Expression open brackets do. How a math tutor teaches the topic “multiplication of polynomials”

In this lesson you will learn how to transform an expression containing parentheses into an expression without parentheses. You will learn how to open parentheses preceded by a plus sign and a minus sign. We will remember how to open brackets using the distributive law of multiplication. The considered examples will allow you to connect new and previously studied material into a single whole.

Topic: Solving equations

Lesson: Expanding Parentheses

How to expand parentheses preceded by a “+” sign. Using the associative law of addition.

If you need to add the sum of two numbers to a number, you can first add the first term to this number, and then the second.

To the left of the equal sign is an expression with parentheses, and to the right is an expression without parentheses. This means that when moving from the left side of the equality to the right, the opening of the parentheses occurred.

Let's look at examples.

Example 1.

By opening the brackets, we changed the order of actions. It has become more convenient to count.

Example 2.

Example 3.

Note that in all three examples we simply removed the parentheses. Let's formulate a rule:

Comment.

If the first term in brackets is unsigned, then it must be written with a plus sign.

You can follow the example step by step. First, add 445 to 889. This action can be performed mentally, but it is not very easy. Let's open the brackets and see that the changed procedure will significantly simplify the calculations.

If you follow the indicated procedure, you must first subtract 345 from 512, and then add 1345 to the result. By opening the brackets, we will change the procedure and significantly simplify the calculations.

Illustrating example and rule.

Let's look at an example: . You can find the value of an expression by adding 2 and 5, and then taking the resulting number with the opposite sign. We get -7.

On the other hand, the same result can be obtained by adding the opposite numbers of the original ones.

Let's formulate a rule:

Example 1.

Example 2.

The rule does not change if there are not two, but three or more terms in brackets.

Example 3.

Comment. The signs are reversed only in front of the terms.

In order to open the brackets, in this case we need to remember the distributive property.

First, multiply the first bracket by 2, and the second by 3.

The first bracket is preceded by a “+” sign, which means that the signs must be left unchanged. The second sign is preceded by a “-” sign, therefore, all signs need to be changed to the opposite

Bibliography

1. Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemosyne, 2012.
2. Merzlyak A.G., Polonsky V.V., Yakir M.S. Mathematics 6th grade. - Gymnasium, 2006.
3. Depman I.Ya., Vilenkin N.Ya. Behind the pages of a mathematics textbook. - Enlightenment, 1989.
4. Rurukin A.N., Tchaikovsky I.V. Assignments for the mathematics course grades 5-6 - ZSh MEPhI, 2011.
5. Rurukin A.N., Sochilov S.V., Tchaikovsky K.G. Mathematics 5-6. A manual for 6th grade students at the MEPhI correspondence school. - ZSh MEPhI, 2011.
6. Shevrin L.N., Gein A.G., Koryakov I.O., Volkov M.V. Mathematics: Textbook-interlocutor for 5-6 grades of secondary school. Math teacher's library. - Enlightenment, 1989.

1. Online tests in mathematics ().
2. You can download those specified in clause 1.2. books().

Homework

1. Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemosyne, 2012. (link see 1.2)
2. Homework: No. 1254, No. 1255, No. 1256 (b, d)
3. Other tasks: No. 1258(c), No. 1248

In this article we will take a detailed look at the basic rules of such an important topic in a mathematics course as opening parentheses. You need to know the rules for opening parentheses in order to correctly solve equations in which they are used.

How to open parentheses correctly when adding

Expand the brackets preceded by the “+” sign

This is the simplest case, because if there is an addition sign in front of the brackets, the signs inside them do not change when the brackets are opened. Example:

(9 + 3) + (1 - 6 + 9) = 9 + 3 + 1 - 6 + 9 = 16.

How to expand parentheses preceded by a "-" sign

In this case, you need to rewrite all terms without brackets, but at the same time change all the signs inside them to the opposite ones. The signs change only for terms from those brackets that were preceded by the sign “-”. Example:

(9 + 3) - (1 - 6 + 9) = 9 + 3 - 1 + 6 - 9 = 8.

How to open parentheses when multiplying

Before the brackets there is a multiplier number

In this case, you need to multiply each term by a factor and open the brackets without changing the signs. If the multiplier has a “-” sign, then during multiplication the signs of the terms are reversed. Example:

3 * (1 - 6 + 9) = 3 * 1 - 3 * 6 + 3 * 9 = 3 - 18 + 27 = 12.

How to open two parentheses with a multiplication sign between them

In this case, you need to multiply each term from the first brackets with each term from the second brackets and then add the results. Example:

(9 + 3) * (1 - 6 + 9) = 9 * 1 + 9 * (- 6) + 9 * 9 + 3 * 1 + 3 * (- 6) + 3 * 9 = 9 - 54 + 81 + 3 - 18 + 27 = 48.

How to open parentheses in a square

If the sum or difference of two terms is squared, the brackets should be opened according to the following formula:

(x + y)^2 = x^2 + 2 * x * y + y^2.

In the case of a minus inside the brackets, the formula does not change. Example:

(9 + 3) ^ 2 = 9 ^ 2 + 2 * 9 * 3 + 3 ^ 2 = 144.

How to expand parentheses to another degree

If the sum or difference of terms is raised, for example, to the 3rd or 4th power, then you just need to break the power of the bracket into “squares”. The powers of identical factors are added, and when dividing, the power of the divisor is subtracted from the power of the dividend. Example:

(9 + 3) ^ 3 = ((9 + 3) ^ 2) * (9 + 3) = (9 ^ 2 + 2 * 9 * 3 + 3 ^ 2) * 12 = 1728.

How to open 3 brackets

There are equations in which 3 brackets are multiplied at once. In this case, you must first multiply the terms of the first two brackets together, and then multiply the sum of this multiplication by the terms of the third bracket. Example:

(1 + 2) * (3 + 4) * (5 - 6) = (3 + 4 + 6 + 8) * (5 - 6) = - 21.

These rules for opening parentheses apply equally to solving both linear and trigonometric equations.

The main function of parentheses is to change the order of actions when calculating values. For example, in the numerical expression \(5·3+7\) the multiplication will be calculated first, and then the addition: \(5·3+7 =15+7=22\). But in the expression \(5·(3+7)\) the addition in brackets will be calculated first, and only then the multiplication: \(5·(3+7)=5·10=50\).

Example. Expand the bracket: \(-(4m+3)\).
Solution : \(-(4m+3)=-4m-3\).

Example. Open the bracket and give similar terms \(5-(3x+2)+(2+3x)\).
Solution : \(5-(3x+2)+(2+3x)=5-3x-2+2+3x=5\).

Example. Expand the brackets \(5(3-x)\).
Solution : In the bracket we have \(3\) and \(-x\), and before the bracket there is a five. This means that each member of the bracket is multiplied by \(5\) - I remind you that The multiplication sign between a number and a parenthesis is not written in mathematics to reduce the size of entries.

Example. Expand the brackets \(-2(-3x+5)\).
Solution : As in the previous example, the \(-3x\) and \(5\) in the parenthesis are multiplied by \(-2\).

Example. Simplify the expression: \(5(x+y)-2(x-y)\).
Solution : \(5(x+y)-2(x-y)=5x+5y-2x+2y=3x+7y\).

It remains to consider the last situation.

When multiplying a bracket by a bracket, each term of the first bracket is multiplied with each term of the second:

\((c+d)(a-b)=c·(a-b)+d·(a-b)=ca-cb+da-db\)

Example. Expand the brackets \((2-x)(3x-1)\).
Solution : We have a product of brackets and it can be expanded immediately using the formula above. But in order not to get confused, let's do everything step by step.
Step 1. Remove the first bracket - multiply each of its terms by the second bracket:

Step 2. Expand the products of the brackets and the factor as described above:
- First things first...

Then the second.

Step 3. Now we multiply and present similar terms:

It is not necessary to describe all the transformations in such detail; you can multiply them right away. But if you are just learning how to open parentheses, write in detail, there will be less chance of making mistakes.

Note to the entire section. In fact, you don't need to remember all four rules, you only need to remember one, this one: \(c(a-b)=ca-cb\) . Why? Because if you substitute one instead of c, you get the rule \((a-b)=a-b\) . And if we substitute minus one, we get the rule \(-(a-b)=-a+b\) . Well, if you substitute another bracket instead of c, you can get the last rule.

Parenthesis within a parenthesis

Sometimes in practice there are problems with brackets nested inside other brackets. Here is an example of such a task: simplify the expression \(7x+2(5-(3x+y))\).

To successfully solve such tasks, you need:
- carefully understand the nesting of brackets - which one is in which;
- open the brackets sequentially, starting, for example, with the innermost one.

It is important when opening one of the brackets don't touch the rest of the expression, just rewriting it as is.
Let's look at the task written above as an example.

Example. Open the brackets and give similar terms \(7x+2(5-(3x+y))\).
Solution:

Example. Open the brackets and give similar terms \(-(x+3(2x-1+(x-5)))\).
Solution :

\(-(x+3(2x-1\)\(+(x-5)\) \())\)		There is triple nesting of parentheses here. Let's start with the innermost one (highlighted in green). There is a plus in front of the bracket, so it simply comes off.
\(-(x+3(2x-1\)\(+x-5\) \())\)		Now you need to open the second bracket, the intermediate one. But before that, we will simplify the expression of the ghost-like terms in this second bracket.
\(=-(x\)\(+3(3x-6)\) \()=\)		Now we open the second bracket (highlighted in blue). Before the bracket is a factor - so each term in the bracket is multiplied by it.
\(=-(x\)\(+9x-18\) \()=\)
		And open the last bracket. There is a minus sign in front of the bracket, so all signs are reversed.

Expanding parentheses is a basic skill in mathematics. Without this skill, it is impossible to have a grade above a C in grades 8 and 9. Therefore, I recommend that you understand this topic well.

I am continuing the series of methodological articles on the topic of teaching. It's time to consider the features of individual work math tutor for 7th grade students. It is with great pleasure that I will share my thoughts on the forms of presentation of one of the most important topics in the 7th grade algebra course - “opening parentheses.” In order not to try to grasp the immensity, let’s stop at its initial stage and analyze the tutor’s method of working with multiplying a polynomial by a polynomial. How math tutor acts in difficult situations when weak student does not accept the classical form of explanation? What tasks should be prepared for a strong seventh grader? Let's consider these and other questions.

It would seem, what’s so complicated about this? “Brackets are as easy as shelling pears,” any excellent student will say. “There is a distribution law and properties of powers for working with monomials, a general algorithm for any number of terms. Multiply each by each and bring similar ones.” However, not everything is so simple when working with laggards. Despite the efforts of the math tutor, students manage to make errors of all sizes even in the simplest transformations. The nature of the errors is striking in its diversity: from small omissions of letters and signs to serious dead-end “stop errors”.

What prevents a student from completing the transformations correctly? Why is misunderstanding possible?

There are a huge number of individual problems, and one of the main obstacles to the assimilation and consolidation of material is the difficulty in timely and quick switching of attention, the difficulty in processing a large amount of information. It may seem strange to some that I am talking about a large volume, but a weak 7th grade student may not have enough memory and attention resources even for four terms. Coefficients, variables, degrees (indicators) interfere. The student confuses the order of operations, forgets which monomials have already been multiplied and which remained untouched, cannot remember how they are multiplied, etc.

Numerical Approach for Math Tutor

Of course, you need to start with an explanation of the logic behind the construction of the algorithm itself. How to do it? We need to pose a problem: how to change the order of actions in an expression so that the result does not change? I quite often give examples that explain how certain rules work using specific numbers. And only then I replace them with letters. The technique for using the numerical approach will be described below.

Motivation problems.
At the beginning of a lesson, it is difficult for a math tutor to gather a student if he does not understand the relevance of what is being studied. Within the syllabus for grades 6–7, it is difficult to find examples of using the rule for multiplying polynomials. I would emphasize the need to learn change the order of actions in expressions The student should know that this helps solve problems from experience in adding similar terms. He had to add them together when solving equations. For example, in 2x+5x+13=34 he uses that 2x+5x=7x. A math tutor simply needs to focus the student’s attention on this.

Math teachers often refer to the technique of opening parentheses as "fountain" rule.

This image is well remembered and should definitely be used. But how is this rule proven? Let us recall the classical form, which uses obvious identity transformations:

(a+b)(c+d)=(a+b) c+(a+b) d=ac+bc+ad+bd

It is difficult for a math tutor to comment on anything here. The letters speak for themselves. And a strong 7th grade student doesn’t need detailed explanations. However, what to do with the weak, who point-blank does not see any content in this “literal jumble”?

The main problem that interferes with the perception of the classical mathematical justification of the “fountain” is the unusual form of writing the first factor. Neither in the 5th grade nor in the 6th grade did the student have to drag the first bracket to each term of the second. Children dealt only with numbers (coefficients), most often located to the left of the brackets, for example:

By the end of 6th grade, the student has formed a visual image of an object - a certain combination of signs (actions) associated with brackets. And any deviation from the usual view towards something new can disorient a seventh grader. It is the visual image of the “number + bracket” pair that the math tutor uses when explaining.

The following explanation can be offered. The tutor reasons: “If there was some number in front of the bracket, for example 5, then we could change the procedure in this expression? Certainly. Then let's do it . Think about whether his result will change if instead of the number 5 we enter the sum 2+3 enclosed in brackets? Any student will tell the tutor: “What difference does it make how you write: 5 or 2+3.” Wonderful. You will get a recording. The math tutor takes a short break so that the student visually remembers the picture-image of the object. Then he draws his attention to the fact that the bracket, like the number, “distributed” or “jumped” to each term. What does this mean? This means that this operation can be performed not only with a number, but also with a parenthesis. We got two pairs of factors and . Most students easily cope with them on their own and write the result to the tutor. It is important to compare the resulting pairs with the contents of the brackets 2+3 and 6+4 and it will become clear how they open.

If necessary, after the example with numbers, the math tutor conducts a letter proof. It turns out to be a cakewalk through the same parts of the previous algorithm.

Formation of the skill of opening brackets

Forming the skill of multiplying parentheses is one of the most important stages of a math tutor’s work with a topic. And even more important than the stage of explaining the logic of the “fountain” rule. Why? The rationale for the changes will be forgotten the very next day, but the skill, if it is formed and consolidated in time, will remain. Students perform the operation mechanically, as if retrieving a multiplication table from memory. This is what needs to be achieved. Why? If every time a student opens parentheses he remembers why it is opened this way and not otherwise, he will forget about the problem he is solving. That is why the math tutor devotes the remaining time of the lesson to transforming understanding into rote memorization. This strategy is often used in other topics.

How can a tutor develop the skill of opening parentheses in a student? To do this, a 7th grade student must complete a number of exercises in sufficient quantities to consolidate. This raises another problem. A weak seventh grader cannot cope with the increased number of transformations. Even small ones. And mistakes fall one after another. What should a math tutor do? Firstly, it is recommended to draw arrows from each term to each one. If a student is very weak and is not able to quickly switch from one type of work to another, or loses concentration when following simple commands from the teacher, then the math tutor himself draws these arrows. And not all at once. First, the tutor connects the first term in the left parenthesis with each term in the right parenthesis and asks them to perform the corresponding multiplication. Only after this the arrows are directed from the second term to the same right bracket. In other words, the tutor divides the process into two stages. It is better to maintain a short time pause (5-7 seconds) between the first and second operations.

1) One set of arrows should be drawn above the expressions, and the other below them.
2) It is important to skip between lines at least a couple of cells. Otherwise, the recording will be very dense, and the arrows will not only climb onto the previous line, but will also mix with the arrows from the next exercise.

3) In the case of multiplying brackets in the format 3 by 2, arrows are drawn from the short bracket to the long one. Otherwise, there will be not two, but three of these “fountains”. The implementation of the third is noticeably more complicated due to the lack of free space for the arrows.
4) arrows always point from the same point. One of my students kept trying to put them side by side and this is what he came up with:

This arrangement does not allow selecting and recording the current term with which the student works at each stage.

Tutor's finger work

4) To keep attention on a separate pair of multiplied terms, the math tutor puts two fingers on them. This must be done in such a way as not to block the student’s view. For the most inattentive students, you can use the “pulsation” method. The math tutor moves his first finger to the beginning of the arrow (to one of the terms) and fixes it, and with the second he “knocks” at its end (to the second term). Ripple helps to focus attention on the term by which the student is multiplying. After the first multiplication by the right parenthesis is completed, the math tutor says: “Now we work with the other term.” The tutor moves the “fixed finger” towards it, and runs the “pulsating” finger over the terms from the other bracket. The pulsation works like a “turn signal” in a car and allows you to focus the attention of an absent-minded student on the operation he is performing. If the child writes small, then two pencils are used instead of fingers.

Repetition optimization

As when studying any other topic in an algebra course, multiplying polynomials can and should be integrated with previously covered material. To do this, the math tutor uses special bridge tasks that allow you to find the application of what you are studying in various mathematical objects. They not only connect topics into a single whole, but also very effectively organize the repetition of the entire mathematics course. And the more bridges the tutor builds, the better.

Traditionally, 7th grade algebra textbooks integrate opening parentheses with solving linear equations. At the end of the list of numbers there are always tasks of the following order: solve the equation. When opening the brackets, the squares are reduced and the equation is easily solved using 7th grade tools. However, for some reason, the authors of textbooks conveniently forget about constructing a graph of a linear function. In order to correct this shortcoming, I would advise mathematics tutors to include parentheses in analytical expressions of linear functions, for example. In such exercises, the student not only trains the skills of carrying out identical transformations, but also repeats graphs. You can ask to find the point of intersection of two “monsters”, determine the relative position of the lines, find the points of their intersection with the axes, etc.

Kolpakov A.N. Mathematics tutor in Strogino. Moscow

In this video we will analyze a whole set of linear equations that are solved using the same algorithm - that’s why they are called the simplest.

First, let's define: what is a linear equation and which one is called the simplest?

A linear equation is one in which there is only one variable, and only to the first degree.

The simplest equation means the construction:

All other linear equations are reduced to the simplest using the algorithm:

1. Expand parentheses, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Give similar terms to the left and right of the equal sign;
4. Divide the resulting equation by the coefficient of the variable $x$.

Of course, this algorithm does not always help. The fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. In the video below we will look at several reasons why this situation is possible.
2. The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.

Now let's see how all this works using real-life examples.

Examples of solving equations

Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.

Such constructions are solved in approximately the same way:

1. First of all, you need to expand the parentheses, if there are any (as in our last example);
2. Then combine similar
3. Finally, isolate the variable, i.e. move everything connected with the variable—the terms in which it is contained—to one side, and move everything that remains without it to the other side.

Then, as a rule, you need to give similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “x”, and we will get the final answer.

In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Typically, errors are made either when opening brackets or when calculating the “pluses” and “minuses”.

In addition, it happens that a linear equation has no solutions at all, or that the solution is the entire number line, i.e. any number. We will look at these subtleties in today's lesson. But we will start, as you already understood, with the simplest tasks.

Scheme for solving simple linear equations

First, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the brackets, if any.
2. We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
3. We present similar terms.
4. We divide everything by the coefficient of “x”.

Of course, this scheme does not always work; there are certain subtleties and tricks in it, and now we will get to know them.

Solving real examples of simple linear equations

Task No. 1

The first step requires us to open the brackets. But they are not in this example, so we skip this step. In the second step we need to isolate the variables. Please note: we are talking only about individual terms. Let's write it down:

We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:

\[\frac(6x)(6)=-\frac(72)(6)\]

So we got the answer.

Task No. 2

We can see the parentheses in this problem, so let's expand them:

Both on the left and on the right we see approximately the same design, but let's act according to the algorithm, i.e. separating the variables:

Here are some similar ones:

At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.

Task No. 3

The third linear equation is more interesting:

\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]

There are several brackets here, but they are not multiplied by anything, they are simply preceded by different signs. Let's break them down:

We perform the second step already known to us:

\[-x+x+2x=15-6-12+3\]

Let's do the math:

We carry out the last step - divide everything by the coefficient of “x”:

\[\frac(2x)(x)=\frac(0)(2)\]

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, I would like to say the following:

• As I said above, not every linear equation has a solution - sometimes there are simply no roots;
• Even if there are roots, there may be zero among them - there is nothing wrong with that.

Zero is the same number as the others; you shouldn’t discriminate against it in any way or assume that if you get zero, then you did something wrong.

Another feature is related to the opening of brackets. Please note: when there is a “minus” in front of them, we remove it, but in parentheses we change the signs to opposite. And then we can open it using standard algorithms: we will get what we saw in the calculations above.

Understanding this simple fact will help you avoid making stupid and hurtful mistakes in high school, when doing such things is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complex and when performing various transformations a quadratic function will appear. However, we should not be afraid of this, because if, according to the author’s plan, we are solving a linear equation, then during the transformation process all monomials containing a quadratic function will certainly cancel.

Example No. 1

Obviously, the first step is to open the brackets. Let's do this very carefully:

Now let's take a look at privacy:

\[-x+6((x)^(2))-6((x)^(2))+x=-12\]

Here are some similar ones:

Obviously, this equation has no solutions, so we’ll write this in the answer:

\[\varnothing\]

or there are no roots.

Example No. 2

We perform the same actions. First step:

Let's move everything with a variable to the left, and without it - to the right:

Here are some similar ones:

Obviously, this linear equation has no solution, so we’ll write it this way:

\[\varnothing\],

or there are no roots.

Nuances of the solution

Both equations are completely solved. Using these two expressions as an example, we were once again convinced that even in the simplest linear equations, everything may not be so simple: there can be either one, or none, or infinitely many roots. In our case, we considered two equations, both simply have no roots.

But I would like to draw your attention to another fact: how to work with parentheses and how to open them if there is a minus sign in front of them. Consider this expression:

Before opening, you need to multiply everything by “X”. Please note: multiplies each individual term. Inside there are two terms - respectively, two terms and multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can you open the bracket from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the brackets, which means that everything below simply changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.

We do the same with the second equation:

It is not by chance that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and again learn to solve such simple equations.

Of course, the day will come when you will hone these skills to the point of automaticity. You will no longer have to perform so many transformations each time; you will write everything on one line. But while you are just learning, you need to write each action separately.

Solving even more complex linear equations

What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.

Task No. 1

\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]

Let's multiply all the elements in the first part:

Let's do some privacy:

Here are some similar ones:

Let's complete the last step:

\[\frac(-4x)(4)=\frac(4)(-4)\]

Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, they canceled each other out, which makes the equation linear and not quadratic.

Task No. 2

\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]

Let's carefully perform the first step: multiply each element from the first bracket by each element from the second. There should be a total of four new terms after the transformations:

Now let’s carefully perform the multiplication in each term:

Let’s move the terms with “X” to the left, and those without - to the right:

\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]

Here are similar terms:

Once again we have received the final answer.

Nuances of the solution

The most important note about these two equations is the following: as soon as we begin to multiply brackets that contain more than one term, this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we will have four terms.

About the algebraic sum

With this last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean a simple construction: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.

As soon as, when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.

Finally, let's look at a couple more examples that will be even more complex than the ones we just looked at, and to solve them we will have to slightly expand our standard algorithm.

Solving equations with fractions

To solve such tasks, we will have to add one more step to our algorithm. But first, let me remind you of our algorithm:

1. Open the brackets.
2. Separate variables.
3. Bring similar ones.
4. Divide by the ratio.

Alas, this wonderful algorithm, for all its effectiveness, turns out to be not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on both the left and the right in both equations.

How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be done both before and after the first action, namely, getting rid of fractions. So the algorithm will be as follows:

1. Get rid of fractions.
2. Open the brackets.
3. Separate variables.
4. Bring similar ones.
5. Divide by the ratio.

What does it mean to “get rid of fractions”? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numerical in their denominator, i.e. Everywhere the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, we will get rid of fractions.

Example No. 1

\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]

Let's get rid of the fractions in this equation:

\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]

Please note: everything is multiplied by “four” once, i.e. just because you have two parentheses doesn't mean you have to multiply each one by "four." Let's write down:

\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]

Now let's expand:

We seclude the variable:

We perform the reduction of similar terms:

\[-4x=-1\left| :\left(-4 \right) \right.\]

\[\frac(-4x)(-4)=\frac(-1)(-4)\]

We have received the final solution, let's move on to the second equation.

Example No. 2

\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]

Here we perform all the same actions:

\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]

\[\frac(4x)(4)=\frac(4)(4)\]

The problem is solved.

That, in fact, is all I wanted to tell you today.

Key points

Key findings are:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Don't worry if you have quadratic functions somewhere; most likely, they will be reduced in the process of further transformations.
• There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and no roots at all.

I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site and solve the examples presented there. Stay tuned, many more interesting things await you!