Notable Products: Algebra Explained Step-by-Step

You notable products they receive this nomenclature because they need attention. I wonder why? Simply because they make calculations easier, reduce resolution time and speed up learning.

Back in the past, the Greeks used procedures. algebraic and geometric exactly the same as modern remarkable products. At. Euclid's work of Alexandria, Elements, the remarkable products were. used and recorded in the form of geometric representations.

In algebra, polynomials appear quite frequently and can be called remarkable products. In this article we will learn a little about some algebraic operations often associated with notable products, such as the square of the sum of two terms, o square of the difference of two terms, the product of the sum by the difference of two terms, the cube of the sum of two terms, and finally the cube of the difference of two terms.

See too: Roman numbers.

Expert. math comments on notable products

Also according to the explanation of Naysa Oliveira, graduating from. Mathematics, the remarkable products present five distinct cases. According to her, before we understand what remarkable products are, we must know what they are. algebraic expressions, that is, equations that have letters and numbers.

See some examples:

2x + 3 = 4

-y + 2x + 1 = 0

z2 + ax + 2y = 3

Notable products have general formulas, which, on their own. instead, they are the simplification of algebraic products. Look:

(x + 2). (x + 2) =

(y – 3). (y – 3) =

(z + 4). (z – 4) =

Five cases of Notable Products

There are five distinct cases of notable products, namely:

First Case: Square of the sum of two terms.

square = exponent 2;

Sum of two terms = a + b;

Hence, the square of the sum of two terms is: (a + b) 2

Making the product of the square of the sum, we obtain:

(a + b) 2 = (a + b). (a + b) = a2 + a. b + a. b + b2 = a2. + 2. The. b+b2

All this expression, when reduced, forms the product. remarkable, which is given by:

(a + b) 2 = a2 + 2. The. b+b2

Thus, the square of the sum of two terms is equal to. square of the first term, plus twice the first term by the second, plus. the square of the second term.

Examples:

(2 + a) 2 = 22 + 2. 2. a + a2 = 4 + 4. a + a2

(3x + y) 2 = (3 x) 2 + 2. 3x. y + y2 = 9×2 +6. x. y + y2

Second Case: Square. of the difference of two terms.

Square = exponent 2;

Difference of two terms = a – b;

Hence, the square of the difference of two terms is: (a – b) 2.

We will carry the products through the property. distributive:

(a – b) 2 = (a – b). (a – b) = a2 – a. b - a. b + b2 = a2. – 2nd. b+b2

Reducing this expression, we get the remarkable product:

(a – b) 2 = a2 – 2 .a. b+b2

So we have what the square of the difference of two terms is. equal to the square of the first term, minus twice the first term by. second, plus the square of the second term.

Examples:

(a – 5c) 2 = a2 – 2. The. 5c + (5c) 2 = a2 – 10. The. c + 25c2

(p - 2s) = p2 - 2. P. 2s + (2s) 2 = p2 - 4. P. s + 4s2

Third Case: Product. of the sum by the difference of two terms.

Product = multiplication operation;

Sum of two terms = a + b;

Difference of two terms = a – b;

The product of the sum and the difference of two terms is: (a + b). (a - b)

Solving the product of (a + b). (a – b), we obtain:

(a + b). (a – b) = a2 – ab + ab – b2 = a2 + 0 + b2 = a2 – b2

Reducing the expression, we get the remarkable product:

(a + b). (a - b) = a2 - b2

We can therefore conclude that the product of the sum by the. difference of two terms is equal to the square of the first term minus the square. of the second term.

Examples:

(2 - c). (2 + c) = 22 - c2 = 4 - c2

(3×2 – 1). (3×2 + 1) = (3×2)2 – 12 =9×4 – 1

Fourth case: Cube. of the sum of two terms

Cube = exponent 3;

Sum of two terms = a + b;

Hence, the cube of the sum of two terms is: (a + b) 3

Making the product through the distributive property, we obtain:

(a + b) 3 = (a + b). (a + b). (a + b) = (a2 + a. b + a. B. + b2). (a + b) = (a2 + 2. The. b+b2). ( a + b ) = a3 +2. a2. b + a. b2. + a2. b + 2. The. b2 + b3 = a3 +3. a2. b + 3. The. b2 + b3

Reducing the expression, we get the remarkable product:

(a + b) 3 = a3 + 3. a2. b + 3. The. b2 + b3

The cube of the sum of two terms is given by the cube of the first, plus three times the first term squared by the second term, plus three. times the first term by the second squared, plus the cube of the second term.

Examples

(3c + 2a) 3 = (3c) 3 + 3. (3c) 2 .2a + 3. 3c. (2a) 2 + (2a) 3 = 27c3 + 54. c2. to +36. ç. a2 + 8a3

Fifth case: Cube of the. two-term difference

Cube = exponent 3;

Difference of two terms = a – b;

Hence, the cube of the difference of two terms is: ( a – b )3.

Making the products, we obtain:

(a – b) 3 = (a – b). (a – b). (a – b) = (a2 – a. b - a. B. + b2). (a – b) = (a2 – 2. The. b + b2). (a – b) =a3 – 2. a2. b + a. b2 – a2. b + 2. The. b2 – b3 = a3 – 3. a2. b + 3. The. b2 - b3

Reducing the expression, we get the remarkable product:

(a – b) 3 = a3 – 3. a2. b + 3. The. b2 - b3

The cube of the difference of two terms is given by the cube of. first, minus three times the first term squared for the second term, plus three times the first term for the second squared, minus the cube of. second term.

Example:

(x – 2y) 3 = x3 – 3. x2. 2y + 3. x. (2y) 2 - (2y) 3 =x3 - 6. x2. y + 12. x. y2 – 8y3

Tips to Solve Some Notable Products

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