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In the study of algebraic calculus we learned how to operate polynomials, do their factorization and find their mmc. And with this information, it is possible to make some demonstrations such as:

• The sum of two consecutive whole numbers will always be the difference of their squares.

Consider x to be any integer, its successor can be represented by the polynomial x + 1. Adding these two polynomials we will arrive at the following algebraic expression:

x + (x + 1) = x + x + 1 = 2x + 1

The difference of the squares of these two consecutive numbers will be represented by the following algebraic expression:

(x+1)^{2} - x^{2} = (x^{2} + 2x + 1) - x^{2} = x^{2} + 2x + 1 -x^{2} = 2x + 1

Comparing the two algebraic expressions found, we can confirm that

x + (x + 1) = (x +1)^{2} - x^{2}

• The sum of five consecutive integers will always be a multiple of 5.

Consider the polynomials as five consecutive integers: x-2; x-1; x; x + 1; x + 2.

A number to be a multiple of five can be written as follows: 5x, where x is any integer, that is, any number that multiplied by 5 will be a multiple of five.

Adding the five consecutive numbers we will have:

x - 2 + x - 1 + x + x + 1 + x + 2 = 5x -3 + 3 = 5x, so it is true to say that the sum of 5 consecutive integers will have a multiple of 5.

• The sum of two odd integers will always be an even number.

For a number to be even, it must be written as follows: 2x, where x represents any integer. So an odd number would equal 2x +1.

Adding two odd numbers would be the same as:

(2x +1) + (2x + 1) = 2 (2x + 1). The algebraic expression (2x + 1) will have a numeric value equal to any integer, when multiplied by 2 (2x + 1) will result in an even number.

by Danielle de Miranda

Graduated in Mathematics

Brazil School Team

**Polynomial - Math - Brazil School**

**Source:** Brazil School - https://brasilescola.uol.com.br/matematica/demonstracoes-atraves-calculo-algebrico.htm