Abbreviated multiplication formulas (FSU) are needed in order to multiply and raise to a power of numbers, expressions, including polynomials. That is, with the help of formulas, you can work with numbers much faster and easier. Thus, it is possible to make an ordinary equation out of a complex equation, which will simplify the task.

Table with abbreviated multiplication formulas

Pay attention to the first four formulas. Thanks to them, you can square or cube the sum (difference) of two expressions. As for the fifth formula, it must be used to briefly multiply the difference or sum of two expressions.

The last two formulas (6 and 7) are used to multiply the sums of both expressions by their incomplete squared difference or sum.

The above formulas are quite often needed in practice. That is why it is desirable to know them by heart.

If you come across an example of factoring a polynomial, then in many cases you need to swap the left and right sides.

For example, take the same first formula:

and put the left side to the right, and the right side to the left:

The same procedure can be done with the rest of the formulas.

FSU Proof

Let us dwell on the proofs of the abbreviated multiplication formulas. This is not difficult. You just need to open the brackets. Consider the first formula - the square of the sum:.

Step one.

Raise a + b to the second power. To do this, we will not touch the degree, but perform a banal multiplication: = x.

Step two. Now we take it out of brackets: x + x.

Step Three. Expand the brackets: x + x + x + x .

Step Four. We multiply, not forgetting about the signs: x + x +.

Step five. We simplify the expression: .

In the same way, absolutely any abbreviated multiplication formula can be proved.

Examples and solutions using FSO

As a rule, these seven formulas are used when you need to simplify the expression in order to solve any equation and even a common example.

Example 1

Exercise

Simplify the expression:

As you can see, the first abbreviated multiplication formula, Sum Square, fits this example.

Solution

Based on the first formula, it is necessary to decompose the example into factors. To do this, we look at the formula and substitute numbers instead of letters. In our case, "a" is 3x and "b" is 5:

We consider the right side and write the result. We get:

In the example, you need to multiply everything that is multiplied and immediately get the answer:

Of course, there are examples with fractions. But, if you learn how to solve simple examples, then you will not be afraid of other types.

Example 2

Exercise

Simplify the expression

Solution

= – x x + =

The double product of these expressions is , which coincides with the second member of the trinomial (with the plus sign), which means

So, as you can see, there is nothing complicated in the examples. The main thing is to know the formulas, where they can be applied, and where you can do without them.

Useful sources

1. Arefieva I. G., Piryutko O. N. Algebra: textbook manual for the 7th grade of institutions of general secondary education: Minsk “Narodnaya Asveta”, 2017 - 304 p.
2. Nikolsky S. M., Potapov M. K. Algebra Grade 7: M: 2015 - 287 p.
3. Rubin A. G., Chulkov P. V. Algebra. 7th grade. M: 2015 - 224 p.

FSU - formulas for abbreviated multiplication in algebra for grade 7 with examples updated: November 22, 2019 by: Scientific Articles.Ru

Abbreviated multiplication formulas (FSU) are used to exponentiate and multiply numbers and expressions. Often these formulas allow you to make calculations more compactly and quickly.

In this article, we will list the main formulas for abbreviated multiplication, group them into a table, consider examples of using these formulas, and also dwell on the principles for proving abbreviated multiplication formulas.

For the first time, the topic of FSU is considered within the course "Algebra" for the 7th grade. Below are 7 basic formulas.

Abbreviated multiplication formulas

1. sum square formula: a + b 2 = a 2 + 2 a b + b 2
2. difference square formula: a - b 2 \u003d a 2 - 2 a b + b 2
3. sum cube formula: a + b 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
4. difference cube formula: a - b 3 \u003d a 3 - 3 a 2 b + 3 a b 2 - b 3
5. difference of squares formula: a 2 - b 2 \u003d a - b a + b
6. formula for the sum of cubes: a 3 + b 3 \u003d a + b a 2 - a b + b 2
7. cube difference formula: a 3 - b 3 \u003d a - b a 2 + a b + b 2

The letters a, b, c in these expressions can be any numbers, variables or expressions. For ease of use, it is better to learn the seven basic formulas by heart. We summarize them in a table and give them below, circling them with a box.

The first four formulas allow you to calculate, respectively, the square or cube of the sum or difference of two expressions.

The fifth formula calculates the difference of squares of expressions by multiplying their sum and difference.

The sixth and seventh formulas are, respectively, the multiplication of the sum and difference of expressions by the incomplete square of the difference and the incomplete square of the sum.

The abbreviated multiplication formula is sometimes also called the abbreviated multiplication identities. This is not surprising, since every equality is an identity.

When solving practical examples, abbreviated multiplication formulas are often used with rearranged left and right parts. This is especially convenient when factoring a polynomial.

Additional abbreviated multiplication formulas

We will not limit ourselves to the 7th grade course in algebra and add a few more formulas to our FSU table.

First, consider Newton's binomial formula.

a + b n = C n 0 a n + C n 1 a n - 1 b + C n 2 a n - 2 b 2 + . . + C n n - 1 a b n - 1 + C n n b n

Here C n k are the binomial coefficients that are in line number n in pascal's triangle. Binomial coefficients are calculated by the formula:

C nk = n ! k! · (n - k) ! = n (n - 1) (n - 2) . . (n - (k - 1)) k !

As you can see, the FSU for the square and cube of the difference and the sum is a special case of Newton's binomial formula for n=2 and n=3, respectively.

But what if there are more than two terms in the sum to be raised to a power? The formula for the square of the sum of three, four or more terms will be useful.

a 1 + a 2 + . . + a n 2 = a 1 2 + a 2 2 + . . + a n 2 + 2 a 1 a 2 + 2 a 1 a 3 + . . + 2 a 1 a n + 2 a 2 a 3 + 2 a 2 a 4 + . . + 2 a 2 a n + 2 a n - 1 a n

Another formula that may come in handy is the formula for the difference of the nth powers of two terms.

a n - b n = a - b a n - 1 + a n - 2 b + a n - 3 b 2 + . . + a 2 b n - 2 + b n - 1

This formula is usually divided into two formulas - respectively for even and odd degrees.

For even exponents 2m:

a 2 m - b 2 m = a 2 - b 2 a 2 m - 2 + a 2 m - 4 b 2 + a 2 m - 6 b 4 + . . + b 2 m - 2

For odd exponents 2m+1:

a 2 m + 1 - b 2 m + 1 = a 2 - b 2 a 2 m + a 2 m - 1 b + a 2 m - 2 b 2 + . . + b 2 m

The formulas for the difference of squares and the difference of cubes, you guessed it, are special cases of this formula for n = 2 and n = 3, respectively. For the difference of cubes, b is also replaced by - b .

How to read abbreviated multiplication formulas?

We will give the corresponding formulations for each formula, but first we will deal with the principle of reading formulas. The easiest way to do this is with an example. Let's take the very first formula for the square of the sum of two numbers.

a + b 2 = a 2 + 2 a b + b 2 .

They say: the square of the sum of two expressions a and b is equal to the sum of the square of the first expression, twice the product of the expressions and the square of the second expression.

All other formulas are read similarly. For the squared difference a - b 2 \u003d a 2 - 2 a b + b 2 we write:

the square of the difference of two expressions a and b is equal to the sum of the squares of these expressions minus twice the product of the first and second expressions.

Let's read the formula a + b 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3. The cube of the sum of two expressions a and b is equal to the sum of the cubes of these expressions, three times the product of the square of the first expression and the second, and three times the product of the square of the second expression and the first expression.

We proceed to reading the formula for the difference of cubes a - b 3 \u003d a 3 - 3 a 2 b + 3 a b 2 - b 3. The cube of the difference of two expressions a and b is equal to the cube of the first expression minus three times the square of the first expression and the second, plus three times the square of the second expression and the first expression, minus the cube of the second expression.

The fifth formula a 2 - b 2 \u003d a - b a + b (difference of squares) reads like this: the difference of the squares of two expressions is equal to the product of the difference and the sum of the two expressions.

Expressions like a 2 + a b + b 2 and a 2 - a b + b 2 for convenience are called, respectively, the incomplete square of the sum and the incomplete square of the difference.

With this in mind, the formulas for the sum and difference of cubes are read as follows:

The sum of the cubes of two expressions is equal to the product of the sum of these expressions and the incomplete square of their difference.

The difference of the cubes of two expressions is equal to the product of the difference of these expressions by the incomplete square of their sum.

FSU Proof

Proving FSU is quite simple. Based on the properties of multiplication, we will carry out the multiplication of the parts of the formulas in brackets.

For example, consider the formula for the square of the difference.

a - b 2 \u003d a 2 - 2 a b + b 2.

To raise an expression to the second power, the expression must be multiplied by itself.

a - b 2 \u003d a - b a - b.

Let's expand the brackets:

a - b a - b \u003d a 2 - a b - b a + b 2 \u003d a 2 - 2 a b + b 2.

The formula has been proven. The other FSOs are proved similarly.

Examples of application of FSO

The purpose of using reduced multiplication formulas is to quickly and concisely multiply and exponentiate expressions. However, this is not the entire scope of the FSO. They are widely used in reducing expressions, reducing fractions, factoring polynomials. Let's give examples.

Example 1. FSO

Let's simplify the expression 9 y - (1 + 3 y) 2 .

Apply the sum of squares formula and get:

9 y - (1 + 3 y) 2 = 9 y - (1 + 6 y + 9 y 2) = 9 y - 1 - 6 y - 9 y 2 = 3 y - 1 - 9 y 2

Example 2. FSO

Reduce the fraction 8 x 3 - z 6 4 x 2 - z 4 .

We notice that the expression in the numerator is the difference of cubes, and in the denominator - the difference of squares.

8 x 3 - z 6 4 x 2 - z 4 \u003d 2 x - z (4 x 2 + 2 x z + z 4) 2 x - z 2 x + z.

We reduce and get:

8 x 3 - z 6 4 x 2 - z 4 = (4 x 2 + 2 x z + z 4) 2 x + z

FSUs also help to calculate the values ​​of expressions. The main thing is to be able to notice where to apply the formula. Let's show this with an example.

Let's square the number 79. Instead of cumbersome calculations, we write:

79 = 80 - 1 ; 79 2 = 80 - 1 2 = 6400 - 160 + 1 = 6241 .

It would seem that a complex calculation was carried out quickly with just the use of abbreviated multiplication formulas and a multiplication table.

Another important point is the selection of the square of the binomial. The expression 4 x 2 + 4 x - 3 can be converted to 2 x 2 + 2 2 x 1 + 1 2 - 4 = 2 x + 1 2 - 4 . Such transformations are widely used in integration.

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One of the first topics studied in an algebra course are the formulas for abbreviated multiplication. In grade 7, they are used in the simplest situations, where it is required to recognize one of the formulas in the expression and factorize the polynomial or, conversely, quickly square or cube the sum or difference. In the future, the FSU is used to quickly solve inequalities and equations, and even to calculate some numerical expressions without a calculator.

What does the list of formulas look like?

There are 7 basic formulas that allow you to quickly multiply polynomials in brackets.

Sometimes this list also includes a fourth-degree expansion, which follows from the identities presented and has the form:

a⁴ - b⁴ = (a - b)(a + b)(a² + b²).

All equalities have a pair (sum - difference), except for the difference of squares. There is no formula for the sum of squares.

The rest of the equalities are easy to remember.:

It should be remembered that FSOs work in any case and for any values. a and b: it can be both arbitrary numbers and integer expressions.

In a situation where you suddenly cannot remember which sign is in the formula in front of one or another term, you can open the brackets and get the same result as after using the formula. For example, if a problem arose when applying the FSU of the difference cube, you need to write the original expression and do the multiplication one by one:

(a - b)³ = (a - b)(a - b)(a - b) = (a² - ab - ab + b²)(a - b) = a³ - a²b - a²b + ab² - a²b + ab² + ab² - b³ = a³ - 3a²b + 3ab² - b³.

As a result, after reducing all such terms, the same polynomial was obtained as in the table. The same manipulations can be carried out with all other FSOs.

Application of FSO to solve equations

For example, you need to solve an equation containing 3rd degree polynomial:

x³ + 3x² + 3x + 1 = 0.

The school curriculum does not consider universal techniques for solving cubic equations, and such tasks are most often solved by simpler methods (for example, factorization). If you notice that the left side of the identity resembles the cube of the sum, then the equation can be written in a simpler form:

(x + 1)³ = 0.

The root of such an equation is calculated orally: x=-1.

Inequalities are solved in a similar way. For example, we can solve the inequality x³ - 6x² + 9x > 0.

First of all, it is necessary to decompose the expression into factors. First you need to take out the brackets x. After that, you should pay attention that the expression in brackets can be converted to the square of the difference.

Then you need to find the points at which the expression takes zero values, and mark them on the number line. In a particular case, these will be 0 and 3. Then, using the interval method, determine in what intervals x will meet the inequality condition.

FSOs can be helpful in carrying out some calculations without the help of a calculator:

703² - 203² = (703 + 203)(703 - 203) = 906 ∙ 500 = 453000.

In addition, by factoring expressions, you can easily reduce fractions and simplify various algebraic expressions.

Examples of tasks for grades 7-8

In conclusion, we will analyze and solve two tasks for the application of abbreviated multiplication formulas in algebra.

Task 1. Simplify the expression:

(m + 3)² + (3m + 1)(3m - 1) - 2m (5m + 3).

Solution. In the condition of the assignment, it is required to simplify the expression, i.e. open the brackets, perform the operations of multiplication and exponentiation, and also bring all such terms. We conditionally divide the expression into three parts (according to the number of terms) and open the brackets one by one, using the FSU where possible.

• (m + 3)² = m² + 6m + 9(squared sum);
• (3m + 1)(3m - 1) = 9m² - 1(difference of squares);
• In the last term, you need to perform multiplication: 2m (5m + 3) = 10m² + 6m.

Substitute the results in the original expression:

(m² + 6m + 9) + (9m² - 1) - (10m² + 6m).

Taking into account the signs, we open the brackets and give like terms:

m² + 6m + 9 + 9m² 1 - 10m² - 6m = 8.

Task 2. Solve the equation containing the unknown k to the power of 5:

k⁵ + 4k⁴ + 4k³ - 4k² - 4k = k³.

Solution. In this case, it is necessary to use the FSO and the grouping method. We need to transfer the last and penultimate terms to the right side of the identity.

k⁵ + 4k⁴ + 4k³ = k³ + 4k² + 4k.

The common multiplier is taken from the right and left parts (k² + 4k +4):

k³(k² + 4k + 4) = k(k² + 4k + 4).

Everything is transferred to the left side of the equation so that 0 remains on the right side:

k³(k² + 4k + 4) - k(k² + 4k + 4) = 0.

Again, you need to take out the common factor:

(k³ - k)(k² + 4k + 4) = 0.

From the first factor obtained, we can derive k. According to the short multiplication formula, the second factor will be identically equal to (k + 2)²:

k (k² - 1)(k + 2)² = 0.

Using the difference of squares formula:

k (k - 1)(k + 1)(k + 2)² = 0.

Since the product is 0 if at least one of its factors is zero, it will not be difficult to find all the roots of the equation:

1. k = 0;
2. k - 1 = 0; k = 1;
3. k + 1 = 0; k = -1;
4. (k + 2)² = 0; k = -2.

Based on illustrative examples, one can understand how to remember the formulas, their differences, and also solve several practical problems using FSU. The tasks are simple and should not be difficult to complete.

When, etc. Below we will look at the most popular formulas and analyze how they are obtained.

sum square

Let us square the sum of two monomials, like this: \((a+b)^2\). Squaring is the multiplication of a number or expression by itself, that is, \((a+b)^2=(a+b)(a+b)\). Now we can simply open the brackets, multiply them as we did, and bring like terms. We get:

And if we omit intermediate calculations and write only the initial and final expressions, we get the final formula:

The square of the sum:\((a+b)^2=a^2+2ab+b^2\)

Most students learn it by heart. And now you know how to derive this formula, and if you suddenly forget, you can always do it.
Okay, but how to use it and why is this formula needed? The square of the sum allows you to quickly write the result of squaring the sum of two terms. Let's look at an example.

Example . Open brackets: \((x+5)^2\)
Solution :

Notice how much faster and with less effort the result is obtained in the second case. And when you master this and other formulas to automatism, it will be even faster: you can simply write the answer right away. Therefore, they are called abbreviated multiplication formulas. So, knowing them and learning how to apply them is definitely worth it.

Just in case, we note that as \(a\) and \(b\) there can be any expressions - the principle remains the same. For example:

If you suddenly did not understand some transformations in the last two examples, repeat the topic.

Example . Convert the expression \((1+5x)^2-12x-1 \) to standard form.

Solution :

Answer: \(25x^2-2x\).

Important! It is necessary to learn how to use formulas not only in the "forward" but also in the "reverse" direction.

Example . Calculate the value of the expression \((368)^2+2 368 132+(132)^2\) without a calculator.

Solution :

Answer: \(250 000\).

The square of the difference

Above we found a formula for the sum of monomials. Let's now find the formula for the difference, that is, for \((a-b)^2\):

In a shorter notation, we have:

The square of the difference: \((a-b)^2=a^2-2ab+b^2\)

It is applied in the same way as the previous one.

Example . Simplify the expression \((2a-3)^2-4(a^2-a)\) and find its value when \(a=\frac(17)(8)\).

Solution :

Answer: \(8\).

Difference of squares

So, we have dealt with the situations of the product of two brackets with a plus in them and two brackets with a minus. There remains the case of the product of identical brackets with different signs. Let's see what happens:

We got the formula:

Difference of squares \(a^2-b^2=(a+b)(a-b)\)

This formula is one of the most commonly used when and working with.

Example . Reduce the fraction \(\frac(x^2-9)(x-3)\) .

Solution :

Answer: \(x+3\).

Example .Factorize \(25x^4-m^(10) t^6\).
Solution :

These are the three basic formulas you need to know necessarily! There are also formulas with cubes (see above), it is also desirable to remember them or be able to quickly derive them. We also note that in practice several such formulas are often encountered at once in one problem - this is normal. Just learn to notice the formulas and apply them carefully and you'll be fine.

An example (increased complexity!) .Reduce the fraction.
Solution :

Algebra

Short multiplication formulas are used to transform expressions. Identities are used to represent the whole expression as a polynomial and factorize polynomials.

• 1 sum square(a + b) 2 = a 2 + 2ab + b 2
• 2 The square of the difference(a - b) 2 = a 2 - 2ab + b 2
• 3 Difference of squares a 2 - b 2 = (a - b) (a + b)
• 4 sum cube(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 2
• 5 difference cube(a - b) 3 = a 3 - 3a 2 b + 3ab 2 - b 2
• 6 Sum of cubes a 3 + b 3 = (a + b)(a 2 - ab + b 2)
• 7 Difference of cubes a 3 - b 3 \u003d (a - b) (a 2 + ab + b 2)

Formulas for squares

\((a + b)^2 = a^2 + 2ab + b^2\)

\((a - b)^2 = a^2 - 2ab + b^2\)

\(a^2 - b^2 = (a + b)(a - b)\)

Cube formulas

\((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)

\((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\)

\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

Formulas for the fourth degree

\((a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\)

\((a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4\)

\(a^4 - b^4 = (a - b)(a + b)(a^2 + b^2)\);
follows from \(a^2 - b^2 = (a + b)(a - b)\).

Abbreviated multiplication formulas

1. Square the sum

2. Square difference

3. Sum and difference of squares

4. Sum to the third power (cube of the sum)

5. Difference to the third degree (difference cube)

6. Sum and difference of cubes

7. Formulas for abbreviated multiplication for the fourth degree

8. Formulas for abbreviated multiplication for the fifth degree

9. Abbreviated multiplication formulas for the sixth degree

10. Abbreviated multiplication formulas for degree n, where n- any natural number

11. Abbreviated multiplication formulas for degree n, where n- even positive number

12. Abbreviated multiplication formulas for degree n, where n- odd positive number