Remarkable products are mathematical expressions that frequently arise in various situations and are essential for simplifying calculations and solving problems. In this context, understanding and mastering notable products is essential for the study of algebra and mathematics in general. In this article, we'll explain the concept of notable products, present key examples, and propose solved exercises to help you grasp and understand this important topic.
Simplifying the explanation of remarkable products in simple and practical steps.
Remarkable products are mathematical expressions that have a specific, recurring form, facilitating calculations and simplifying equations. To better understand this concept, let's break it down into simple, practical steps.
First, it's important to understand that notable products are composed of algebraic expressions that follow a predefined pattern. The main notable products are: square of the sum, square of the difference, product of the sum and the difference e square of a binomial.
To calculate these remarkable products, simply apply the corresponding mathematical properties to each case. For example, in the case of square of the sum, we use the formula (a + b)² = a² + 2ab + b². In the square of the difference, we have (a – b)² = a² – 2ab + b².
To make it easier to understand, let's solve a practical exercise: calculate the square of the sum between 3x and 2y. Applying the formula (a + b)², we have (3x + 2y)² = (3x)² + 2(3x)(2y) + (2y)².
By simplifying the expression, we obtain: 9x² + 12xy + 4y². In this way, we find the remarkable product corresponding to the square of the sum of 3x and 2y.
In short, notable products are mathematical expressions with standardized forms that facilitate calculation and simplification of equations. With practice and knowledge of the appropriate formulas, it's possible to solve problems with ease and precision.
Tips for solving notable product problems effectively and practically.
Solving problems involving notable products can be challenging for many students, but with the right tips, it's possible to make this process easier and more effective. Here are some tips for solving notable product problems effectively and practically:
1. Identify the type of notable product: Before you begin solving the problem, identify whether it's a square of the sum, a square of the difference, a product of the sum and the difference, or a square of a binomial. Knowing the type of product will guide you toward the correct solution.
2. Use specific formulas: Each type of notable product has a specific formula for its solution. Make sure you know them and apply them correctly to the problem at hand.
3. Simplify the expressions: Problems involving notable products can often seem complex at first glance. Therefore, it's important to simplify expressions and identify patterns that facilitate resolution.
4. Practice with varied exercises: Practice is essential to mastering remarkable products. Solve various exercises, varying the types of problems and difficulties, to hone your skills and understanding of the subject.
5. Consult supporting material: If you have questions or difficulties troubleshooting a product, consult textbooks, explanatory videos, or instructors for assistance and clarification.
Now that you know some tips for solving remarkable product problems effectively and practically, put them into practice and strengthen your math skills. With dedication and persistence, you'll be able to master this content and succeed in your studies.
Solving remarkable products: a simple step-by-step guide to solving these special mathematical expressions.
Remarkable products are special mathematical expressions that facilitate the solution of equations and the simplification of polynomials. To solve remarkable products, it's important to understand the formulas and apply them correctly. In this article, we'll explain simply and clearly how to solve these special mathematical expressions.
One of the most common notable products is the square of the sum of two terms, which can be represented by the formula: (a + b)² = a² + 2ab + b². To solve this expression, simply substitute the values of a e b in the formula and perform the necessary mathematical operations.
Another example of a notable product is the square of the difference of two terms, which follows the formula: (a – b)² = a² – 2ab + b². To solve this expression, simply substitute the values of a e b in the formula and perform the corresponding mathematical operations.
In addition to these, there are other notable products that can be useful in solving more complex mathematical problems. It's important to practice solving exercises to familiarize yourself with these formulas and ensure good performance on tests and entrance exams.
Now that you understand how to solve remarkable products, practice solving the following exercises:
1) Calculate the value of (3 + 4)²
2) Simplify the expression (5 – 2)²
With these examples and constant practice, you'll be able to solve any notable product with ease. Remember to review the formulas and practice regularly to keep your math skills sharp!
Discover the three remarkable product types in just one simple and straightforward explanation.
Remarkable products are mathematical expressions that have special characteristics and can be easily simplified. There are three main types of remarkable products: square of the sum, square of the difference e product of the sum and the difference.
Notable products: explanation and solved exercises
Products Notable are algebraic operations, in which multiplications of polynomials are expressed, which do not need to be solved traditionally, but with the help of certain rules you can find their results.
Polynomials are multiplied if, therefore, they can have a large number of terms and variables. To shorten the process, remarkable product rules are used, which allow multiplications to be performed without having to go term by term.
Notable products and examples
Each notable product is a formula that results from a factorization, composed of polynomials of several terms, such as binomials or trinomials, called factors.
Factors are the base of a power and have an exponent. When factors are multiplied, the exponents must be added.
There are several notable product formulas, some more used than others, depending on the polynomials, and they are as follows:
Squared binomial
It is the multiplication of a binomial by itself, expressed in power form, where the terms are added or subtracted:
a. Binomial sum of squares: is equal to the square of the first term, plus twice the product of the terms, plus the square of the second term. It is expressed as follows:
(a+b) 2 =(a+b) * (a + b).
The following figure shows how the product develops according to the aforementioned rule. The result is called a perfect square trinomial.
1 Example
(x + 5)² = x² + 2 (x * 5) + 5²
(x + 5) ² = x² + 2 (5x) + 25
(x + 5)² = x² + 10x + 25.
2 Example
(4a + 2b) = (4a) 2 + 2 (4th * 2b) + (2b) 2
(4a + 2b) = 8a 2 + 2 (8ab) + 4b 2
(4a + 2b) = 8a 2 + 16 ab + 4b 2 .
b. Binomial of a squared subtraction: the same rule for the binomial sum applies, only in this case the second term is negative. Its formula is as follows:
(a - b) 2 = [(a) + (- b)] 2
(a - b) 2 = A 2 + 2a * (-b) + (-b) 2
(a - b) 2 = A 2 - 2ab + b 2 .
1 Example
(2x - 6) 2 = (2x) 2 – 2 (2x * 6) + 6 2
(2x - 6) 2 = 4x 2 – 2 (12x) + 36
(2x - 6) 2 = 4x 2 – 24x + 36.
Product of conjugate binomials
Two binomials are conjugate when the second terms of each have different signs, i.e., the first is positive and the second is negative, or vice versa. This is solved by squaring and subtracting each monomial. The formula is as follows:
(a+b) * (a - b)
In the following figure, the product of two conjugate binomials is developed, where it can be seen that the result is a difference of squares.
1 Example
(2a + 3b) (2a – 3b) = 4a 2 + (-6ab) + (6ab) + (-9b 2 )
(2a + 3b) (2a – 3b) = 4a 2 - 9b 2 .
Product of two binomials with a common term
It's one of the most complex and rarely used notable products because it's a multiplication of two binomials that have a common term. The rule states the following:
- The square of the common term.
- Also, add the terms that are not common and then multiply them by the common term.
- Plus the sum of the multiplication of terms that are not common.
It is represented in the formula: (x + a) * (x + b) and is expanded as shown in the image. The result is a non-perfect square trinomial.
(x+6) * (x + 9) = x 2 + (6 + 9) * x + (6 * 9)
(x+6) * (x + 9) = x 2 +15x +54.
There is a possibility that the second term (the different term) is negative and its formula is as follows: (x + a) * (x – b).
2 Example
(7x+4) * (7x – 2) = (7x * 7x) + ( 4-2 ) * 7x + (4 * -2)
(7x+4) * (7x – 2) = 49x 2 + (2) * 7x - 8
(7x+4) * (7x – 2) = 49x 2 +14x – 8.
It could also be that both terms are negative. Your formula will be: (x – a) * (x – b).
3 Example
(3b – 6) * (3b – 5) = (3b * 3b) + (-6-5) * (3b) + (-6 * -5)
(3b – 6) * (3b – 5) = 9b 2 + (-11) * (3b) + (30)
(3b – 6) * (3b – 5) = 9b 2 – 33b + 30.
Squared polynomial
In this case, there are more than two terms and, to develop it, each one is squared and added to twice the multiplication of one term by the other; Its formula is: (a + b + c) 2 and the result of the operation is a square trinomial.
1 Example
(3x + 2y + 4z) 2 = (3x) 2 + (2y) 2 + (4z) 2 + 2 (6xy + 12xz + 8yz)
(3x + 2y + 4z) 2 = 9x 2 +4y 2 +16z 2 + 12xy + 24xz + 16yz.
Binomial to the cube
It's a remarkable complex product. To develop it, multiply the binomial by its square, as follows:
a. For the binomial in the cube of a sum:
- The cube of the first term, plus three times the square of the first term times the second.
- Plus three times the first term, for the second square.
- Plus the cube of the second term.
(a+b) 3 =(a+b) * (a+b) 2
(a+b) 3 =(a+b) * (a 2 +2ab+b 2 )
(a+b) 3 = A 3 + 2a 2 b+ab 2 + ba 2 +2ab 2 + b 3
(a+b) 3 = A 3 + 3a 2 b+3ab 2 + b 3 .
1 Example
(to + 3) 3 = A 3 + 3 (a) 2 * (3) + 3 (a) * (3) 2 + (3) 3
(to + 3) 3 = A 3 + 3 (a) 2 * (3) + 3 (a) * (9) + 27
(to + 3) 3 = A 3 + 9 to 2 + 27a + 27.
b. For the binomial in the cube of a subtraction:
- The cube of the first term, minus three times the square of the first term times the second.
- Plus three times the first term, for the second square.
- Minus the cube of the second term.
(a - b) 3 = (a – b) * (a - b) 2
(a - b) 3 = (a – b) * (a 2 - 2ab + b 2 )
(a - b) 3 = A 3 – 2th 2 b+ab 2 –ba 2 +2ab 2 - b 3
(a - b) 3 = a 3 – 3th 2 b+3ab 2 - b 3 .
2 Example
(b – 5) 3 = b 3 + 3 (b) 2 * (-5) + 3 (b) * (-5) 2 + (-5) 3
(b – 5) 3 = b 3 + 3 (b) 2 * (-5) + 3 (b) * (25) -125
(b – 5) 3 = b 3 - 15b 2 + 75b – 125.
Cube of a trinomial
It's multiplied by its square. It's a very extensive product, because there are three terms cubed, plus three times each term squared, multiplied by each of the terms, plus six times the product of the three terms. A better way to look at it is:
(a+b+c) 3 = (a+b+c) * (a+b+c) 2
(a+b+c) 3 = (a+b+c) * (a 2 + b 2 + c 2 + 2ab + 2ac + 2bc)
(a+b+c) 3 = A 3 + b 3 + c 3 + 3a 2 b+3ab 2 + 3a 2 c + 3ac 2 +3b 2 c+3bc 2 + 6abc.
1 Example
Solved exercises on notable products
1 Exercise
Develop the following binomial for the cube: (4x – 6) 3 .
Solution
Remembering that a binomial for the cube is equal to the first term cubed, minus three times the square of the first term by the second; plus three times the first term, for the second square, minus the cube of the second term.
(4x - 6) 3 = (4x) 3 – 3 (4x) 2 (6) + 3 (4x) * (6) 2 - (6) 2
(4x - 6) 3 = 64x 3 – 3 (16x 2 ) (6) + 3 (4x) * (36) - 36
(4x - 6) 3 = 64x 3 - 288x 2 +432x – 36.
2 Exercise
Develop the following binomial: (x + 3) (x + 8).
Solution
There is a binomial in which there is a common term, which is x, and the second term is positive. To develop it, simply square the common term, plus the sum of the non-common terms (3 and 8), and then multiply them by the common term, plus the sum of the multiplication of the non-common terms.
(x + 3) (x + 8) = x 2 + (3 + 8) x + (3 * 8)
(x + 3) (x + 8) = x 2 +11x +24.
References
- Angel, AR (2007). Elementary Algebra Education at Pearson.
- Arthur Goodman, L.H. (1996). Algebra and trigonometry with analytic geometry. Pearson Education.
- Das, S. (n.d.). Maths Plus 8. United Kingdom: Ratna Sagar.
- Jerome E. Kaufmann, K. L. (2011). Elementary and Intermediate Algebra: A Combined Approach. Florida: Cengage Learning.
- Pérez, C. D. (2010). Pearson Education.