Mathematics is made up of a set of branches, each specialized in a specific area:
- Arithmetic referring to processes and operations involving rational and integer numbers, powers, radicals, and logarithms.
- Algebra processes of polynomials, equations, systems of equations, inequalities and matrices.
- Analysis, which deals with functions, construction of graphs, derivatives, integrals and integration methods.
- Geometry study of figures, trigonometry, vectors and lines in the plane and space. Statistics study of discrete and continuous variables, regression lines, sampling, probability, mode, mean and median, deviation.
Algebra contemplates the use of numbers, signs and letters to carry out mathematical processes and operations, which over the centuries have been complemented with procedures such as Ruffini’s Rule. Here we will explain what it is about and the different variables, processes and concepts that this topic includes, which can sometimes be complicated but with care you will be able to understand.
What do you need:
We will show you some examples with mathematical exercises that you will need to develop once explained, so you will need:
- Paper.
- Pencil.
- Eraser to practice what is shown.
- Quite patience and desire.
So with a lot of calm and confidence let’s start fully with the subject.
Instructions
Paolo Ruffini Italian mathematician and physician born in 1765 and died in 1822, creator of Ruffini’s Rule, designed this method to perform the divisive calculation between a polynomial and a binomial, to obtain the resolution of equations of third degree, fourth degree and fifth degree, Calculate roots of polynomials of degree 3 and up and the factorization of polynomials of third degree or higher.
Let’s start with some concepts that will progressively lead us to understand everything related to Ruffini’s Rule.
What are polynomials?
- Polynomials: They are mathematical expressions made up of whole numbers and fractions that represent constants, variables that are usually represented with the letters “X”, “Y” or “Z” and exponents (only positive integers). They must be formed by finite terms that can include one or more of the three aforementioned elements (constants, variables and exponents) and within the expression they are separated by addition and subtraction operations. To carry out operations with polynomials, these must be simplified so that the terms that comprise them are grouped with their same variables and constants. Observe the following polynomial. Example No. 1: 5x + 4y + 4xy + 3y +1; The terms X, Y, XY and 1 are identified, having to group equal terms simplifying the expression: 5x = 5x; 4y + 3y = 7y ; 4xy = 4xy ; 1 = 1. NOTE: Care must be taken when in the grouping of equal terms, there are terms with different signs and the addition is made to simplify. Remaining then the polynomial totally grouped in similar terms: 5x + 7y + 4xy + 1; Let’s see another example this time with like terms but with opposite signs: 27x + 12y – 31x – 7y + 6x + 8 – 2 – 3; We separate all the terms and add them separately: X terms: 27 + 6 – 31 = 2x; Y terms: 12 – 7 = 5y; Constant terms: 8 – 2 – 3 = 3. Finally, after separating and adding all the similar terms, we organize them again: 2x + 5y + 3. Let’s see another example this time with like terms but with opposite signs: 27x + 12y – 31x – 7y + 6x + 8 – 2 – 3; We separate all the terms and add them separately: X terms: 27 + 6 – 31 = 2x; Y terms: 12 – 7 = 5y; Constant terms: 8 – 2 – 3 = 3. Finally, after separating and adding all the similar terms, we organize them again: 2x + 5y + 3. Let’s see another example this time with like terms but with opposite signs: 27x + 12y – 31x – 7y + 6x + 8 – 2 – 3; We separate all the terms and add them separately: X terms: 27 + 6 – 31 = 2x; Y terms: 12 – 7 = 5y; Constant terms: 8 – 2 – 3 = 3. Finally, after separating and adding all the similar terms, we organize them again: 2x + 5y + 3.
- Types of Polynomials: First we are going to detail the structure of the minimum expression of a polynomial that is the monomial: A monomial is made up of 3 parts: The coefficient: which is given by a numerical value that multiplies the next element of the monomial; Literal value: representative letter of a quantity; The Exponent: represents the value of the times that the coefficient and the literal value must be multiplied by themselves to find the real monomial value. Example No. 2: 6×3. Where: The coefficient is the number “6”. The literal value is the letter “X”. The exponent the number “3” also called, index or power, which for values with exponent “2” is usually called “Squared” and indicates that the base must be multiplied 2 times by itself, values with exponent “3” are They can be called “Al Cubo” which implies multiplying the base 3 times by its same value. Monomials. 7xy; Polynomials made up of two terms: Binomial. 7xy–5y; Polynomials formed by three terms: trinomials. 7xy – 5y + 4; We can conclude that a polynomial is the mathematical expression formed by 3 or more terms.
- Degrees of a polynomial: The equationsof polynomials are made up of monomials, binomials, trinomials that indicate the type of polynomial with which we are working, now we also have the concept of Degrees of a polynomial. To explain this definition once again we go from understanding a simple expression to a more complex one: Degree of a Monomial: the degree of a monomial is obtained by adding all the values of the exponents that make up the monomial, including the exponents of all literal values if more than one occurs. Example No. 3. Calculation of the degree of the following Monomial: 8x2y3z. The values of the exponents of the literal terms X, Y, Z are added; 2 + 3+ 1 = 4. Resulting in a monomial of degree 4. Note that the literal “Z” does not have any value written, being that for these cases it is understood that the value in question is equal to 1. To determine the degree of a polynomial we must analyze and calculate the degree of each of the monomials that make it up and the monomial with the highest degree will indicate the degree of the polynomial in general. Example No. 4. Calculation of the degree for the following Trinomial: 6x + 4xy + 7x2y. We determine separately the degree of each monomial. The 6x monomial: the term X has exponent value 1, it is degree 1. The 4xy monomial: the terms X and Y add their exponents 1 + 1 resulting in a monomial of degree 2. The 7x2y monomial: the terms X and Y have exponents 2 + 1 resulting in a degree 3 monomial. Having as a result that the polynomial 6x + 4y + 7x2y is of degree 3. To determine the degree of a polynomial we must analyze and calculate the degree of each of the monomials that make it up and the monomial with the highest degree will indicate the degree of the polynomial in general. Example No. 4. Calculation of the degree for the following Trinomial: 6x + 4xy + 7x2y. We determine separately the degree of each monomial. The 6x monomial: the term X has exponent value 1, it is degree 1. The 4xy monomial: the terms X and Y add their exponents 1 + 1 resulting in a monomial of degree 2. The 7x2y monomial: the terms X and Y have exponents 2 + 1 resulting in a degree 3 monomial. Having as a result that the polynomial 6x + 4y + 7x2y is of degree 3. To determine the degree of a polynomial we must analyze and calculate the degree of each of the monomials that make it up and the monomial with the highest degree will indicate the degree of the polynomial in general. Example No. 4. Calculation of the degree for the following Trinomial: 6x + 4xy + 7x2y. We determine separately the degree of each monomial. The 6x monomial: the term X has exponent value 1, it is degree 1. The 4xy monomial: the terms X and Y add their exponents 1 + 1 resulting in a monomial of degree 2. The 7x2y monomial: the terms X and Y have exponents 2 + 1 resulting in a degree 3 monomial. Having as a result that the polynomial 6x + 4y + 7x2y is of degree 3. We determine separately the degree of each monomial. The 6x monomial: the term X has exponent value 1, it is degree 1. The 4xy monomial: the terms X and Y add their exponents 1 + 1 resulting in a monomial of degree 2. The 7x2y monomial: the terms X and Y have exponents 2 + 1 resulting in a degree 3 monomial. Having as a result that the polynomial 6x + 4y + 7x2y is of degree 3. We determine separately the degree of each monomial. The 6x monomial: the term X has exponent value 1, it is degree 1. The 4xy monomial: the terms X and Y add their exponents 1 + 1 resulting in a monomial of degree 2. The 7x2y monomial: the terms X and Y have exponents 2 + 1 resulting in a degree 3 monomial. Having as a result that the polynomial 6x + 4y + 7x2y is of degree 3.
- Ruffini’s Rule: Ruffini‘s Rule allows us to solve equations greater than the second degree for when these expressions have integer solutions, for complex or real solutions this method is not feasible. Among the methods for solving equationswe have: Reduction Method; substitution method; Equalization Method. To solve quadratic equations, it is previously required to carry out a factorization in which we will use the Ruffini Method. Step 1: In the following equation we will identify the numbers (coefficients) that go before each unknown. Example No. 4: X3 + 2×2 – x – 2 = 0; 1 + 2 – 1 – 2 = 0. Step 2: Two lines are drawn, two perpendicular lines that will be the “format” where we will develop the procedures. Step 3: The coefficients are placed according to their degree in descending order. 1 2 – 1 – 2. The coefficients were placed in order of the degree that x3 , x2, x, 0 have, if for the case in which there is no progressive correlation between the degrees, that is, for example, the absence of X2 this value with number 2, it would be replaced with the number 0. Step 4: With this rule we can factor polynomials into binomials of the form (X – 1), (X + 1), in this case we will start by placing the number 1 on the horizontal line that would represent the binomial (X – 1). For explanatory reasons we will call this value 1, a multiplicative term since we will use it in throughout the procedure. Step 5: We perform an addition operation in the first column, then we multiply the result of the addition of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them. With this rule we can factor polynomials into binomials of the form (X – 1), (X + 1), in this case we will start by placing the number 1 on the horizontal line that would represent the binomial (X – 1). For explanatory reasons we will call this value 1, a multiplicative term since we will use it in throughout the procedure. Step 5: We perform an addition operation in the first column, then we multiply the result of the addition of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them. With this rule we can factor polynomials into binomials of the form (X – 1), (X + 1), in this case we will start by placing the number 1 on the horizontal line that would represent the binomial (X – 1). For explanatory reasons we will call this value 1, a multiplicative term since we will use it in throughout the procedure. Step 5: We perform an addition operation in the first column, then we multiply the result of the addition of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them. (X + 1), in this case we will start by placing the number 1 on the horizontal line that would represent the binomial (X – 1). For explanatory reasons we will call this value 1, a multiplicative term since we will use it in throughout the procedure. Step 5: We perform an addition operation in the first column, then we multiply the result of the addition of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them. (X + 1), in this case we will start by placing the number 1 on the horizontal line that would represent the binomial (X – 1). For explanatory reasons we will call this value 1, a multiplicative term since we will use it in throughout the procedure. Step 5: We perform an addition operation in the first column, then we multiply the result of the addition of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them. For explanatory reasons we will call this value 1, a multiplicative term since we will use it in throughout the procedure. Step 5: We perform an addition operation in the first column, then we multiply the result of the addition of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them. For explanatory reasons we will call this value 1, a multiplicative term since we will use it in throughout the procedure. Step 5: We perform an addition operation in the first column, then we multiply the result of the addition of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them. then we multiply the result of the sum of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them. then we multiply the result of the sum of the first column by the multiplicative term (1) and the result will be written in the second column. Step 6: The second column is added and we repeat the same procedure of multiplying the multiplicative term with the result of the sum of column no. 2 and placing the value obtained in the third column. We repeat this procedure until we finish all of them.
- The objective is that at the end of the last column we have the value of 0 as a result of the summation. If this value is not obtained, the “multiplicative term” must be changed from 1 to – 1, to comply with the binomial rule (x – 1) (x + 1), and so on until reaching the final value of 0 in the last column. The new resulting equation is the following: X2 + 3x + 2. As the result of the last column is a value of 0, the term to its left will have a zero degree and the degrees will increase progressively to the left. Step 7: We must continue factoring successively with the values 1, -1, 2, -2 until the equation is reduced to the minimum, remembering that we will change the value when the final result is different from 0. In this case we will test directly with the value -2, since trying again with the values 1 and + 1 the result is different from 0. The product of the last row is the result of the factorization using the Ruffini method. Where: The multiplicand term “1” represents the binomial x – 1. The multiplicand “-2” represents the binomial x – (-2) = x + 2. The final value 1 the binomial x + 1. The resulting equation is the following: X3 + 2×2 – x – 2 = (x + 1) (x + 2) (x – 1).
Tips
Ruffini’s rule is a mathematical tool for factoring polynomials in order to simplify them, it is a procedure that requires organization and order at the time of writing it. At the same time, we must have clear concepts such as clearing up equations, types of characteristic polynomials and, of course, a lot of patience, since it is a process that can be long and frustrating.
The key is to practice it over and over again, at first it may seem long and it will require a lot of detail and concentration, but over time we will be able to develop the technique of even being able to predict the resulting equations, just by looking at the original expression, choosing a priori the correct pairing to start the development of the exercise.