MMW Finals PDF

CHAPTER I POLYNOMIALS OBJECTIVES: 1. Recall the fundamental operations for signed numbers; 2. Apply the basic rules and concept on rational expressions; and 3. Understand the basic operations polynomials. KEY/CONCEPTS: ALGEBRA is a branch...

CHAPTER I POLYNOMIALS OBJECTIVES: 1. Recall the fundamental operations for signed numbers; 2. Apply the basic rules and concept on rational expressions; and 3. Understand the basic operations polynomials. KEY/CONCEPTS: ALGEBRA is a branch of mathematics that uses mathematical statements to describe relationships between things that vary. These variables include things like the relationship between supply of an object and its price. When we use a mathematical statement to describe a relationship, we often use letters to represent the quantity that varies, since it is not a fixed amount. It is the use of letters and symbols to represent values and their relations, especially for solving equations. The combination of each letters and symbols are called “Algebraic Expressions”. The Real Number System SET A set is a collection of unique objects such as set of positive numbers, a set of drawing, etc. Each of the unique objects making up a set is called an ELEMENT of that set. A set is a given a name, usually an uppercase letter, while their elements by lower-case letter. There are two ways of writing sets: the roster form and the set-builder form. To indicate that an object is an element of a set, the symbol “ϵ”, read as is the element, is used. Thus , if a is an element of set A, we indicate this by writing “a ϵ A”, read as “a is an element of the set A”. Examples: Set of positive numbers Set of counting numbers less than ten Set of vowels in the English Alphabet Examples:  A. Write each set described below in roster form 1. Set A consist of the vowels of the English Alphabet. 2. Set B consist of the founding members of the ASEAN. 3. Set C consist of numbers less than 10. B. Write each set listed below in set-builder form 1. D = {1, 3, 5, 7, 9, 11, 13, 15,…….} 2. E = {+, -, x, ÷ } 3. F = {millimetre, centimetre, meter, kilometre } Types of Numbers The following diagram shows the types of numbers that form the set of real numbers Irrational Numbers Rational Numbers Integers Whole Numbers Natural Numbers The natural numbers are the numbers used for counting. It is a prime number if it is greater than 1 and its only factors are 1 and itself. It is a composite number if it is greater than 1 and it is not prime. The whole numbers are the natural numbers and zero. The integers are all the whole numbers and their additive inverse. No fractions or decimals - an integer is even if it can be written in the form 2ⴄ, where ⴄ is an integer (if 2 is a factor). - an integer is odd if it can be written in the form 2ⴄ - 1, where ⴄ is an integer(if 2 is not a factor). The rational numbers are the numbers that can be written as the ratio of two integers. All rational numbers when written in their equivalent decimal form will have terminating or repeating decimals. The irrational numbers are any real numbers that can not be represented as a ratio of two integers. The numbers usually are imperfect roots. Pi is also an irrational number. Irrational numbers when written in their equivalent decimal form have non-terminating and non-repeating decimals. The square root of a prime number is irrational. A real number is either a rational or an irrational number. A real number is positive if it is greater than zero, negative if it is less than zero.  𝑘 Undefined numbers are numbers in the form. 0 OPERATION OF SIGNED NUMBERS Rules in Adding Integers 1. In adding integers with the same sign, add their absolute values then affix to the sum the common sign. Example: a. 27 + 41 = 68 b. – 39 + (- 56) = -95 2. In adding integers with different signs, subtract their absolute values then affix to the difference the sign of the added having the greater absolute value. Example: a. -65 + 87 = 22 b. 79 + (-90) = -11 Rules in Subtracting Integers In subtracting an integer from another integer, change the sign of the subtrahend and perform the rules for addition. Examples: Give the difference. a. 43 -17 = 26 b. – 6 – 18 = -24 c. 14 – (- 8) = 22 d. – 34 – (- 43) = 9 Rules in Multiplying Integers Multiplying like signs is positive and unlike sign is negative Examples: Give the product. a. (4) (5) = 20 b. (7) (- 3) = -21 c. (- 14) ( 8 ) = - 112 d. (- 11) (-10) = 110 e. (-11) (7) (- 4) = 308 Rules in Dividing Integers Dividing like sign is positive and unlike sign is negative Examples: Give the quotient of each. a. 128 ÷ 16 = 8 b. (-51) ÷ (-3) = 17  c. (-57) ÷ 19 = -3 d. 252 ÷ (-21) = - 12 ALGEBRAIC EXPRESSIONS: An expression is a meaningful collection of numbers, variables and signs (positive or negative) of operations that must make mathematical and logical sense. Example of an algebraic expression is 8x² where 8 is called the constant or numerical coefficient, x is the variable or literal coefficient and 2 exponent or power THE LAWS OF EXPONENT The EXPONENT tells the number of times the base is to be used as a factor. Base → 3²← exponent 1. The Product of Powers: If a, m and n are real numbers, 𝑎𝑚. 𝑎𝑛 = 𝑎𝑚+𝑛 Example: Find the product of the following. 1. 37. 38 = 37+8 = 315 2. 𝑥 5. 𝑥 2. 𝑥 4 = 𝑥 5+2+4 = 𝑥 11 3. (𝑎 + 𝑏)4 ( 𝑎 + 𝑏)² = (𝑎 + 𝑏)⁶ 2. The Power of a Powers: If a, m, and n are real numbers, (aᵐ)ⁿ = aᵐⁿ Example: Simplify the following expressions. 1. (5⁴)² = 54(2) = 58 2. (x⁷)⁹ = 𝑥 63 3. (y⁵)⁴ = 𝑦 5(4) = 𝑦 20 3. Power of a Product: If a, m, and n are real numbers, (ab)ᵐ = aᵐbᵐ Example: Simplify the following. 1. (𝑥𝑦)4 = 𝑥 4 𝑦 4 2. (3𝑎)2 = 32 𝑎² = 9𝑎2 3. (𝑢 2 𝑣 3 )⁴ = 𝑢 8 𝑣12  4. [ (2𝑥 2 ) (3𝑦 4)]2 = (2𝑥 2 )2 (3𝑦⁴)2 = 4𝑥 4 9𝑦 8 4. Quotient of Powers: If a ≠ 0 and a, m, and n are real numbers, 𝑎ᵐ = 𝑎𝑚−𝑛 , if m>n ; 𝑎ⁿ 1 = 𝑎𝑛−𝑚 , if m˂n = 𝑎0 = 1, if m = n Example: Simplify the following. 𝑥 12 1. = 𝑥 12−7 = 𝑥 5 𝑥7 33 1 1 1 2. = 37−3 = = 81 37 34 (𝑦+3)5 1 3. (𝑦+3 )6 = (𝑦 + 3)5−6 = (𝑦 + 3)−1 = (𝑦+3) 4. (5𝑥)0 = 1 5. 8𝑥 0 = 8 5. Power of a Quotient: If a, b, and m are real numbers and b ≠ 0, 𝑎 𝑚 [ 𝑏 ]ᵐ = 𝑎𝑏𝑚 Examples: 3 𝑥2 (3𝑥²)3 27𝑥 6 1. ( 2𝑦 )³ = (2𝑦)3 = 8𝑦 3 −2 (−2)3 −8 2. ( 5 )ᶟ = = 125 53 3𝑎²𝑏 (3𝑎2 𝑏)² 9𝑎⁴𝑏² 3. ( 4𝑧3 )² = (4𝑧 3 )² = 16𝑧⁶  POLYNOMIALS Polynomial is a special type of algebraic expression that contains a specific number of terms each of which is of the form axⁿ or bxᵐyⁿ where a and b are real numbers, and m and n are whole numbers. Examples: 1. 2x 5𝑥 2. +𝑦 4 3. x² + 3x – 1 4. √2 x⁴ + 3xᶟy - x²y² - xyᶟ +y⁴ The following are algebraic expressions but are not polynomials: 1. 3x ̄ ² + 4 2. 5ˣ 3. √7𝑥ᶟ 4 4. 𝑥 5. 3x²𝑦1/2 No variables in the denominator No variables under radical sign No variables as an exponent No negative exponent No fractional exponent A polynomial with one term is called a monomial, a polynomial with two terms is called a binomial, and a three terms is a trinomial. -7xᶟ ; 5x -4y ; x² -8x + 12 Addition and Subtraction of Polynomials - Only similar terms can be added or subtracted Examples: Simplify by combining similar terms. a. 9x + 2y -5x + 3 – 8y = 4x -6y +3 b. 3x² - 4x – 8 - x² + 4x – 11 = 2x² - 19 c. 3b – 2a – (7b + 4) + (2a – 7) = 3b – 2a – 7b – 4 +2a – 7 = - 4b – 11 d. Add 2x² + 6x – 1 ; - xᶟ + 5x² - x ; 6 – 4xᶟ + x⁴ Solution: 2x² + 6x – 1 Add - xᶟ + 5x² - x x⁴ - 4xᶟ +6 x⁴ - 5xᶟ + 7x² + 5x + 5 e. Subtract 8x² + 7x – 9 from - 3xᶟ - 2x² + 4 - 3xᶟ - 2x² +4 ̅ 8x² + 7x – 9 (change the signs then add) -3xᶟ - 10x² - 7x +13 Multiplication of Polynomials - Product of a monomial by another monomial, get the product of the numerical coefficient following the rule of signs for multiplication, and for its literal factors follow the exponential law of multiplication. [(aⁿ)(aᵐ)] = 𝑎𝑛+𝑚 ] Example: (3a²y)(-5ay²) = - 15aᶟyᶟ (-7mn)(-n²)(-5m²) = - 35mᶟnᶟ - For product of a monomial to a polynomial apply the distributive law Example: (2ab²)(3a² - 4ab + b²) = 6aᶟb² - 8a²bᶟ + 2ab⁴ or 3aᶟb² - 4a²bᶟ + ab⁴ (-4x²yz)(2xy² - 3xyz + 4yz²) = - 8xᶟyᶟz + 12xᶟy² - 16x²y²zᶟ = - 2xᶟyᶟz + 3xᶟy² - 4x²y²zᶟ - Multiplication of a polynomial to another polynomial is actually a series of multiplication of a polynomial by a monomial. For, convenience, arrange the terms in the descending or ascending power of one letter.  Example: Multiply 2x² - 3x + 5 by 3x – 2 Solution 1: (3x – 2)( 2x² - 3x + 5) = 6xᶟ - 9x² + 15x – 4x² + 6x – 10 = 6xᶟ - 13x² + 21x – 10 Solution 2: 2x² - 3x + 5 3x – 2_____ 6xᶟ - 9x² + 15x -4x² + 6x – 10_ 6xᶟ - 13x² + 21x – 10 Division of Polynomials: 1. Quotient of monomial to another monomial – is a monomial with a numerical coefficient equal to the quotient of the numerical coeff. and literal coeff. equal to the quotient of the coeff., observing the rule of 𝑎𝑚 signs for division and the exponential laws for division ( 𝑛 ) = 𝑎𝑚−𝑛 ; 𝑎 a ≠ 0. Rule: the quotient of two powers with the same base is equal to the same base with an exponent equal to the difference of their exponents. Example: 4𝑥 3 ÷ 2𝑥 4𝑥3 = 2𝑥 = 2𝑥 3−1 = 2x² 2. Quotient of a polynomial with another polynomial : Steps: a. Arrange both the dividend and the divisor in decreasing power of one letter. b. Divide the first term of the dividend by the first term of the divisor; the result thus obtained will be referred to as the partial quotient. c. Multiply the partial quotient obtained in (b) by all the terms of the divisor and subtract the result from the dividend. The remainder is the new dividend.  d. Repeat (b) and (c) until the new dividend (remainder) is either zero or of a degree lower than the first term of the divisor. e. The complete quotient consist of the partial quotients plus the remainder indicated as a ratio of the divisor. Example 1: Divide 3x + xᶟ - 3x² + 1 by x -1 __x² - 2x + 1______ X – 1 ǀ xᶟ - 3x² + 3x + 1 ̅ xᶟ - x² - 2x² + 3x + 1 ̅ - 2x² + 2x X+1 ̅ X – 1__ 2 Answer: x² - 2x + 1 + 2___ X–1 Example 2: Divide xᶟ - yᶟ by x – y _x² - xy + y²___ X – y ǀ xᶟ + 0 + 0 - yᶟ - xᶟ - x²y _ x²y x²y - xy² xy² - yᶟ xy² - yᶟ 0 Answer: x² - xy + y² Activity#1: Polynomial I. Combine like terms:  1. 7ax – 16ay + 5ax – 2ax + 11 ay –ay =10ax – 6ay or 2a(5x – 3y) 2. (-12m²n) + ( 16 m²n) − ( 5 m²n) = − m²n 3. 4(a -2b) + 3(a -2b) – (a – 2b) = 4+3−1(a – 2b) = 6(a – 2b) 4. (7xy) + (-7xy) – (5xy) = − 5xy II. Add the following polynomials: 1. a²y – 3ay + 5y² ; 6ay - y² + 2a²y ; 7a²y - y² - 5ay ; and -10a²y – 2y² a²y – 3ay + 5y² 2a²y + 6ay − y² 7a²y – 5ay − y² -10a²y − 2y² - 2ay + y² 2. 5e – 2f – 3g + 3h ; 3f – g + e ; 4h – 2f + 5g ; and 2e – 2h + g 5e – 2f – 3g + 3h e + 3f − g - 2f + 5g + 4h 2e + g −2h 8e − f + 2g + 5h 3. 25m² - 3mn – 5n² ; 3m² - 5mn - n² ; 6mn - n² + m² ; 3m² - n² ; and 2mn – 27m² 25m² - 3mn – 5n² 3m² − 5mn − n² m² + 6mn − n²  3m² − n² − 27m² + 2mn_____ 5m² − 8n² III. Subtract the following: 1. Subtract (3x – 8y + 5z) from 7x – 4y – 3z 7x – 4y – 3z 3x – 8y + 5z 4x + 4y – 8z or 4(x + y – 2z) 2. Subtract ( 5x² - 6x + 4) from the sum of (-3x² + 3x + 1) and (2x² - 5x +7) -3x² + 3x +1 2x² - 5x +7 − x² − 2x + 8 5x² - 6x + 4 ( change the sign) − 6x² + 4x + 4 IV. Multiply the following: 1. (-8a)(-3b)(-5c)(-abc) = 120a²b²c² 2. (-4mᶟn)(3mᶟn² - 5m²n +9m) = - 12m⁶n³ + 20m⁵n² - 36m⁴n 3. (3xᶟ - 2x²y – 7xy² - 5yᶟ)(3x – 4y) = 9x⁴ − 18x³y – 13x²y² + 13xy² + 20y⁴ V. Divide the following: 1. (12xᶟy⁶ + 36x⁴y⁶) ÷ ( 6xᶟy⁶) = 2 + 6x 2. 6xᶟ - 11x² + 26x + 10 by 3x + 5 = 2x² − 7x + 3 3. 4aᶟ - 7a² -45 ÷ a – 3 = 4a² + 15a + 15  CHAPTER II SPECIAL PRODUCTS OBJECTIVES: 1. To know the different methods of multiplying polynomials; and 2. Find the product polynomials by just inspection. KEY/CONCEPTS Product of Two binomials of the form: ( ax + b) (cx + d) Example: (2a + b) ( 3a + 2b) = 2a (3a + 2b) + b (3a + 2b) = 6a² + 4ab + 3ab + 2b² = 6a² + 7ab + 2b² A shorter method (FOIL method) could be made: Rules: The product of a binomial by another binomial is obtained as follows: 1. The 1st term is the product of the first terms of the multiplicand and the multiplier. From the example: (2a) (3a) = 6a² 2. The middle term is obtained by adding the product of the extreme terms to the product of the adjacent terms. From the example: Extreme terms: 2a & 2b Adjacent term: b & 3a Therefore, the middle term is: (2a) (2b) + (b) (3a) = 7ab 3. The last term is the product of the last terms of the multiplicand and the multiplier. From the example: (b) (2b) = 2b² Therefore, (2a + b) (3a + 2b) = 6a² + 7ab + 2b² Examples: Solve just by inspection 1. (2x – 3y) (x + 5y) = 2x² + 10xy – 3xy – 15y² = 2x² + 7xy – 15y² 2. (5x + 2) (2x – 1) = 10x² - x – 2 3. (4x – 3a) (3x + 2a) = 12x² - ax – 6a²  4. (7a + 2) (3a + 1) 5. (2b² + 4y²) (6b² - 9y²) Product of the Sum and Difference of Two Numbers: Example: (2x + 5) (2x – 5) = 4x² - 10x + 10x – 25 = 4x² - 25 Shortcut Solution: Rule: The product of the sum and difference of two numbers is equal to the square of the first number minus the square of the second number. In symbols, (x + y) (x –y) = x² - y² Examples: Solve just by inspection: 1. (3x + y) (3x – y) = 9x² - y² 2. (2xᶟ + 3y²) (2xᶟ - 3y²) = 4x⁶ - 9y⁴ 3. (a + b + 2) (a + b -2) = which can be expressed as sum and difference [(a + b) + 2] [(a + b) – 2] = (a + b)² - 4 = a² + 2ab + b² - 4 4. (3b² - 2x) (3b² + 2x) 5. (bx² - dy²) (bx² + dy²) The Square of a Binomial: Example: (2x + 5)² = (2x + 5) (2x + 5) = 2x (2x + 5) + 5 (2x + 5) = 4x² + 10x + 10x + 25 = 4x² + 20x + 25 Shortcut Method: Rules: The square of a binomial is a trinomial whose terms are obtained as follows: 1. The 1st term is the square of the 1st term of the binomial From the example: 1st term: (2x)² = 4x² 2. The middle term is twice the product of the two terms of the binomial. From the example: Middle term: 2 (2x) (5) = 20x 3. The third or last term is the square of the last term of the binomial. From the example: Last term: (5)² = 25 In symbols: (x + y)² = x² + 2xy + y² Examples: 1. (3a + 5)² = 9a² + 30a + 25 2. (2x – 3y²)² = 4x² - 12xy² + 9y⁴ 3. (7m + 9n)² = 49m² + 126mn + 81n² 1 4. ( p² - y)² 4 5. (4 + 3a²) (4 + 3a²) Activity#2: Special Products Find the products by inspection. 1. (a – 3b) ( a + 3b) = a² - 9b² 2. (x + 6y)² = x² + 12xy + 36y² 3. (3m – 7n)² = 9m² - 42mn + 49n² 4. (x -9) (4x – 7) = 4x² - 43x + 63 5. (2x + 3)² 6. (3x – 4y) (2x + 5y) 7. (x + 6y)² 1 2 8. ( a + b)² 3 5 9. (x + 8) (x + 7) 10. (5p – 3q)² 11. (2x + 3y)(6x – 7y) 12. (3x +5)(x – 2) + (5x -8)(2x – 3) 3𝑎 2𝑏 3𝑎 2𝑏 13. ( 4 + )(4 - ) 5 5 14. (5a²bᶟ + 1)² 15. (cᶟ - 4dᶟ) (cᶟ + 4dᶟ) 16. (8 + b)(8 + b) 17. (5m²n² − 2rs)(4m²n² − 3rs) 3 3 18. (x - 4) (x + 4) 19. (xy + 6)(xy − 6) 20. (a -5) (a + 30)  CHAPTER III FACTORING OBJECTIVES: 1. Know the factoring methods of binomials, trinomials and other polynomials; 2. Apply the different methods in finding the factors of polynomials. KEY/CONCEPTS: Factoring - The process of expressing a given number in terms of its prime factors. - Is the reverse process of multiplication. 1. Removal of the Common Factor: - If all the terms of a given algebraic expression contain a common factor, this should be removed by dividing the polynomial by the highest common factor. Examples: a. Factor 7x² - 7y The highest common factor: 7 Therefore, 7x² - 7y = 7 (x² - y) b. Factor 2bx – 6by + 4bz The highest common factor: 2b Therefore, = 2b (x – 3y + 2z) c. Factor 3a⁵b² - 16aᶟb⁴ Common factor: aᶟb² Hence, = aᶟb²(3a² - 16b²) d. Factor 9x⁴ + 27x²y – 63x⁵y² Common factor: 9x² = 9x² (x² + 3y – 7xᶟy²) e. 8bc – 2cd = 2c(4b – d) f. 12a³ - 24a²- 48 a = 12a(a² - 2a – 4) g. 8x³- 16 x = 8x(x² - 2)  2. Factoring the Difference of Two Squares: - The product of the sum and difference of two numbers is the difference of their squares. - Since factoring is the reverse of finding the product, the difference of two squares is then factored as the sum and difference of the square roots of the two squares. That is: (x² - y²) = (x + y) (x – y) Examples: a. Factor x² - 36y² The given binomial is obviously the difference of two squares. Hence, x² - 36y² = (x + 6y) (x – 6y) b. Factor 4a²x² - 25b²x² It is readily seen that x² is a common factor. Hence, 4a²x² - 25b²x² = x² (4a² - 25b²) = x² (2a + 5b) (2a – 5b) c. Factor 25x² - 144y² = (5x + 12y) (5x – 12y) d. Factor 18aᶟ - 72ab² = 18a(a²- 4b²) = 18a (a + 2b)(a – 2b) e. Factor 81m² - 49n² = (9m + 7n) (9m – 7n) 3. Factoring a Perfect Trinomial Square: - A perfect trinomial square is a polynomial which has two of its terms as perfect squares and the other term as twice the product of the square roots of the two squares. - Its factors are two identical binomials which are either positive or negative depending on the sign of the middle term on the given trinomial. In symbols: X² + 2xy + y² = (x + y)² = (x + y) (x + y) X² - 2xy + y² = (x – y)² = (x – y) (x – y) Examples: a. Factor x² - 4x + 4 = (x – 2) (x – 2) = (x – 2 )² b. Factor 8x² + 40xy + 50y² - Removing the common monomial factor, (2) and factoring the expression as a perfect trinomial square, = 2 (4x² + 20xy + 25y²) = 2 (2x + 5y) (2x + 5y) = 2 (2x + 5y)² Note: Inspect 1st and last term if it is a perfect square. c. 25x² + 70xy + 49y² = (5x + 7y) (5x + 7y) = (5x + 7y)² d. 144w² - 264wc + 121c² = (12w – 11c)(12w – 11c) = (12w – 11c)² e. 1 – 28ab + 196a²b² = (1 – 14ab)(1-14ab) = (1-14ab)² 4. Factoring Trinomials of the Form x² + bx + c. - It is immediately apparent that the factors of x² + bx + c are two binomials whose first term are x and whose last two terms are two factors of the last term of the trinomial which add to produce the coefficient of x in the given trinomial. - That is : x² + bx + c = (x + p)(x + q) Where: c = pq b=p+q Examples: a. x² - 2x – 15 = (x – 5)(x+3) b. y² + xy – 42x² = (y + 7x)(y – 6x) c. (a + 2b)² - 2(a + 2b) - 35 = [(a + 2b) + 5][(a + 2b) – 7 ] = (a + 2b + 5)(a + 2b – 7) d. x² + 5x – 24 = (x + 8) (x - 3) e. x² - 10x + 9 = (x – 1) (x-9) f. w⁴ - 4w² - 5 = (w² - 5)(w² + 1) g. y² - 11y + 18 = (y – 2)(y – 9) h. m²+ 10m – 24 = (m – 2)(m+ 12) i. a²- 5a – 14 = (a + 2)(a - 7) 5. Factoring Trinomials of the form: ax² + bx + c, |a| > 1 - This expression is commonly referred to as a general quadratic trinomial. - Trinomial of this form is normally factored by trial and error method. Method: ax² + bx + c = (px + r)(qx + s) Where: (p)(q) = a (r)(s) = c (p)(s) + (q)(r) = b Examples: 1. Factor 6x² - 7x – 20 Note: To test whether the trinomial is factorable or not, substitute values to the discriminant, b² - 4ac, and if the result is a perfect square then the trinomial is factorable. Solution: a = 6; b = -7; c = -20 b² - 4ac = (-7)² - 4 (6) (-20) = 529 √529 = 23 Therefore, factorable, by trial and error 6x² - 7x – 20 = (3x + 4) (2x – 5) 2. Factor 4x² - 11x + 6 a = 4; b = -11; c = 6 b² - 4ac = (-11)² - 4(4)(6) = 121 – 96 = 25 and √25 = 5 therefore, factorable: 4x² - 11x + 6 = (4x – 3) (x -2) 3. Factor 6m² + 13m – 5 b² - 4ac = (13)² - 4(6)(-5) = 169 – 120 = 289 √289 = 17 factorable 6m² + 13m – 5 = (3m – 1) (2m + 5) 4. Factoring the Sum and Difference of Two Cubes In symbols: xᶟ + yᶟ = (x + y) ( x² - xy + y²) xᶟ - yᶟ = (x – y) ( x² + xy + y²) Rules: 1. The 1st factor is the sum of the cube roots of the two terms of the given expression. 3 3 √𝑥 3 = x ; √𝑦 3 = y 2. The other factor is a trinomial whose terms are derived from the terms of the first factor. The square of the 1st factor (x² & y²) make up two of the three terms while their product (xy) multiplied by -1 is the middle term. Examples: a. Factor 8xᶟ - 27bᶟ Solution: Cube roots of the terms are: 3 3 √8𝑥 3 = 2x ; - √27𝑏3 = - 3b Therefore: 8xᶟ - 27bᶟ = (2x – 3b) (4x² + 6bx + 9b²) b. Factor axᶟ - 8ayᶟ = a (xᶟ - 8yᶟ) Common factor: a  axᶟ - 8ayᶟ = a (xᶟ - 8yᶟ) Cube roots of the terms are: 3 3 √𝑥 3 = x ; √8𝑦 3 = 2y Therefore, axᶟ - 8ayᶟ = a (xᶟ - 8yᶟ) = a (x – 2y) (x² + 2xy + 4y²) c. 2x²yᶟ + 16x² Common factor: 2x² 2x²yᶟ + 16x² = 2x² (yᶟ + 8) Cube roots of the terms are: 3 3 √𝑦 3 = y ; √8 = 2 Therefore: 2x² (yᶟ + 8) = 2x² (Y + 2) ( y² - 2y + 4) 𝑥3 d. - 64 𝑦3 Cube roots of the terms are: 3 𝑥3 𝑥 √𝑦3 = 𝑦 ; -3√64 = - 4 𝑥3 𝑥 𝑥2 𝑥 Therefore: - 64 = ( - 4) (𝑦 2 + 4𝑦 + 16) 𝑦3 𝑦 e. m⁶ - n⁶ Cube roots of the terms are: 3 3 √𝑚6 = m² ; - √𝑛 6 = n² f. Therefore: m⁶ - n⁶ = (m² - n²) ( m⁴ + m²n² + n⁴) 5. Factoring by Grouping - Occasionally the need arises to factor a polynomial of more than three terms. In such situation, grouping certain terms together may reduce the polynomial to a form factorable in one of the ways discussed. Example: a. Factor 2a² + 4ab – 3ad – 6bd = 2a (a + 2b) – 3d (a + 2b)  = (a + 2b) (2a – 3d) b. Factor 3ax + 2bx – 3ay – 2by = x ( 3a + 2b) – y (3a + 2b) = (x – y) (3a + 2b) or = 3a (x – y) + 2b (x – y) = (x –y) (3a + 2b) Activity: Factor the following expressions completely: 1. 8am + a + 8bm +b 2. 45xᶟyᶟ + 75xᶟyᶟ - 30xy⁴ 3. 5 (a + b) – x (a + b) 4. y² + 8y + 16 5. x² + 18x + 81 6. m² +5m – 6 7. 8yᶟ - 27 8. 16a²b² - 25c² 9. p⁶ + 1 10. 2x²yᶟ + 16x² 𝑎3 11. + 125 𝑏3 12. 64 – 48xy + 9x²y² 13. (a + b)³ - 1 14. 2x² - xy + 2cx – cy 15. 6m² + 13m - 5  CHAPTER IV RADICALS OBJECTIVES: 1. Familiarize with the properties of Radicals; and 2. Solve and simplify radicals. KEY/ CONCEPTS: 𝑛 √𝑎 a –> radicand √ -> radical sign n –> index 1. Properties of Radicals: 𝑛 a. a = √𝑎𝑛 𝑛 b. am/n = √𝑎𝑚 𝑛 c. a1/n = √𝑎1 d. n√a n√b = n√ab n e. √a / n√b = n√a/b Power and Radical Relationship Element Power (am/n) Radical (n√am) a Base Radicand m Numerator of Exponent Exponent n Denominator of Exponent Index Terms: a. Like or Similar – radicals having same index and radicand Example: 3√2x , -5 3√2x and 7 3√2x b. Radicals of same order are radicals with same index 3√2x; 3 √x2 c. Root of a number is one of the equal factors of the given number. Examples: √25 = ± 5, since (± 5)2 =25 3 √−8 = -2, since (-2)3 = -8 5 √32 x10 = 2x2, since (2x2)5 = 32x10 By Convention, however n√a will stand for the positive nth root. Notes: a. If n is even and a (positive number), it has two real nth roots one + and – b. If n is odd, every number, (a) has one and only one real nth root, the sign follows the sign of the radicand. c. If n is even and (a) is a negative number it has no real n th root, its roots are imaginary numbers. 2. Operations of Radicals: a. Removal of factors from the radicand From: n√ab = n√a n√b Examples: a. √24 = √(6)(4) = √6 √4 = 2√6 b. 3 √−16x7 = 3√(−8𝑥 6 )(2𝑥) = 3√−8x6 3√2x = -2x2 3√2x 4 c. √81a5b8 = 4√81a4b8 (a) = 3ab2 4√a d. √50x3 = √(25)(2)(x2)(x) = 5x√2x b. Reduction of the Radicand to an Integral form: From: 𝑛 𝑛 𝑎 √𝑎 √𝑏 = 𝑛 √𝑏  Examples: 1 1 2 2 √2 √2 1 a. √ = √2 ▪ 2 = √4 = = or 2 √2 2 √4 2 3 −𝑥³ = 3 −𝑥³ 3 4𝑦 = 3 −𝑥 3.4𝑦 = 𝑥 3 b. √ 2𝑦² √ ∙ √4𝑦 √ − √4𝑦 2𝑦² 8𝑦³ 2𝑦 √3 √2 √6 c.. = = √6 / 4 √8 √2 √16 c. Reduction of the Index of a Radical Examples : Reduce the order: 4 a. √9 = 4√32 = (3) 2/4 = 31/2 = √3 6 b. √81x2y2 = 6√(9 xy)2 = (9xy)2/6 = 3√9xy d. Addition and Subtraction of Radicals - Only similar radicals (same index and same radicand) can be combined. Examples: 1. Reduce the given radicals, whenever feasible, to similar radicals by any of the procedures of simplification. 2. Combine similar radicals according to the distributive law of addition. a. Add: √50 + 2√1/2 - √18 √50 = √25▪2 = + 5√2 √2 2√1/2 = 2√1/2▪2/2 = 1√2 2 = 1√2 √4 √18 = √9▪2 = -3√2 3√2 3 3 3 b. Combine: √2𝑎⁴ - √16𝑎4 + 2 √54𝑎⁴ 3 √2𝑎⁴ = 3√𝑎3. 2𝑎 3 = a √2𝑎 3 3 3 √16𝑎4 = √8𝑎³. 2𝑎 = -2a √2𝑎  3 3 3 2 √54𝑎4 = 2 √27𝑎3. 2𝑎 = 6a √2𝑎 3 5a √2𝑎 e. Multiplication of Radicals From: n√a n√b = n√ab − Radicals of same order (same index) can only be multiplied. Examples: 1. (3 √2)(-2 √8 ) = -6 √(2)(8) = - 6 √16 = - 6(4) = - 24 2. (4√5 + √3) (√5 - √3) = 4√25 - 4√15 + √15 - √9 = 4(5) - 3√15 - 3 = 17 - 3 √15 3. (3√2x) (4√3x2) = (2x)1/3 (3x2)1/4 = (2x)4/12 (3x2)3/12 = 12√(2x)4. 12√(3x2)3 = 12√16x4. 12√27x6 = 12√432 x 10 f. Division of Radicals From : n√a / n√b = n√a/b - Same ordered radicals can be divided. Examples √24 1. = √8 = √4. 2 = 2 √2 √3 3 80 3 3 3 2. √ 5 = √16 = √8. 2 = 2 √2 10 √16𝑎³𝑏² 3. = 5√8𝑎²𝑏 = 5 √4𝑎2. 2𝑏 = 10a √2𝑏 2 √2𝑎𝑏 3 1 1 √4 4. 6 = 43 ÷ 26 √2 2 1 = 46 ÷ 26 6 6 = √4² ÷ √2  6 6 = √16 ÷ √2 6 = √8 Simplify the following radicals a. 3√2 + 4√2 - 5√2 = (3+4-5)√2 = 2√2 b. 3√24 - √54 + √96 3√4. 6 = 3(2)√6 = 6√6 √9. 6= = -3√6 √16. 6 = 4√6 7√6 8 √98 c. = 4√49 = (4)(7) = 28 2√2 ⁶ d. ( √5 )(√6 ) =√5,400 3 :  CHAPTER V LINEAR EQUATION OBJECTIVES: 1. Define equation and understand its principle; and 2. Solve linear equation in one unknown KEY/CONCEPTS: Equation – is a statement denoting that two algebraic expressions are equal. A linear equation in one variable is an equation of the first degree and it can be written in the form; ax + b = c, where a, b and c are real numbers and a ≠ 0. Example: 3x – 5 = x + 3 Left side of the equation Right side of the equation Kinds of Linear equation: 1. Conditional Equation – an equation which is true only for certain values of the variable or variables involved. Example: 3x – 5 = x + 3 this statement is only true if x = 4 ; thus a conditional equation. 2. Identity Equation – an equation which holds to be true for all permissible values of the unknown. Example: (x – 1)² = x² - 2x + 1 substituting (2) or any arbitrary number to x; (2 – 1)² = 2² -2(2) + 1 1 = 1 it follows that the statement is true for all values of x and therefore the equation is identity. Solution to linear equation with one unknown: - A linear equation in one unknown, (x) is one which is of the first degree in x. Any equation which can be expressed or reduced to the form: ax + b = c where a, b, and c real number and a ≠ 0 Method of solution: Simple Transposition. Transposition – the process of transferring a term from one side of the equation to the other side by changing its sign. Steps to follow in solving linear equation in one unknown: 1. If the equation is fractional, clear the equation of fractions by multiplying both sides of the equation by the L.C.M. of all the denominators. 2. Simplify both sides of the equation by carrying out indicated operation of multiplication. If there are any and by collecting similar terms. 3. Put all the terms containing the unknown to one side of the equation and all other terms to the other side applying the axiom of addition and/or subtraction. 4. Divide both sides of the equation by the coefficients of the unknown. 5. Check the solution by substituting the value obtained in the original equation. Reject the extraneous roots. Examples: 1. Solve the unknown and check: 7x – 20 = 2x Solution: 7x – 20 = 2x 7x – 2x = 20 5x = 20 5𝑥 20 = ; cross multiply 5 5 x= 4 Check: 7(4) – 20 = 2(4) 28 – 20 =8 8=8 2. Solve for x and check: 7x + 6 = 5 + 4x Solution: 7x + 6 = 5 + 4x 7x – 4x = 5 - 6 3x = -1 −1 X= 3 −1 −1 Check: 7( 3 ) + 6 = 5 + 4( 3 ) −7 4 +6=5- 3 3 −7+18 15−4 -= 3 3  11 11 = 3 3 3. 3(y + 1) = 12 + 4(y – 1) 3y + 3 = 12 + 4y – 4 3y – 4y = 12 – 4 -3 - y = 12 – 7 Y=-5 Check: 3(-5 + 1) = 12 +4 (-5 − 1) 3(-4) = 12 + 4(-6) -12 = 12 – 24 -12 = -12 𝑏+6 2𝑏−3 3𝑏+4 4. Solve for unknown and check: - = 2 5 4 Solution: 𝑏+6 2𝑏−3 3𝑏+4 - = L.C.M. = 20 2 5 4 10 (b + 6) – 4 (2b – 3) = 5 (3b + 4) 10b + 60 – 8b + 12 = 15b + 20 10b – 8b – 15b = 20 – 60 – 12 - 13b = - 52 b=4 4+6 2(4)−3 3𝑏+4 check: - = 2 5 4 10 5 16 - = 2 5 4 5–1=4 4=4 2 3 14 1 5. Solve for a: + 4 = 6𝑎 - 3𝑎 2 L.C.M. = 12a 4(2) + 9a = 28 – 6a 9a + 6a = 28 - 8 15a = 20 a = 20/15 or 4/3 Solve the following linear equations and check 1. 6 + 3y = - 4 ( y + 2) 𝑥 𝑥 2. + =2 𝑥+1 𝑥+3 7𝑥+5 3𝑥+15 3. –2= 8 3