Abbreviated multiplication formulas.

Studying the formulas for abbreviated multiplication: the square of the sum and the square of the difference of two expressions; difference of squares of two expressions; the cube of the sum and the cube of the difference of two expressions; sums and differences of cubes of two expressions.

Application of abbreviated multiplication formulas when solving examples.

To simplify expressions, factorize polynomials, and reduce polynomials to a standard form, abbreviated multiplication formulas are used. Abbreviated multiplication formulas you need to know by heart.

Let a, b R. Then:

1. The square of the sum of two expressions is the square of the first expression plus twice the product of the first expression and the second plus the square of the second expression.

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

2. The square of the difference of two expressions is the square of the first expression minus twice the product of the first expression and the second plus the square of the second expression.

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

3. Difference of squares two expressions is equal to the product of the difference of these expressions and their sum.

a 2 - b 2 \u003d (a - b) (a + b)

4. sum cube of two expressions is equal to the cube of the first expression plus three times the square of the first expression times the second plus three times the product of the first expression times the square of the second plus the cube of the second expression.

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

5. difference cube of two expressions is equal to the cube of the first expression minus three times the product of the square of the first expression and the second plus three times the product of the first expression and the square of the second minus the cube of the second expression.

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

6. Sum of cubes two expressions is equal to the product of the sum of the first and second expressions by the incomplete square of the difference of these expressions.

a 3 + b 3 \u003d (a + b) (a 2 - ab + b 2)

7. Difference of cubes of two expressions is equal to the product of the difference of the first and second expressions by the incomplete square of the sum of these expressions.

a 3 - b 3 \u003d (a - b) (a 2 + ab + b 2)

Application of abbreviated multiplication formulas when solving examples.

Example 1

Calculate

a) Using the formula for the square of the sum of two expressions, we have

(40+1) 2 = 40 2 + 2 40 1 + 1 2 = 1600 + 80 + 1 = 1681

b) Using the formula for the squared difference of two expressions, we obtain

98 2 \u003d (100 - 2) 2 \u003d 100 2 - 2 100 2 + 2 2 \u003d 10000 - 400 + 4 \u003d 9604

Example 2

Calculate

Using the formula for the difference of the squares of two expressions, we obtain

Example 3

Simplify Expression

(x - y) 2 + (x + y) 2

We use the formulas for the square of the sum and the square of the difference of two expressions

(x - y) 2 + (x + y) 2 \u003d x 2 - 2xy + y 2 + x 2 + 2xy + y 2 \u003d 2x 2 + 2y 2

Abbreviated multiplication formulas in one table:

(a + b) 2 = a 2 + 2ab + b 2
(a - b) 2 = a 2 - 2ab + b 2
a 2 - b 2 = (a - b) (a+b)
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
(a - b) 3 = a 3 - 3a 2 b + 3ab 2 - b 3
a 3 + b 3 \u003d (a + b) (a 2 - ab + b 2)
a 3 - b 3 \u003d (a - b) (a 2 + ab + b 2)

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- 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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Difference of squares

We derive the formula for the difference of squares $a^2-b^2$.

To do this, remember the following rule:

If any monomial is added to the expression and the same monomial is subtracted, then we get the correct identity.

Let's add to our expression and subtract from it the monomial $ab$:

In total, we get:

That is, the difference of the squares of two monomials is equal to the product of their difference and their sum.

Example 1

Express as a product of $(4x)^2-y^2$

\[(4x)^2-y^2=((2x))^2-y^2\]

\[((2x))^2-y^2=\left(2x-y\right)(2x+y)\]

Sum of cubes

We derive the formula for the sum of cubes $a^3+b^3$.

Let's take the common factors out of brackets:

Let's take $\left(a+b\right)$ out of brackets:

In total, we get:

That is, the sum of the cubes of two monomials is equal to the product of their sum by the incomplete square of their difference.

Example 2

Express as a product $(8x)^3+y^3$

This expression can be rewritten in the following form:

\[(8x)^3+y^3=((2x))^3+y^3\]

Using the difference of squares formula, we get:

\[((2x))^3+y^3=\left(2x+y\right)(4x^2-2xy+y^2)\]

Difference of cubes

We derive the formula for the difference of cubes $a^3-b^3$.

To do this, we will use the same rule as above.

Let's add to our expression and subtract from it the monomials $a^2b\ and\ (ab)^2$:

Let's take the common factors out of brackets:

Let's take $\left(a-b\right)$ out of brackets:

In total, we get:

That is, the difference of the cubes of two monomials is equal to the product of their difference by the incomplete square of their sum.

Example 3

Express as a product of $(8x)^3-y^3$

This expression can be rewritten in the following form:

\[(8x)^3-y^3=((2x))^3-y^3\]

Using the difference of squares formula, we get:

\[((2x))^3-y^3=\left(2x-y\right)(4x^2+2xy+y^2)\]

An example of tasks for using the formulas for the difference of squares and the sum and difference of cubes

Example 4

Multiply.

a) $((a+5))^2-9$

c) $-x^3+\frac(1)(27)$

Solution:

a) $((a+5))^2-9$

\[(((a+5))^2-9=(a+5))^2-3^2\]

Applying the difference of squares formula, we get:

\[((a+5))^2-3^2=\left(a+5-3\right)\left(a+5+3\right)=\left(a+2\right)(a +8)\]

Let's write this expression in the form:

Let's apply the formula of cubes of cubes:

c) $-x^3+\frac(1)(27)$

Let's write this expression in the form:

\[-x^3+\frac(1)(27)=(\left(\frac(1)(3)\right))^3-x^3\]

Let's apply the formula of cubes of cubes:

\[(\left(\frac(1)(3)\right))^3-x^3=\left(\frac(1)(3)-x\right)\left(\frac(1)( 9)+\frac(x)(3)+x^2\right)\]