Polynomials and Types of Equations PDF

Mathematics Chapter 2 Polynomials and Types of Equations Batterjee Medical College For Sciences & Technology 1st Semester 2023/24 Mathematics Department Chapter Outlines  Variables and Constants  Algebraic Expressions and Polynomials  Classification of Po...

Mathematics Chapter 2 Polynomials and Types of Equations Batterjee Medical College For Sciences & Technology 1st Semester 2023/24 Mathematics Department Chapter Outlines  Variables and Constants  Algebraic Expressions and Polynomials  Classification of Polynomials  Evaluation of Polynomials / Zero of a Polynomial  Linear Equations in one variable  Consecutive Integers  Quadratic Equations  Self-Evaluation Variables and Constants A variable is literal number which can have different values whereas a constant has a fixed value. In algebra, we usually denote constants by a, b, c and variables by x, y, z. All the real numbers are constants 5 6, −3.5, , 15, … …. 11 𝟏 While 𝟒𝒙, 𝒚, 𝟗𝒓 , ….. contain unknown x, y and r respectively and therefore 𝟐 do not have fixed values. Their values depend on x, y and r respectively. Therefore, x, y and z are variables. Algebraic Expressions and Polynomials An algebraic expression is a combination of numbers, variables and arithmetical operations. Signs + or – separate an algebraic expression into several parts. Each part along with its sign is called a term of the expression. Often, the plus sign of the first term is omitted in writing an algebraic expression. For example, we write 2𝑥 − 5𝑦 + 6 Here 𝟐𝒙, −𝟓𝒚 and 6 are the three terms of the expression. Algebraic Expressions and Polynomials The coefficients are the numbers in front of the letters. In writing 𝟐𝒙, the number 2 is called the coefficient of 𝒙. 𝟐 2 The coefficient of 𝒚 is in 𝑦. 𝟑 3 If there is no number written with the variable, then it will be 1. For example the coefficient of 𝑥 is 1. Algebraic Expressions and Polynomials An algebraic expression is called a polynomial, in which a) No variables in the denominator, b) Exponents of variables are whole numbers c) All Coefficients are real numbers, Examples of Polynomials 1 2 4 3 3 2 𝑥 − 3𝑦, 𝑥, −3𝑥 + 2𝑥 − 1, 2𝑥 𝑦 − 9𝑥 𝑦 + 6𝑥𝑦 + 8 3 NOT Polynomials 1 −2 2 2 3𝑥 + 4, + 𝑥 − 7, 2 𝑥 − 5𝑦 + 𝑧, 𝑥 +1 4 𝑥 Algebraic Expressions and Polynomials A polynomial can have many variables 2 −3𝑥 + 2𝑥 − 1 one variable 4 3 3 2 2𝑥 𝑦 − 9𝑥 𝑦 + 6𝑥𝑦 + 8 two variables 2 2 2 3 2 2𝑥 𝑦 + 3𝑥 𝑦 𝑧 − 8𝑥𝑦𝑧 + 2 three variables Classification of Polynomials A polynomial may have many terms and many variables. Every term has a degree. The degree of the term is defined as: The sum of the exponents of the variables in a term is called the degree of that term. Examples (Term Degree) Look at the following polynomial. 𝟐 𝟐 𝟐 𝟑 𝟐𝒙 𝒚 + 𝟑𝒙 𝒚 − 𝟖𝒙𝒚 + 𝟐 It has two variables and four terms. The Degree of the: First term 𝟐 2𝑥 𝑦 𝟐 is 2+2=4 𝟐 𝟑 Second term 3𝑥 𝑦 is 2+3=5 𝟏 𝟏 Third term 8𝑥𝑦 = −8𝑥 𝑦 is 1 + 1 =2 𝟎 Fourth term 2 = 2𝑥 is 0 Classification of Polynomials Look at the following polynomial. 𝟒 𝟑 𝟓 𝟐 𝟑𝒙 − 𝟐𝒙 + 𝒙 − 𝟖𝒙 − 𝟐 𝟔 It has One variable and five terms. The Degree of the First term is 4 The Degree of the Second term is 3 The Degree of the Third term is 2 The Degree of the Fourth term is 1 The Degree of the Fifth term is 0 Classification of Polynomials Every term in a polynomial has a degree. The highest of all the term degrees is called as the degree of that polynomial. “the highest term degree” In the above examples, The degree of the polynomial 2 2𝑥 𝑦 2 𝟐 + 𝟑𝒙 𝒚𝟑 − 8𝑥𝑦 + 2 is 5 Note: All the non-zero constants have degree “0” Classification of Polynomials A polynomial of degree one is called a linear polynomial. e.g. 5𝑥 + 4, 𝑥 + 1, 2𝑥 − 3, ….. A polynomial of degree two is called a quadratic polynomial. e.g. 3𝑥 2 − 5𝑥 + 4, 𝑥 2 −𝑥+1 A polynomial of degree three is called a cubic polynomial. e.g. 3 2 2𝑥 − 𝑥 + 𝑥 − 3 Classification of Polynomials (1) Monomial: A polynomial consisting of only one term: 5𝑥, 2 2𝑥 , 6 −3𝑥 , 8, 3 5𝑥 𝑦 8 (2) Binomial: A polynomial consisting of two terms: 1 3 2 2 𝑥 − 1, 3𝑦 − , 2𝑥 + 3𝑥 , 2𝑥𝑦 +4 2 (3) Trinomial: A polynomial consisting of three terms: 2 3 2 𝑥 + 𝑦 + 2, 3𝑥 − 5𝑥 + 4, 2𝑥 + 3𝑥 + 9 Evaluation of a Polynomial We can evaluate a polynomial for given value of the variable occurring in it. Example: Find the value of the polynomial 𝑃 𝑥 = 3𝑥 2 +𝑥−5 for 𝑥 = 2. Solution: Replace the variable x by its value and evaluate the answer. 2 𝑃 𝑥 = 3𝑥 + 𝑥 − 5 2 𝑃 2 =3 𝟐 + (𝟐) − 5 = 12 + 2 − 5 =9 Zero of a Polynomial The value(s) of the variable for which the value of a polynomial in one variable is zero is (are) called zero(s) of the polynomial. For example, if we replace 𝒙 by 1 or 2 in the following polynomial 2 𝑃 𝑥 = 𝑥 − 3𝑥 + 2 We get zero. 𝑃 𝑥 = 𝑥 2 − 3𝑥 + 2 𝑃 𝟏 = 2 (𝟏) −3 𝟏 +2=𝟎 2 𝑃 𝟐 = (𝟐) −3 𝟐 +2=𝟎 Thus, 𝒙 = 𝟏, 𝟐 are the zeroes of the above polynomial. Linear Equations in one variable Example: Solve the following equation: 2𝑥 − 3 = 13 Solution: 2𝑥 = 13 + 𝟑 2𝑥 = 16 𝑥=8 Linear Equations in one variable In the previous example, we used the following principle. The first technique for solving linear equation in one variable is the Addition- Subtraction Rule which states: If c is any real number, the following equations are equivalent. (They have the same solution set) 𝒂=𝒃 𝒂±𝒄=𝒃±𝒄 The second line states that adding or subtracting the same number to BOTH sides of an equation will not change the solution set of that equation. Linear Equations in one variable If c is any real number, the following equations are equivalent. (They have the same solution set.) 𝑎=𝑏 𝒂×𝒄=𝒃×𝒄 𝒂 𝒃 = ,𝒄 ≠ 𝟎 𝒄 𝒄 If we multiply or divide both sides of an equation by any real number, it does not affect the solution of the equation. Linear Equations in one variable Example: Solve the following equations: (1) 𝒙 − 𝟓 = 𝟏𝟐 Solution: 𝒙 − 𝟓 + 𝟓 = 𝟏𝟐 + 𝟓 𝒙 = 𝟏𝟕 (2) 𝒙 + 𝟐𝟓 = 𝟏𝟑 Solution: 𝒙 + 𝟐𝟓 − 𝟐𝟓 = 𝟏𝟑 − 𝟐𝟓 𝒙 = −𝟏𝟐 Linear Equations in one variable Example: Solve the following equation: 𝑥 = −4 5 Solution: 𝑥 × 𝟓 = −4 × 𝟓 5 𝑥 = −20 Linear Equations in one variable Example: Solve the following equation: 10𝑥 + 9 = 3𝑥 − 5 Solution: 10𝑥 − 3𝑥 = −5 − 9 7𝑥 = −14 𝑥 = −2 Consecutive Integers Link one variable to another using given information. Consecutive Integers, Consecutive Odd Integers or Consecutive Even Integers The condition of being consecutive allows us to get the necessary relationships among the variables to be able to get one equation in one unknown. The three relations are... Consecutive Integers 1, 2, 3, 4, 5, 6, 7,... the difference is 1 Let n = first consecutive integer n + 1 = second consecutive integer n + 2 = third consecutive integer and so on. Consecutive Integers Find four consecutive integers whose sum is 66. Solution: Let n = first consecutive integer Sum = 66 n+1 = second consecutive integer 𝑛 + 𝑛 + 1 + 𝑛 + 2 + (𝑛 + 3) = 66 n+2 = third consecutive integer n+3 = fourth consecutive integer 4𝑛 + 6 = 66 Now the remaining integers will be 4𝑛 = 60 n + 1 = 15 + 1 = 16 𝒏 = 𝟏𝟓 n + 2 = 15 + 2 = 17 n + 3 = 15 + 3 = 18 Thus the four consecutive integers whose sum is 66 are 15, 16, 17 and 18. Consecutive Integers 2, 4, 6, 8, 10, 12... the difference is 2 Let n = first consecutive even integer n+2 = second consecutive even integer n+4 = third consecutive even integer etc. Consecutive Integers Find three consecutive even integers whose sum is 162. Solution: Let Sum = 162 n = first consecutive even integer 𝑛 + 𝑛 + 2 + 𝑛 + 4 = 162 n+2 = second consecutive even integer n+4 = third consecutive even integer 3𝑛 + 6 = 162 Now the remaining even integers will be 3𝑛 = 156 n + 2 = 52 + 2 = 54 𝒏 = 𝟓𝟐 and n + 4 = 52 + 4 = 56 Thus the three consecutive even integers whose sum is 162 are 52, 54 and 56. Consecutive Integers 1, 3, 5, 7, 9, 11, 13, … the difference is 2 Let n = first consecutive odd integer n+2 = second consecutive odd integer n+4 = third consecutive odd integer etc. Consecutive Integers Find three consecutive odd integers whose sum is 159. Solution: Let Sum = 159 n = first consecutive odd integer 𝑛 + 𝑛 + 2 + 𝑛 + 4 = 159 n+2 = second consecutive odd integer n+4 = third consecutive odd integer 3𝑛 + 6 = 159 Now the remaining odd integers will be 3𝑛 = 153 n + 2 = 51 + 2 = 53 𝒏 = 𝟓𝟏 and n + 4 = 51 + 4 = 55 Thus the three consecutive odd integers whose sum is 159 are 51, 53 and 55. Quadratic Equations The general form of a quadratic equation in one variable is: 𝟐 𝒂𝒙 + 𝒃𝒙 + 𝒄 = 𝟎 Where a, b and c are real numbers and a is not zero because if 𝒂 = 𝟎, then the equation is not quadratic. 𝒃𝒙 + 𝒄 = 𝟎 Linear Unlike linear equations which have exactly one solution (in almost all cases), quadratic equations will have 0, 1 or 2 real solutions. Quadratic Equations Many techniques can be used for solving quadratic equations. The most common is the quadratic formula. The quadratic formula will allow us to find the real solutions of all quadratic equations. We use the following formula to find the solution/s of a quadratic equation in one variable. −𝒃± 𝟐 𝒃 −𝟒𝒂𝒄 𝒙= 𝟐𝒂 Quadratic Equations 2 Example: Solve: 2𝑥 − 3𝑥 + 1 = 0 Solution: Here 𝑎 = 2, 𝑏 = −3, 𝑐 = 1 Putting the values in the formula 𝟐 −(−3) ± (−𝟑) − 𝟒(𝟐)(𝟏) 𝑥= 2(2) 3± 𝟗−𝟖 3± 𝟏 3±1 𝑥= = = 4 4 4 3+1 3−1 𝑥= 𝑜𝑟 𝑥= 4 4 𝟏 𝒙 = 𝟏 𝑜𝑟 𝒙 = 𝟐 Here we get two real solutions. Quadratic Equations 2 Example: Solve: 𝑥 + 2𝑥 + 1 = 0 Solution: Here 𝑎 = 1, 𝑏 = 2, 𝑐 = 1 Putting the values in the formula 𝟐 −(2) ± (𝟐) − 𝟒(𝟏)(𝟏) 𝑥= 2(1) −2 ± 𝟒 − 𝟒 −2 ± 𝟎 −2 ± 0 𝑥= = = 2 2 2 −2 + 0 −2 − 0 𝑥= 𝑜𝑟 𝑥= 2 2 𝒙 = −𝟏 𝑜𝑟 𝒙 = −𝟏 In this example, we get only one real solution. Quadratic Equations 2 Example: Solve: 2𝑥 − 2𝑥 + 5 = 0 Solution: Here 𝑎 = 2, 𝑏 = −2, 𝑐 = 5 Putting the values in the formula −(−2) ± (−𝟐) 𝟐 − 𝟒(𝟐)(𝟓) 𝑥= 2(2) 2 ± 𝟒 − 𝟒𝟎 2 ± −𝟑𝟔 𝑥= = 4 4 Since −36 has no real answer, so in this example, we do not get any real solution. Self-Evaluation (1) Choose the correct answer: i) Find 2 consecutive EVEN integers whose sum is 34. A) 14, 20 B) 15, 19 C) 16, 18 D) None of the above ii) Find 4 consecutive integers whose sum is 46. A) 10, 11, 12, 13 B) 11, 12, 13, 14 C) 12, 13, 14, 15 D) None of the above Self-Evaluation 2) Solve the following: a) 3𝑥 + 6 = 4𝑥 − 2 Ans. 8 b) 2𝑥 − 3 𝑥 + 2 = 0 Ans. -2, 3/2 or 1.5 c) 2𝑥 2 − 5𝑥 + 2 = 0. Ans. 2, ½ or 0.5 3) How many solutions does the following equations have? 2 (1) 4𝑥 − 12𝑥 + 9 = 0 Ans. One solution 2 (2) 𝑥 − 𝑥 + 12 = 0. Ans. No solution End of Chapter 2