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 the 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
Decision. 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
Decision. 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
Decision. 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
Decision. 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 times the second, plus three times the product of the first number times 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
Decision. 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
Decision. 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
.
Decision. By the difference of squares formula, we get
Example 15 Factorize
Decision. 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
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 two numbers is equal to the cube of the first number minus three times the square of the first number times the second plus three times the product of the first number times 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 "
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).
Decision. In the task condition, 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.
Decision. 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.
Lesson content
The square of the sum of two expressions
There are a number of cases where the multiplication of a polynomial by a polynomial can be greatly simplified. Such, for example, is the case (2 x+ 3y) 2 .
Expression (2 x+ 3y) 2 is the multiplication of two polynomials, each of which is equal to (2 x+ 3y)
(2x+ 3y) 2 = (2x+ 3y)(2x+ 3y)
We got the multiplication of a polynomial by a polynomial. Let's execute it:
(2x+ 3y) 2 = (2x+ 3y)(2x+ 3y) = 4x 2 + 6xy + 6xy + 9y 2 = 4x 2 + 12xy+ 9y 2
That is, the expression (2 x+ 3y) 2 is equal to 4x 2 + 12xy + 9y 2
(2x+ 3y) 2 = 4x 2 + 12xy+ 9y 2
Let's solve a similar example, which is simpler:
(a+b) 2
Expression ( a+b) 2 is the multiplication of two polynomials, each of which is equal to ( a+b)
(a+b) 2 = (a+b)(a+b)
Let's do this multiplication:
(a+b) 2 = (a+b)(a+b) = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2
That is the expression (a+b) 2 is equal to a 2 + 2ab + b 2
(a+b) 2 = a 2 + 2ab + b 2
It turns out that the case ( a+b) 2 can be extended for any a and b. The first example we solved, namely (2 x+ 3y) 2 can be solved using the identity (a+b) 2 = a 2 + 2ab + b 2 . To do this, you need to substitute instead of variables a and b corresponding terms from expression (2 x+ 3y) 2 . In this case, the variable a match dick 2 x, and the variable b match dick 3 y
a = 2x
b = 3y
And then we can use the identity (a+b) 2 = a 2 + 2ab + b 2 , but instead of variables a and b you need to substitute expressions 2 x and 3 y respectively:
(2x+ 3y) 2 = (2x) 2 + 2 × 2 x× 3 y + (3y) 2 = 4x 2 + 12xy+ 9y 2
Like last time, we got a polynomial 4x 2 + 12xy+ 9y 2 . The solution is usually written shorter, performing all elementary transformations in the mind:
(2x+ 3y) 2 = 4x 2 + 12xy+ 9y 2
Identity (a+b) 2 = a 2 + 2ab + b 2 is called the formula for the square of the sum of two expressions. This formula can be read like this:
The square of the sum of two expressions is equal to the square of the first expression plus twice the product of the first expression and the second plus the square of the second expression.
Consider the expression (2 + 3) 2 . It can be calculated in two ways: perform addition in brackets and square the result, or use the formula for the square of the sum of two expressions.
First way:
(2 + 3) 2 = 5 2 = 25
Second way:
(2 + 3) 2 = 2 2 + 2 × 2 × 3 + 3 2 = 4 + 12 + 9 = 25
Example 2. Convert expression (5 a+ 3) 2 into a polynomial.
Let's use the formula for the square of the sum of two expressions:
(a+b) 2 = a 2 + 2ab + b 2
(5a + 3) 2 = (5a) 2 + 2 × 5 a × 3 + 3 2 = 25a 2 + 30a + 9
Means, (5a + 3) 2 = 25a 2 + 30a + 9.
Let's try to solve this example without using the sum square formula. We should get the same result:
(5a + 3) 2 = (5a + 3)(5a + 3) = 25a 2 + 15a + 15a + 9 = 25a 2 + 30a + 9
The formula for the square of the sum of two expressions has a geometric meaning. We remember that to calculate the area of a square, you need to raise its side to the second power.
For example, the area of a square with a side a will be equal to a 2. If you increase the side of the square by b, then the area will be equal to ( a+b) 2
Consider the following figure:
Imagine that the side of the square shown in this figure is increased by b. A square has all sides equal. If its side is increased by b, then the other sides will also increase by b
The result is a new square, which is larger than the previous one. To see it well, let's complete the missing sides:
To calculate the area of this square, you can separately calculate the squares and rectangles included in it, then add the results.
First, you can calculate a square with a side a- its area will be equal to a 2. Then you can calculate rectangles with sides a and b- they will be equal ab. Then you can calculate a square with a side b
The result is the following sum of areas:
a 2 + ab+ab + b 2
The sum of the areas of identical rectangles can be replaced by multiplying 2 ab, which literally means "repeat two times the area of rectangle ab" . Algebraically, this is obtained by reducing like terms ab and ab. The result is an expression a 2 + 2ab+ b 2 , which is the right side of the formula for the square of the sum of two expressions:
(a+b) 2 = a 2 + 2ab+ b 2
The square of the difference of two expressions
The formula for the square of the difference of two expressions is as follows:
(a-b) 2 = a 2 − 2ab + b 2
The square of the difference of two expressions is equal to the square of the first expression minus twice the product of the first expression and the second plus the square of the second expression.
The formula for the square of the difference of two expressions is derived in the same way as the formula for the square of the sum of two expressions. Expression ( a-b) 2 is the product of two polynomials, each of which is equal to ( a-b)
(a-b) 2 = (a-b)(a-b)
If you perform this multiplication, you get a polynomial a 2 − 2ab + b 2
(a-b) 2 = (a-b)(a-b) = a 2 − ab− ab+ b 2 = a 2 − 2ab + b 2
Example 1. Convert expression (7 x− 5) 2 into a polynomial.
Let's use the formula of the square of the difference of two expressions:
(a-b) 2 = a 2 − 2ab + b 2
(7x− 5) 2 = (7x) 2 − 2 × 7 x × 5 + 5 2 = 49x 2 − 70x + 25
Means, (7x− 5) 2 = 49x 2 + 70x + 25.
Let's try to solve this example without using the difference square formula. We should get the same result:
(7x− 5) 2 = (7x− 5) (7x− 5) = 49x 2 − 35x − 35x + 25 = 49x 2 − 70x+ 25.
The formula for the square of the difference of two expressions also has a geometric meaning. If the area of a square with a side a is equal to a 2 , then the area of the square whose side is reduced by b, will be equal to ( a-b) 2
Consider the following figure:
Imagine that the side of the square shown in this figure is reduced by b. A square has all sides equal. If one side is reduced by b, then the other sides will also decrease by b
The result is a new square, which is smaller than the previous one. It is highlighted in yellow in the figure. Its side is a− b since the old side a decreased by b. To calculate the area of this square, you can use the original area of the square a 2 subtract the areas of the rectangles that were obtained in the process of reducing the sides of the old square. Let's show these rectangles:
Then we can write the following expression: old area a 2 minus area ab minus area ( a-b)b
a 2 − ab − (a-b)b
Expand the brackets in the expression ( a-b)b
a 2 − ab - ab + b 2
Here are similar terms:
a 2 − 2ab + b 2
The result is an expression a 2 − 2ab + b 2 , which is the right side of the formula for the square of the difference of two expressions:
(a-b) 2 = a 2 − 2ab + b 2
The formulas for the square of the sum and the square of the difference are generally called abbreviated multiplication formulas. These formulas allow you to significantly simplify and speed up the process of multiplying polynomials.
Earlier we said that considering a member of a polynomial separately, it must be considered together with the sign that is located in front of it.
But when applying the abbreviated multiplication formulas, the sign of the original polynomial should not be considered as the sign of this term itself.
For example, given the expression (5 x − 2y) 2 , and we want to use the formula (a-b) 2 = a 2 − 2ab + b 2 , then instead of b need to substitute 2 y, not −2 y. This is a feature of working with formulas that should not be forgotten.
(5x − 2y) 2
a = 5x
b = 2y
(5x − 2y) 2 = (5x) 2 − 2 × 5 x×2 y + (2y) 2 = 25x 2 − 20xy + 4y 2
If we substitute −2 y, then this will mean that the difference in the brackets of the original expression has been replaced by the sum:
(5x − 2y) 2 = (5x + (−2y)) 2
and in this case it is necessary to apply not the formula of the square of the difference, but the formula of the square of the sum:
(5x + (−2y) 2
a = 5x
b = −2y
(5x + (−2y)) 2 = (5x) 2 + 2 × 5 x× (−2 y) + (−2y) 2 = 25x 2 − 20xy + 4y 2
An exception may be expressions of the form (x− (−y)) 2 . In this case, using the formula (a-b) 2 = a 2 − 2ab + b 2 instead of b should be substituted (− y)
(x− (−y)) 2 = x 2 − 2 × x× (− y) + (−y) 2 = x 2 + 2xy + y 2
But squaring expressions of the form x − (−y) , it will be more convenient to replace subtraction with addition x+y. Then the original expression will take the form ( x +y) 2 and it will be possible to use the formula of the square of the sum, and not the difference:
(x +y) 2 = x 2 + 2xy + y 2
Sum Cube and Difference Cube
The formulas for the cube of the sum of two expressions and the cube of the difference of two expressions are as follows:
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
(a-b) 3 = a 3 − 3a 2 b + 3ab 2 − b 3
The formula for the cube of the sum of two expressions can be read like this:
The cube of the sum 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.
And the formula for the cube of the difference of two expressions can be read as follows:
The cube of the difference 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.
When solving problems, it is desirable to know these formulas by heart. If you don't remember, don't worry! You can take them out on your own. We already know how.
Let's derive the sum cube formula on our own:
(a+b) 3
Expression ( a+b) 3 is a product of three polynomials, each of which is equal to ( a+ b)
(a+b) 3 = (a+ b)(a+ b)(a+ b)
But the expression ( a+b) 3 can also be written as (a+ b)(a+ b) 2
(a+b) 3 = (a+ b)(a+ b) 2
In this case, the factor ( a+ b) 2 is the square of the sum of the two expressions. This square of the sum is equal to the expression a 2 + 2ab + b 2 .
Then ( a+b) 3 can be written as (a+ b)(a 2 + 2ab + b 2) .
(a+b) 3 = (a+ b)(a 2 + 2ab + b 2)
And this is the multiplication of a polynomial by a polynomial. Let's execute it:
(a+b) 3 = (a+ b)(a 2 + 2ab + b 2) = a 3 + 2a 2 b + ab 2 + a 2 b + 2ab 2 + b 3 = a 3 + 3a 2 b + 3ab 2 + b 3
Similarly, you can derive the formula for the cube of the difference of two expressions:
(a-b) 3 = (a- b)(a 2 − 2ab + b 2) = a 3 − 2a 2 b + ab 2 − a 2 b + 2ab 2 − b 3 = a 3 − 3a 2 b+ 3ab 2 − b 3
Example 1. Convert the expression ( x+ 1) 3 into a polynomial.
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
(x+ 1) 3 = x 3+3× x 2×1 + 3× x× 1 2 + 1 3 = x 3 + 3x 2 + 3x + 1
Let's try to solve this example without using the cube formula of the sum of two expressions
(x+ 1) 3 = (x+ 1)(x+ 1)(x+ 1) = (x+ 1)(x 2 + 2x + 1) = x 3 + 2x 2 + x + x 2 + 2x + 1 = x 3 + 3x 2 + 3x + 1
Example 2. Convert expression (6a 2 + 3b 3) 3 into a polynomial.
Let's use the cube formula for the sum of two expressions:
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
(6a 2 + 3b 3) 3 = (6a 2) 3 + 3 × (6 a 2) 2×3 b 3+3×6 a 2 × (3b 3) 2 + (3b 3) 3 = 216a 6+3×36 a 4×3 b 3+3×6 a 2×9 b 6 + 27b 9
Example 3. Convert expression ( n 2 − 3) 3 into a polynomial.
(a-b) = a 3 − 3a 2 b + 3ab 2 − b 3
(n 2 − 3) 3 = (n 2) 3 − 3 × ( n 2) 2×3 + 3× n 2 × 3 2 − 3 3 = n 6 − 9n 4 + 27n 2 − 27
Example 4. Convert expression (2x 2 − x 3) 3 into a polynomial.
Let's use the cube formula of the difference of two expressions:
(a-b) = a 3 − 3a 2 b + 3ab 2 − b 3
(2x 2 − x 3) 3 = (2x 2) 3 − 3 × (2 x 2) 2× x 3+3×2 x 2×( x 3) 2 − (x 3) 3 =
8x 6 − 3 × 4 x 4× x 3+3×2 x 2× x 6 − x 9 =
8x 6 − 12x 7 + 6x 8 − x 9
Multiplying the difference of two expressions by their sum
There are problems in which it is required to multiply the difference of two expressions by their sum. For example:
(a-b)(a+b)
In this expression, the difference of two expressions a and b multiplied by the sum of the same two expressions. Let's do this multiplication:
(a-b)(a+b) = a 2 + ab − ab − b 2 = a 2 − b 2
That is the expression (a-b)(a+b) equals a 2 − b 2
(a-b)(a+b) = a 2 − b 2
We see that when multiplying the difference of two expressions by their sum, we get the difference of the squares of these expressions.
The product of the difference of two expressions and their sum is equal to the difference of the squares of these expressions.
Happening (a-b)(a+b) can be extended to any a and b. Simply put, if when solving a problem it is necessary to multiply the difference of two expressions by their sum, then this multiplication can be replaced by the difference of the squares of these expressions.
Example 1. Perform multiplication (2x − 5)(2x + 5)
In this example, the expression difference is 2 x and 5 multiplied by the sum of these same expressions. Then according to the formula (a-b)(a+b) = a 2 − b 2 we have:
(2x − 5)(2x + 5) = (2x) 2 − 5 2
We calculate the right side, we get 4 x 2 − 25
(2x − 5)(2x + 5) = (2x) 2 − 5 2 = 4x 2 − 25
Let's try to solve this example without using the formula (a-b)(a+b) = a 2 − b 2 . We will get the same result 4 x 2 − 25
(2x − 5)(2x + 5) = 4x 2 − 10x + 10x − 25 = 4x 2 − 25
Example 2. Perform multiplication (4x − 5y)(4x + 5y)
(a-b)(a+b) = a 2 − b 2
(4x − 5y)(4x + 5y) = (4x) 2 − (5y) 2 = 16x 2 − 25y 2
Example 3. Perform multiplication (2a+ 3b)(2a− 3b)
Let's use the formula for multiplying the difference of two expressions by their sum:
(a-b)(a+b) = a 2 − b 2
(2a + 3b)(2a- 3b) = (2a) 2 − (3b) 2 = 4a 2 − 9b 2
In this example, the sum of terms is 2 a and 3 b located earlier than the difference of these terms. And in the formula (a-b)(a+b) = a 2 − b 2 the difference is located earlier.
It makes no difference how the factors are arranged ( a-b) in ( a+b) in the formula. They can be written as (a-b)(a+b) , and (a+b)(a-b) . The result will still be a 2 − b 2 , since the product does not change from a permutation of the factors.
So in this example, the factors (2 a + 3b) and 2 a- 3b) can be written as (2a + 3b)(2a- 3b) , and (2a- 3b)(2a + 3b) . The result will still be 4. a 2 − 9b 2 .
Example 3. Perform multiplication (7 + 3x)(3x − 7)
Let's use the formula for multiplying the difference of two expressions by their sum:
(a-b)(a+b) = a 2 − b 2
(7 + 3x)(3x − 7) = (3x) 2 − 7 2 = 9x 2 − 49
Example 4. Perform multiplication (x 2 − y 3)(x 2 + y 3)
(a-b)(a+b) = a 2 − b 2
(x 2 − y 3)(x 2 + y 3) = (x 2) 2 − (y 3) 2 = x 4 − y 6
Example 5. Perform multiplication (−5x− 3y)(5x− 3y)
In the expression (−5 x− 3y) we take out −1, then the original expression will take the following form:
(−5x− 3y)(5x− 3y) = −1(5x + 3y)(5x − 3y)
Work (5x + 3y)(5x − 3y) replace by the difference of squares:
(−5x− 3y)(5x− 3y) = −1(5x + 3y)(5x − 3y) = −1((5x) 2 − (3y) 2)
The difference of squares was enclosed in brackets. If this is not done, then it will turn out that −1 is multiplied only by (5 x) 2 . And this will lead to an error and change the value of the original expression.
(−5x− 3y)(5x− 3y) = −1(5x + 3y)(5x − 3y) = −1((5x) 2 − (3y) 2) = −1(25x 2 − 9x 2)
Now multiply −1 by the parenthesized expression and get the final result:
(−5x− 3y)(5x− 3y) = −1(5x + 3y)(5x − 3y) = −1((5x) 2 − (3y) 2) =
−1(25x 2 − 9y 2) = −25x 2 + 9y 2
Multiplying the difference of two expressions by the incomplete square of their sum
There are problems in which it is required to multiply the difference of two expressions by the incomplete square of their sum. This piece looks like this:
(a-b)(a 2 + ab + b 2)
First polynomial ( a-b) is the difference of two expressions, and the second polynomial (a 2 + ab + b 2) is the incomplete square of the sum of these two expressions.
The incomplete square of the sum is a polynomial of the form a 2 + ab + b 2 . It is similar to the usual square of the sum a 2 + 2ab + b 2
For example, the expression 4x 2 + 6xy + 9y 2 is an incomplete square of the sum of expressions 2 x and 3 y .
Indeed, the first term of the expression 4x 2 + 6xy + 9y 2 , namely 4 x 2 is the square of expression 2 x, since (2 x) 2 = 4x 2. The third term of the expression 4x 2 + 6xy + 9y 2 , namely 9 y 2 is the square of 3 y, since (3 y) 2 = 9y 2. mid dick 6 xy, is the product of expressions 2 x and 3 y.
So let's multiply the difference ( a-b) by an incomplete square of the sum a 2 + ab + b 2
(a-b)(a 2 + ab + b 2) = a(a 2 + ab + b 2) − b(a 2 + ab + b 2) =
a 3 + a 2 b + ab 2 − a 2 b − ab 2 − b 3 = a 3 − b 3
That is the expression (a-b)(a 2 + ab + b 2) equals a 3 − b 3
(a-b)(a 2 + ab + b 2) = a 3 − b 3
This identity is called the formula for multiplying the difference of two expressions by the incomplete square of their sum. This formula can be read like this:
The product of the difference of two expressions and the incomplete square of their sum is equal to the difference of the cubes of these expressions.
Example 1. Perform multiplication (2x − 3y)(4x 2 + 6xy + 9y 2)
First polynomial (2 x − 3y) is the difference of two expressions 2 x and 3 y. Second polynomial 4x 2 + 6xy + 9y 2 is the incomplete square of the sum of two expressions 2 x and 3 y. This allows us to use the formula without making lengthy calculations (a-b)(a 2 + ab + b 2) = a 3 − b 3 . In our case, the multiplication (2x − 3y)(4x 2 + 6xy + 9y 2) can be replaced by the difference of cubes 2 x and 3 y
(2x − 3y)(4x 2 + 6xy + 9y 2) = (2x) 3 − (3y) 3 = 8x 3 − 27y 3
(a-b)(a 2 + ab+ b 2) = a 3 − b 3 . We get the same result, but the solution becomes longer:
(2x − 3y)(4x 2 + 6xy + 9y 2) = 2x(4x 2 + 6xy + 9y 2) − 3y(4x 2 + 6xy + 9y 2) =
8x 3 + 12x 2 y + 18xy 2 − 12x 2 y − 18xy 2 − 27y 3 = 8x 3 − 27y 3
Example 2. Perform multiplication (3 − x)(9 + 3x + x 2)
The first polynomial (3 − x) is the difference of the two expressions, and the second polynomial is the incomplete square of the sum of these two expressions. This allows us to use the formula (a-b)(a 2 + ab + b 2) = a 3 − b 3
(3 − x)(9 + 3x + x 2) = 3 3 − x 3 = 27 − x 3
Multiplying the sum of two expressions by the incomplete square of their difference
There are problems in which it is required to multiply the sum of two expressions by the incomplete square of their difference. This piece looks like this:
(a+b)(a 2 − ab + b 2)
First polynomial ( a+b (a 2 − ab + b 2) is an incomplete square of the difference of these two expressions.
The incomplete square of the difference is a polynomial of the form a 2 − ab + b 2 . It is similar to the usual squared difference a 2 − 2ab + b 2 except that in it the product of the first and second expressions is not doubled.
For example, the expression 4x 2 − 6xy + 9y 2 is an incomplete square of the difference of expressions 2 x and 3 y .
(2x) 2 − 2x× 3 y + (3y) 2 = 4x 2 − 6xy + 9y 2
Let's go back to the original example. Let's multiply the sum a+b by the incomplete square of the difference a 2 − ab + b 2
(a+b)(a 2 − ab + b 2) = a(a 2 − ab + b 2) + b(a 2 − ab + b 2) =
a 3 − a 2 b + ab 2 + a 2 b − ab 2 + b 3 = a 3 + b 3
That is the expression (a+b)(a 2 − ab + b 2) equals a 3 + b 3
(a+b)(a 2 − ab + b 2) = a 3 + b 3
This identity is called the formula for multiplying the sum of two expressions by the incomplete square of their difference. This formula can be read like this:
The product of the sum of two expressions and the incomplete square of their difference is equal to the sum of the cubes of these expressions.
Example 1. Perform multiplication (2x + 3y)(4x 2 − 6xy + 9y 2)
First polynomial (2 x + 3y) is the sum of two expressions 2 x and 3 y, and the second polynomial 4x 2 − 6xy + 9y 2 is the incomplete square of the difference of these expressions. This allows us to use the formula without making lengthy calculations (a+b)(a 2 − ab + b 2) = a 3 + b 3 . In our case, the multiplication (2x + 3y)(4x 2 − 6xy + 9y 2) can be replaced by the sum of cubes 2 x and 3 y
(2x + 3y)(4x 2 − 6xy + 9y 2) = (2x) 3 + (3y) 3 = 8x 3 + 27y 3
Let's try to solve the same example without using the formula (a+b)(a 2 − ab+ b 2) = a 3 + b 3 . We get the same result, but the solution becomes longer:
(2x + 3y)(4x 2 − 6xy + 9y 2) = 2x(4x 2 − 6xy + 9y 2) + 3y(4x 2 − 6xy + 9y 2) =
8x 3 − 12x 2 y + 18xy 2 + 12x 2 y − 18xy 2 + 27y 3 = 8x 3 + 27y 3
Example 2. Perform multiplication (2x+ y)(4x 2 − 2xy + y 2)
First polynomial (2 x+ y) is the sum of two expressions, and the second polynomial (4x 2 − 2xy + y 2) is an incomplete square of the difference of these expressions. This allows us to use the formula (a+b)(a 2 − ab+ b 2) = a 3 + b 3
(2x+ y)(4x 2 − 2xy + y 2) = (2x) 3 + y 3 = 8x 3 + y 3
Let's try to solve the same example without using the formula (a+b)(a 2 − ab+ b 2) = a 3 + b 3 . We get the same result, but the solution becomes longer:
(2x+ y)(4x 2 − 2xy + y 2) = 2x(4x 2 − 2xy + y 2) + y(4x 2 − 2xy + y 2) =
8x 3 − 4x 2 y + 2xy 2 + 4x 2 y − 2xy 2 + y 3 = 8x 3 + y 3
Tasks for independent solution
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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 is equal to 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 is added the triple product of 1 value multiplied by the square of 2 value, 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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