Let us now consider the squaring of a binomial and, applying an arithmetic point of view, we will speak of the square of the sum, i.e. (a + b)², and the square of the difference of two numbers, i.e. (a – b)².
Since (a + b)² = (a + b) ∙ (a + b),
then we find: (a + b) ∙ (a + b) = a² + ab + ab + b² = a² + 2ab + b², i.e.
(a + b)² = a² + 2ab + b²
It is useful to remember this result both in the form of the above-described equality and in words: the square of the sum of two numbers is equal to the square of the first number plus the product of two by the first number and the second number, plus the square of the second number.
Knowing this result, we can immediately write, for example:
(x + y)² = x² + 2xy + y²
(3ab + 1)² = 9a² b² + 6ab + 1
(x n + 4x)² = x 2n + 8x n+1 + 16x 2
Let's look at the second of these examples. We need to square the sum of two numbers: the first number is 3ab, the second 1. The result should be: 1) the square of the first number, i.e. (3ab)², which is equal to 9a²b²; 2) the product of two by the first number and the second, i.e. 2 ∙ 3ab ∙ 1 = 6ab; 3) the square of the 2nd number, i.e. 1² = 1 - all these three terms must be added together.
We also obtain a formula for squaring the difference of two numbers, i.e. for (a – b)²:
(a – b)² = (a – b) (a – b) = a² – ab – ab + b² = a² – 2ab + b².
(a – b)² = a² – 2ab + b²,
i.e. the square of the difference of two numbers is equal to the square of the first number, minus the product of two by the first number and the second, plus the square of the second number.
Knowing this result, we can immediately perform the squaring of binomials, which, from an arithmetic point of view, represent the difference of two numbers.
(m – n)² = m² – 2mn + n²
(5ab 3 – 3a 2 b) 2 = 25a 2 b 6 – 30a 3 b 4 + 9a 4 b 2 
(a n-1 – a) 2 = a 2n-2 – 2a n + a 2, etc.
Let's explain the 2nd example. Here we have in brackets the difference of two numbers: the first number is 5ab 3 and the second number is 3a 2 b. The result should be: 1) the square of the first number, i.e. (5ab 3) 2 = 25a 2 b 6, 2) the product of two by the 1st and the 2nd number, i.e. 2 ∙ 5ab 3 ∙ 3a 2 b = 30a 3 b 4 and 3) the square of the second number, i.e. (3a 2 b) 2 = 9a 4 b 2 ; The first and third terms must be taken with a plus, and the 2nd with a minus, we get 25a 2 b 6 – 30a 3 b 4 + 9a 4 b 2. To explain the 4th example, we only note that 1) (a n-1)2 = a 2n-2 ... the exponent must be multiplied by 2 and 2) the product of two by the 1st number and by the 2nd = 2 ∙ a n-1 ∙ a = 2a n .
If we take the point of view of algebra, then both equalities: 1) (a + b)² = a² + 2ab + b² and 2) (a – b)² = a² – 2ab + b² express the same thing, namely: the square of the binomial is equal to the square of the first term, plus the product of the number (+2) by the first term and the second, plus the square of the second term. This is clear because our equalities can be rewritten as:
1) (a + b)² = (+a)² + (+2) ∙ (+a) (+b) + (+b)²
2) (a – b)² = (+a)² + (+2) ∙ (+a) (–b) + (–b)²
In some cases, it is convenient to interpret the resulting equalities in this way:
(–4a – 3b)² = (–4a)² + (+2) (–4a) (–3b) + (–3b)²
Here we square a binomial whose first term = –4a and second = –3b. Next we get (–4a)² = 16a², (+2) (–4a) (–3b) = +24ab, (–3b)² = 9b² and finally:
(–4a – 3b)² = 6a² + 24ab + 9b²
It would also be possible to obtain and remember the formula for squaring a trinomial, a quadrinomial, or any polynomial in general. However, we will not do this, because we rarely need to use these formulas, and if we need to square any polynomial (except a binomial), we will reduce the matter to multiplication. For example:
31. Let us apply the obtained 3 equalities, namely:
(a + b) (a – b) = a² – b²
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²
to arithmetic.
Let it be 41 ∙ 39. Then we can represent this in the form (40 + 1) (40 – 1) and reduce the matter to the first equality - we get 40² – 1 or 1600 – 1 = 1599. Thanks to this, it is easy to perform multiplications like 21 ∙ 19; 22 ∙ 18; 31 ∙ 29; 32 ∙ 28; 71 ∙ 69, etc.
Let it be 41 ∙ 41; it’s the same as 41² or (40 + 1)² = 1600 + 80 + 1 = 1681. Also 35 ∙ 35 = 35² = (30 + 5)² = 900 + 300 + 25 = 1225. If you need 37 ∙ 37, then this is equal to (40 – 3)² = 1600 – 240 + 9 = 1369. Such multiplications (or squaring two-digit numbers) are easy to perform, with some skill, in your head.
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.
In the previous lesson we dealt with factorization. We mastered two methods: putting the common factor out of brackets and grouping. In this lesson - the following powerful method: abbreviated multiplication formulas. In short - FSU.
Abbreviated multiplication formulas (sum and difference square, sum and difference cube, difference of squares, sum and difference of cubes) are extremely necessary in all branches of mathematics. They are used in simplifying expressions, solving equations, multiplying polynomials, reducing fractions, solving integrals, etc. and so on. In short, there is every reason to deal with them. Understand where they come from, why they are needed, how to remember them and how to use them.
Do we understand?)
Where do abbreviated multiplication formulas come from?
Equalities 6 and 7 are not written in a very familiar way. It's kind of the opposite. This is on purpose.) Any equality works both from left to right and from right to left. This entry makes it clearer where the FSUs come from.
They are taken from multiplication.) For example:
(a+b) 2 =(a+b)(a+b)=a 2 +ab+ba+b 2 =a 2 +2ab+b 2
That's it, no scientific tricks. We simply multiply the brackets and give similar ones. This is how it turns out all abbreviated multiplication formulas. Abbreviated multiplication is because in the formulas themselves there is no multiplication of brackets and reduction of similar ones. Abbreviated.) The result is immediately given.
FSU needs to be known by heart. Without the first three, you can’t dream of a C; without the rest, you can’t dream of a B or A.)
Why do we need abbreviated multiplication formulas?
There are two reasons to learn, even memorize, these formulas. The first is that a ready-made answer automatically reduces the number of errors. But this is not the main reason. But the second one...
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You can get acquainted with functions and derivatives.
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. |
|
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\(-(x+3(2x-1\)\(+x-5\) \())\) |
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. |
|
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\(=-(x\)\(+3(3x-6)\) \()=\) |
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. |
|
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\(=-(x\)\(+9x-18\) \()=\) |
||
|
And open the last bracket. There is a minus sign in front of the bracket, so all signs are reversed. |
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Expanding parentheses is a basic skill in mathematics. Without this skill, it is impossible to have a grade above a C in 8th and 9th grade. Therefore, I recommend that you understand this topic well.
Abbreviated expression formulas are very often used in practice, so it is advisable to learn them all by heart. Until this moment, it will serve us faithfully, which we recommend printing out and keeping before your eyes at all times:
The first four formulas from the compiled table of abbreviated multiplication formulas allow you to square and cube the sum or difference of two expressions. The fifth is intended for briefly multiplying the difference and the sum of two expressions. And the sixth and seventh formulas are used to multiply the sum of two expressions a and b by their incomplete square of the difference (this is what an expression of the form a 2 −a b+b 2 is called) and the difference of two expressions a and b by the incomplete square of their sum (a 2 + a·b+b 2 ) respectively.
It is worth noting separately that each equality in the table is an identity. This explains why abbreviated multiplication formulas are also called abbreviated multiplication identities.
When solving examples, especially in which the polynomial is factorized, the FSU is often used in the form with the left and right sides swapped:
The last three identities in the table have their own names. The formula a 2 −b 2 =(a−b)·(a+b) is called difference of squares formula, a 3 +b 3 =(a+b)·(a 2 −a·b+b 2) - sum of cubes formula, A a 3 −b 3 =(a−b)·(a 2 +a·b+b 2) - difference of cubes formula. Please note that we did not name the corresponding formulas with rearranged parts from the previous table.
Additional formulas
It wouldn’t hurt to add a few more identities to the table of abbreviated multiplication formulas.
Areas of application of abbreviated multiplication formulas (FSU) and examples
The main purpose of abbreviated multiplication formulas (fsu) is explained by their name, that is, it consists in briefly multiplying expressions. However, the scope of application of FSU is much wider, and is not limited to short multiplication. Let's list the main directions.
Undoubtedly, the central application of the abbreviated multiplication formula was found in performing identical transformations of expressions. Most often these formulas are used in the process simplifying expressions.
Example.
Simplify the expression 9·y−(1+3·y) 2 .
Solution.
In this expression, squaring can be performed abbreviated, we have 9 y−(1+3 y) 2 =9 y−(1 2 +2 1 3 y+(3 y) 2). All that remains is to open the brackets and bring similar terms: 9 y−(1 2 +2 1 3 y+(3 y) 2)= 9·y−1−6·y−9·y 2 =3·y−1−9·y 2.
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