When calculating algebraic polynomials, to simplify calculations, we use abbreviated multiplication formulas. There are seven such formulas in total. They all need to be known by heart.
It should also be remembered that instead of "a" and "b" in the formulas, there can be both numbers and any other algebraic polynomials.
Difference of squares
Remember!
Difference of squares two numbers is equal to the product of the difference of these numbers and their sum.
a 2 − b 2 = (a − b)(a + b)- 15 2 − 2 2 = (15 − 2)(15 + 2) = 13 17 = 221
- 9a 2 − 4b 2 with 2 = (3a − 2bc)(3a + 2bc)
sum square
Remember!
The square of the sum of two numbers is equal to the square of the first number plus twice the product of the first number and the second plus the square of the second number.
(a + b) 2 = a 2 + 2ab + b 2
Note that with this reduced multiplication formula, it is easy to find the squares of large numbers without using a calculator or long multiplication. Let's explain with an example:
Find 112 2 .
- Let's decompose 112 into the sum of numbers whose squares we remember well.
112 = 100 + 1 - We write the sum of numbers in brackets and put a square over the brackets.
112 2 = (100 + 12) 2 - Let's use the sum square formula:
112 2 = (100 + 12) 2 = 100 2 + 2 100 12 + 12 2 = 10,000 + 2,400 + 144 = 12,544
Remember that the square sum formula is also valid for any algebraic polynomials.
- (8a + c) 2 = 64a 2 + 16ac + c 2
Warning!
(a + b) 2 is not equal to (a 2 + b 2)The square of the difference
Remember!
The square of the difference between two numbers is equal to the square of the first number minus twice the product of the first and the second plus the square of the second number.
(a − b) 2 = a 2 − 2ab + b 2
It is also worth remembering a very useful transformation:
(a - b) 2 = (b - a) 2The formula above is proved by simply expanding the parentheses:
(a − b) 2 = a 2 −2ab + b 2 = b 2 − 2ab + a 2 = (b − a) 2sum cube
Remember!
The cube of the sum of two numbers is equal to the cube of the first number plus three times the square of the first number times the second plus three times the product of the first times the square of the second plus the cube of the second.
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
How to remember the sum cube
Remembering this "terrible"-looking formula is quite simple.
- Learn that "a 3" comes at the beginning.
- The two polynomials in the middle have coefficients of 3.
- Recall that any number to the zero power is 1. (a 0 = 1, b 0 = 1) . It is easy to see that in the formula there is a decrease in the degree "a" and an increase in the degree "b". You can verify this:
(a + b) 3 = a 3 b 0 + 3a 2 b 1 + 3a 1 b 2 + b 3 a 0 = a 3 + 3a 2 b + 3ab 2 + b 3
Warning!
(a + b) 3 is not equal to a 3 + b 3difference cube
Remember!
difference cube of two numbers is equal to the cube of the first number minus three times the square of the first number and the second plus three times the product of the first number and the square of the second minus the cube of the second.
(a − b) 3 = a 3 − 3a 2 b + 3ab 2 − b 3
This formula is remembered like the previous one, but only taking into account the alternation of the signs "+" and "-". There is a “+” before the first member “a 3” (according to the rules of mathematics, we do not write it). This means that the next member will be preceded by “-”, then again “+”, etc.
(a − b) 3 = + a 3 - 3a 2 b + 3ab 2 - b 3 = a 3 - 3a 2 b + 3ab 2 - b 3Sum of cubes
Not to be confused with the sum cube!
Remember!
Sum of cubes is equal to the product of the sum of two numbers by the incomplete square of the difference.
a 3 + b 3 = (a + b)(a 2 − ab + b 2)The sum of cubes is the product of two brackets.
- The first parenthesis is the sum of two numbers.
- The second bracket is the incomplete square of the difference of numbers. The incomplete square of the difference is called the expression:
(a 2 − ab + b 2)
This square is incomplete, since in the middle, instead of a double product, there is an ordinary product of numbers.
Difference of cubes
Not to be confused with the difference cube!
Remember!
Difference of cubes is equal to the product of the difference of two numbers by the incomplete square of the sum.
a 3 − b 3 = (a − b)(a 2 + ab + b 2)Be careful when writing characters.
Application of abbreviated multiplication formulas
It should be remembered that all the formulas above are also used from right to left.
Many examples in textbooks are designed for you to use formulas to assemble the polynomial back.
- a 2 + 2a + 1 = (a + 1) 2
- (ac − 4b)(ac + 4b) = a 2 c 2 − 16b 2
You can download a table with all the formulas for abbreviated multiplication in the section "
There will also be tasks for an independent solution, to which you can see the answers.
Abbreviated multiplication formulas allow you to perform identical transformations of expressions - polynomials. With their help, polynomials can be factored, and using the formulas in reverse order, the products of binomials, squares and cubes can be represented as polynomials. Let's consider all generally accepted formulas for abbreviated multiplication, their derivation, common tasks for identical transformations of expressions using these formulas, as well as homework assignments (the answers to them are opened by links).
sum square
The formula for the square of the sum is the equality
(the square of the sum of two numbers is equal to the square of the first number plus twice the product of the first number and the second plus the square of the second number).
Instead of a And b any number can be substituted into this formula.
The sum square formula is often used to simplify calculations. For example,
Using the sum square formula, the polynomial can be factorized, namely, represented as a product of two identical factors.
Example 1
.
Example 2 Write as a polynomial expression
Solution. By the formula of the square of the sum, we get
The square of the difference
The formula for the square of the difference is the equality
(the square of the difference between two numbers is equal to the square of the first number minus twice the product of the first number and the second plus the square of the second number).
The squared difference formula is often used to simplify calculations. For example,
Using the difference square formula, the polynomial can be factorized, namely, represented as a product of two identical factors.
The formula follows from the rule for multiplying a polynomial by a polynomial:
Example 5 Write as a polynomial expression
Solution. By the formula of the square of the difference, we get
.
Apply the abbreviated multiplication formula yourself, and then see the solution
Full square selection
Often a polynomial of the second degree contains the square of the sum or difference, but is contained in a hidden form. To get the full square explicitly, you need to transform the polynomial. To do this, as a rule, one of the terms of the polynomial is represented as a double product, and then the same number is added to and subtracted from the polynomial.
Example 7
Solution. This polynomial can be transformed as follows:
Here we have presented 5 x in the form of a double product of 5/2 by x, added to the polynomial and subtracted from it the same number, then applied the sum square formula for the binomial.
So we have proved the equality
,
equals a full square plus the number .
Example 8 Consider a second degree polynomial
Solution. Let's make the following transformations on it:
Here we have presented 8 x in the form of a double product x by 4, added to the polynomial and subtracted from it the same number 4², applied the difference square formula for the binomial x − 4 .
So we have proved the equality
,
showing that a second degree polynomial
equals a full square plus the number −16.
Apply the abbreviated multiplication formula yourself, and then see the solution
sum cube
The sum cube formula is the equality
(The cube of the sum of two numbers is equal to the cube of the first number plus three times the square of the first number and the second, plus three times the product of the first number and the square of the second, plus the cube of the second number).
The sum cube formula is derived as follows:
Example 10 Write as a polynomial expression
Solution. According to the sum cube formula, we get
Apply the abbreviated multiplication formula yourself, and then see the solution
difference cube
The difference cube formula is the equality
(The cube of the difference of two numbers is equal to the cube of the first number minus three times the square of the first number and the second, plus three times the product of the first number and the square of the second minus the cube of the second number).
With the help of the sum cube formula, the polynomial can be decomposed into factors, namely, it can be represented as a product of three identical factors.
The difference cube formula is derived as follows:
Example 12. Write as a polynomial expression
Solution. Using the difference cube formula, we get
Apply the abbreviated multiplication formula yourself, and then see the solution
Difference of squares
The formula for the difference of squares is the equality
(the difference of the squares of two numbers is equal to the product of the sum of these numbers and their difference).
Using the sum cube formula, any polynomial of the form can be factorized.
The proof of the formula was obtained using the multiplication rule for polynomials:
Example 14 Write the product as a polynomial
.
Solution. By the difference of squares formula, we get
Example 15 Factorize
Solution. This expression in an explicit form does not fit any identity. But the number 16 can be represented as a power with base 4: 16=4². Then the original expression will take a different form:
,
and this is the formula for the difference of squares, and applying this formula, we get
In the previous lesson, we dealt with factorization. We mastered two methods: taking the common factor out of brackets and grouping. In this tutorial, the following powerful method: abbreviated multiplication formulas. In a short note - FSU.
Abbreviated multiplication formulas (square of sum and difference, cube of sum and difference, difference of squares, sum and difference of cubes) are essential 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 apply them.
Do we understand?)
Where do abbreviated multiplication formulas come from?
Equalities 6 and 7 are not written in a very usual way. Like the opposite. This is on purpose.) Any equality works both from left to right and from right to left. In such a record, it is clearer where the FSO comes 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 just 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. Reduced.) The result is immediately given.
FSU needs to know by heart. Without the first three, you can not dream of a triple, without the rest - about a four with a five.)
Why do we need abbreviated multiplication formulas?
There are two reasons to learn, even memorize, these formulas. The first - a ready-made answer on the machine dramatically reduces the number of errors. But this is not the main reason. And here's the second one...
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you can get acquainted with functions and derivatives.
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:
- k = 0;
- k - 1 = 0; k = 1;
- k + 1 = 0; k = -1;
- (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.
In order to simplify algebraic polynomials, there are abbreviated multiplication formulas. There are not so many of them and they are easy to remember, but you need to remember them. The notation used in formulas can take any form (number or polynomial).
The first abbreviated multiplication formula is called difference of squares. It lies in the fact that from the square of one number the square of the second number is subtracted equal to the difference between these numbers, as well as their product.
a 2 - b 2 \u003d (a - b) (a + b)
Let's analyze for clarity:
22 2 - 4 2 = (22-4)(22+4)=18 * 26 = 468
9a 2 - 4b 2 c 2 = (3a - 2bc)(3a + 2bc)
The second formula about sum of squares. It sounds like the sum of two values squared equals the square of the first value, the double product of the first value multiplied by the second is added to it, the square of the second value is added to them.
(a + b) 2 = a 2 + 2ab + b 2
Thanks to this formula, it becomes much easier to calculate the square of a large number, without the use of computer technology.
So for example: the square of 112 will be
1) At the beginning, we will analyze 112 into numbers whose squares are familiar to us
112 = 100 + 12
2) We enter the received in brackets squared
112 2 = (100+12) 2
3) Applying the formula, we get:
112 2 = (100+12) 2 = 100 2 + 2 * 100 * 12 + 122 = 10000 + 2400+ 144 = 12544
The third formula is difference squared. Which says that two values subtracted from each other squared are equal to the fact that, from the first value squared, we subtract the double product of the first value multiplied by the second, adding to them the square of the second value.
(a + b) 2 \u003d a 2 - 2ab + b 2
where (a - b) 2 equals (b - a) 2 . To prove this, (a-b) 2 = a 2 -2ab + b 2 = b 2 -2ab + a 2 = (b-a) 2
The fourth abbreviated multiplication formula is called sum cube. Which sounds like: two terms of the value in the cube are equal to the cube of 1 value, the triple product of 1 value squared multiplied by the 2nd value is added to them, the triple product of 1 value multiplied by the square of 2 value is added to them, plus the second value cubed.
(a + b) 3 \u003d a 3 + 3a 2 b + 3ab 2 + b 3
The fifth, as you already understood, is called difference cube. Which finds the differences between the values, as from the first designation in the cube we subtract the triple product of the first designation squared multiplied by the second, the triple product of the first designation multiplied by the square of the second designation is added to them, minus the second designation in the cube.
(a-b) 3 \u003d a 3 - 3a 2 b + 3ab 2 - b 3
The sixth is called sum of cubes. The sum of cubes is equal to the product of two terms multiplied by the incomplete square of the difference, since there is no doubled value in the middle.
a 3 + b 3 \u003d (a + b) (a 2 -ab + b 2)
In another way, you can say the sum of cubes can be called the product in two brackets.
The seventh and final is called difference of cubes(it is easy to confuse it with the difference cube formula, but these are different things). The difference of cubes is equal to the product of the difference of two quantities multiplied by the incomplete square of the sum, since there is no doubled value in the middle.
a 3 - b 3 \u003d (a-b) (a 2 + ab + b 2)
And so there are only 7 formulas for abbreviated multiplication, they are similar to each other and are easy to remember, the only thing is not to get confused in signs. They are also designed to be used in reverse order and there are quite a few such tasks collected in textbooks. Be careful and you will succeed.
If you have any questions about the formulas, be sure to write them in the comments. We will be glad to answer you!
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