `(± 4 ± 1)` (we don't know the signs yet - we need to choose), and it should equal `5`. Obviously, there will be two pluses here $$x^2+5x+4=(x + 4)(x + 1)$$.

   Perform this operation several times (hello, training tasks!) and there will never be more problems with this.

If you need to solve the equation `x^2+5x+4`, then now its solution is not difficult. Its roots are `-4, -1`.

Second example. Factorization of a square trinomial with coefficients of different signs

Let us need to solve the equation `x^2-x-2=0`. Offhand, the discriminant is positive.

We follow the algorithm.

1. $$x^2-x-2=(x \ldots) (x \ldots).$$
2. There is only one integer factorization of 2: `2 · 1`.
3. We skip the point - there is nothing to choose from.
4. $$x^2-x-2=(x \quad 2) (x \quad 1).$$
5. The product of our numbers is negative (`-2` is a free term), which means that one of them will be negative and the other positive.
   Since their sum is equal to `-1` (coefficient of `x`), then `2` will be negative (intuitive explanation - two is the larger of the two numbers, it will "pull" more in the negative direction). We get $$x^2-x-2=(x - 2) (x + 1).$$

Third example. Factorization of a square trinomial

Equation `x^2+5x -84 = 0`.

1. $$x+ 5x-84=(x \ldots) (x \ldots).$$
2. Decomposition of 84 into integer factors: `4 21, 6 14, 12 7, 2 42`.
3. Since we need the difference (or sum) of the numbers to be 5, the pair `7, 12` will do.
4. $$x+ 5x-84=(x\quad 12) (x \quad 7).$$
5. $$x+ 5x-84=(x + 12) (x - 7).$$

Hope, decomposition of this square trinomial into brackets understandably.

If you need a solution to the equation, then here it is: `12, -7`.

Tasks for training

Here are a few examples that are easy to are solved using Vieta's theorem.(Examples taken from Mathematics, 2002.)

1. `x^2+x-2=0`
2. `x^2-x-2=0`
3. `x^2+x-6=0`
4. `x^2-x-6=0`
5. `x^2+x-12=0`
6. `x^2-x-12=0`
7. `x^2+x-20=0`
8. `x^2-x-20=0`
9. `x^2+x-42=0`
10. `x^2-x-42=0`
11. `x^2+x-56=0`
12. `x^2-x-56=0`
13. `x^2+x-72=0`
14. `x^2-x-72=0`
15. `x^2+x-110=0`
16. `x^2-x-110=0`
17. `x^2+x-420=0`
18. `x^2-x-420=0`

A couple of years after the article was written, a collection of 150 tasks appeared for expanding a quadratic polynomial using the Vieta theorem.

Like and ask questions in the comments!