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 the 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, because (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 this lesson, we will get acquainted with the formulas for the square of the sum and the square of the difference and derive them. Let us prove the formula for the square of the sum geometrically. In addition, we will solve many different examples using these formulas.
Consider the formula for the square of the sum:
So, we have derived the formula for the square of the sum:
Verbally, this formula is expressed as follows: the square of the sum is equal to the square of the first number plus twice the product of the first number by the second plus the square of the second number.
This formula is easy to represent geometrically.
Consider a square with side :
Square area.
On the other hand, the same square can be represented differently by dividing the side into a and b (Fig. 1).
Rice. 1. Square
Then the area of the square can be represented as the sum of the areas:
Since the squares were the same, their areas are equal, which means:
So, we have proved geometrically the formula for the square of the sum.
Consider examples:
Comment: the example is solved using the sum square formula.
We derive the formula for the square of the difference:
So, we have derived the formula for the square of the difference:
Verbally, this formula is expressed as follows: the square of the difference is equal to the square of the first number minus twice the product of the first number by the second plus the square of the second number.
Consider examples:
The formulas for the square of the sum and the square of the difference can work both from left to right and from right to left. When used from left to right, these will be abbreviated multiplication formulas, they are used when calculating and transforming examples. And when used from right to left - factorization formulas.
Consider examples in which you need to factorize a given polynomial by applying the formulas for the square of the sum and the square of the difference. To do this, you need to look very carefully at the polynomial and determine exactly how to expand it correctly.
Comment: in order to factorize a polynomial, you need to determine what is represented in this expression. So we see the square and the square of unity. Now we need to find the double product - this is . So, all the necessary elements are there, you just need to determine whether this is the square of the sum or difference. Before the doubled product there is a plus sign, which means that we have the square of the sum.
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)
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).
Using the sum cube formula, the polynomial can be factorized, namely, 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
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