Lecture 2 - Algebraic Terms and Rational Expressions.pdf

LECTURE 2 ALGEBRAIC TERMS AND RATIONAL EXPRESSIONS THE DISTRIBUTIVE PROPERTY A term is a quantity separated by a plus or minus sign. Example: 13x2y + 2xy +22xy2 + 15 The number in a term is called coefficient. A term without a variable is called constant....

LECTURE 2 ALGEBRAIC TERMS AND RATIONAL EXPRESSIONS THE DISTRIBUTIVE PROPERTY A term is a quantity separated by a plus or minus sign. Example: 13x2y + 2xy +22xy2 + 15 The number in a term is called coefficient. A term without a variable is called constant. 2 THE DISTRIBUTIVE PROPERTY Two terms are alike if they have the same variables and we can combine like terms by adding/subtracting coefficients on terms that are alike. Combine like terms: Example: 13x2 + 2x -3y + 2xy2 – 5x +5y – 8x2 = (13x2 - 8x2) + 2xy2 + (2x – 5x) + (5y -3y) = 5x2 + 2xy2 – 3x + 2y 3 THE DISTRIBUTIVE PROPERTY Use the Distributive Property in expanding expressions. Example 1: 10x (5y + 2xy -3x) = 10x(5y) + 10x(2xy) – 10x(3x) = 50xy + 20x2y – 30x2 Example 2: 3(2x2 – 5x + 2y) – 4x(3x -4) = 6x2 – 15x + 6y – 12x2 + 16x = -6x2 + x + 6y Guide: ( a + b + c)(d + e) = ad + bd + cd + ae + be + ce 4 THE DISTRIBUTIVE PROPERTY FOIL METHOD First x First + Outer x Outer + Inside x Inside + Last x Last 5 THE DISTRIBUTIVE PROPERTY FOIL METHOD First x First + Outer x Outer + Inside x Inside + Last x Last EXAMPLE 1: Expand the Expression (2x + 3)(x – 4) F O I L (2x+3)(x – 4) = 2x(x) + 2x(-4) + 3(x) + 3(-4) = 2x2 – 8x + 3x – 12 = 2x2 – 5x -12 6 THE DISTRIBUTIVE PROPERTY FOIL METHOD First x First + Outer x Outer + Inside x Inside + Last x Last EXAMPLE 2: Expand the Expression (3x + 5)(4x – 4) F O I L (3x+5)(4x – 4) = 3x(4x) + 3x(-4) + 5(4x) + 5(-4) = 12x2 – 12x + 20x – 20 = 12x2 + 8x -20 7 FACTORIZATION Factoring is the process of using the distributive Property in reverse. EXAMPLE: 6x2 – 12x + 15xy = 3x (2x – 4 + 5y) We identify what is common to each term then factor them out. 8 FACTORIZATION Many expressions having three terms can be factored so that the FOIL method gives us the original expression. EXAMPLE 1: Factor the expression x2 -4x + 3 x2 - 4x + 3 = (x )(x ) factors either 1,3 or -1, -3 = (x – 1)(x – 3) 9 FACTORIZATION Many expressions having three terms can be factored so that the FOIL method gives us the original expression. EXAMPLE 2: Factor the expression 4x2 - 4x - 15 4x2 – 4x – 15 = Factors of 4x2, (x, 4x), (2x, 2x) Factors of -15, (1,-15),(-1, 15), (-3, 5), (3, -5) = (2x + 3)(2x – 5) 10 FACTORIZATION The difference of two squares can be factored with the formula a2 – b2 = (a – b)(a +b) EXAMPLE 3: Factor the expression x2 - 9 x2 – 9 = (x – 3)(x + 3) EXAMPLE 4: Factor the expression 4x2 - 16 4x2 – 16 = (2x – 4)(2x + 4) 11 FACTORIZATION Special Products and Factoring 1. (x + y)(x – y) = x2 - y2 2. (x + y)2 = x2 + 2xy + y2 Example: (x + 5)2 = x2 + 2x(5) + 52 = x2 + 10x + 25 3. (x - y)2 = x2 - 2xy + y2 Example: (x - 4)2 = x2 - 2x(4) + 42 = x2 - 8x + 16 4. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2xz + 2yz 5. (x3 + y3) = (x + y)(x2 - xy + y2) 6. (x3 - y3) = (x - y)(x2 + xy + y2) 12 RATIONAL EXPRESSIONS A rational expression (or any kind of fraction) is in lowest terms if the numerator and denominator have no common factors (other than 1). We simplify (or reduce) a fraction by dividing the numerator and denominator by their common factor(s). This process is also called cancelling. 𝑥 4𝑥−9 EXAMPLE: , 𝑥−4 2𝑦−𝑥−𝑧 13 RATIONAL EXPRESSIONS BASIC FRACTION OPERATION SUMMARY 1. Multiplying Fractions 𝑎 𝑐 𝑎𝑐 × = 𝑏 𝑑 𝑏𝑑 2. Dividing fractions 𝑎 𝑐 𝑎 𝑑 ÷ = 𝑥 𝑏 𝑑 𝑏 𝑐 3. Adding fractions (with same denominators) 𝑎 𝑐 𝑎+𝑐 + = 𝑏 𝑏 𝑏 4. Adding fractions (with different denominators) 𝑎 𝑐 𝑎𝑑+𝑏𝑐 + = 𝑏 𝑑 𝑏𝑑 5. Simplifying Fractions 𝑎𝑏𝑐 𝑎 = 𝑐𝑏𝑑 𝑑 14 RATIONAL EXPRESSIONS Simplify the rational expressions. 6x2y x2+ 2x −8 Example 1: Example 2: 2 15𝑥𝑦 x − 3x + 2 6x2y (3𝑥𝑦)(2𝑥) = (𝑥+4)(𝑥−2) = (𝑥−1)(𝑥−2) 15𝑥𝑦 (3𝑥𝑦)(5) 𝟐𝒙 𝒙+𝟒 = = 𝟓 𝒙−𝟏 15 RATIONAL EXPRESSIONS Find the product. 15xy2 8 Example 3: x 14 3𝑥𝑦 3𝑥𝑦 5𝑦.(4)(2) = 2 7.(3𝑥𝑦) 𝟐𝟎𝒚 = 𝟕 16 RATIONAL EXPRESSIONS Example 4: Perform the operations in the expression. 1 2 𝑥 + − 2𝑥 3𝑥 6 LCD of 2x, 3x, 6 = 6x 1 3 + 2 2 −(𝑥)(𝑥) = (6𝑥) 3 + 4 −x2 = 6𝑥 −x2+𝟕 = 𝟔𝒙 17 LAWS OF EXPONENTS (INDEX LAW) An exponent refers to the number of times a number is multiplied by itself 1. an = a x a x a x a … (n factors) 2. am x an = am + n Example: 23 x 24 = 23+4 = 27 = 128 xz (xz+2) = xz + (z+2) = x2z+2 am 3. an = am-n 55 Example: = 5 5-3 = 52 = 25 53x−5 x = x x-5 – (2x-2) = x–x-3 x2x −2 18 LAWS OF EXPONENTS (INDEX LAW) 4. (am)n = amn Example: (22)4 = 22 x 4 = 28 = 2256 𝑥 +2x (x ) = x x x+2 x(x+2) =x 5. (abc)m = ambmcm Example: (2xy)3 = 23x3y3 = 8x3y3 𝒂 𝒏 an 6. = 𝒃 bn 2𝑥 2 22x2 4x2 Example: = = 2 3𝑦 3 y 9y 2 2 19 LAWS OF EXPONENTS (INDEX LAW) 𝒏 7. am/n = 𝒂𝒎 Example: x(5+x)/x = 𝑥 x(5+x) 1 1 8. a-m = m and −m = am a a x Example: xy = 2 -2 y 2x = 2xy 2 y−2 9. a0 = 1 10. If am = an then m = n (provided a ≠ 0) 20 PROPERTIES OF RADICALS Radical - The √ symbol that is used to denote square root or nth roots. 𝒏 1. a1/n = 𝒂 2𝑦 Example: x1/2y = 𝑥 𝒏 2. am/n = 𝒂𝒎 Example: xxy/5 = 5 xxy 21 PROPERTIES OF RADICALS 𝒏 3. 𝒂 n =a Example: 5 𝑥 5 =x 𝒏 𝒏 𝒏 4. 𝒂 𝐱 𝒃 = 𝒂𝒃 3 3 𝑛 Example: x2 x 2𝑦 = 2x2𝑦 𝒏 𝒂 𝒏 𝒂 5. 𝒏 = 𝒃 𝒃 𝟒 𝟐𝟒𝒙𝒚 𝟒 𝟐𝟒𝒙𝒚 𝟒 𝟖𝒙𝒚 Example:. 𝟒 = = 𝟑𝒙𝒛 𝟑𝒙𝒛 𝒛 22 OTHER IMPORTANT PROPERTIES OF ALGEBRA 1. a x 0 = 0 2. If a x b = 0, then a = 0 or b = 0 or both a and b is equal to 0. 0 3. = 0, if a ≠ 0 𝑎 𝑎 4. = undefined 0 𝑎 5. =0 ∞ 23 EXPONENTS & RADICALS EXAMPLE PROBLEM 1 Rewrite the expression 3(2x3)4 Solution: 3(2x3)4 = 3(24 x3(4)) = 3 (16) x12 = 48 x12 24 EXPONENTS & RADICALS EXAMPLE PROBLEM 2 Solve for a in the equation a = 64x 4y Solution: a = 64x 4y a = (43)x 4y a = (43x) 4y a = 43x + y 25 EXPONENTS & RADICALS EXAMPLE PROBLEM 3 (ECE BOARD 1993) Find x from the equations: 27x = 9y 81y 3-x = 243 Solution: 81y 3-x = 243 27x = 9y 34y3-x = 35 33x = 32y , If am = an then m = n (provided a ≠ 0) – index law 34(3x/2)3-x = 35 so 3x = 2y and y = 3x/2 36x3-x = 35 36x-x = 35 35x = 35 5x = 5 x=1 26 EXPONENTS & RADICALS EXAMPLE PROBLEM 4 xy−1 4 x2y−2 3 Simplify: −2 3 ÷ −3 3 x y x y x4y−4 x−9y9 𝑥 6 −6 = x4-9-(-8)-6 y-4+9-12-(-6) Solution: x y −8 12 xy xy−1 4 x2y−2 3 = x-3 y-1 ÷ x y −2 3 x−3y3 𝟏 xy−1 4 x−3y3 3 = x−2y3 𝑥 x2y−2 x3 y x4y−4 x−9y9 𝑥 x y −8 12 x6y−6 27 EXPONENTS & RADICALS EXAMPLE PROBLEM 5 Simplify the following: 7a+2 – 8(7a+1) + 5(7a) + 49(7a-2) Solution: 7a+2 – 8(7a+1) + 5(7a) + 49(7a-2) am x an = am + n – index law = 7a72 – 8(7a71) + 5(7a) + 49(7a7-2) = 7a [72 – 8(7) + 5 + 72(7-2)] = 7a [49 – 56 +5 +1] = 7a (-1) = -7a 28 EXPONENTS & RADICALS EXAMPLE PROBLEM 6 Simplify the following: 100x2y4 Solution: 100x2y4 = 102x2(y2)2 = 10xy2 29 EXPONENTS & RADICALS EXAMPLE PROBLEM 7 12+ 80 Simplify the following: 4 Solution: 12+ 80 12+ 16(5) = 4 4 12+ 4 5 = 4 4 3+ 5 = 4 =3 + 5 30 EXPONENTS & RADICALS EXAMPLE PROBLEM 7 3 3 3 Simplify : 2x4 − 16x4 + 2 54x4 Solution: 3 3 3 = 2x4 − 16x4 + 2 54x4 3 3 = 2x4 − (8)(2x4) + 2 3 (27)(2x4) 3 3 3 = 2x4 −2 2x4 + 2(3) 2x4 3 = 2x4 [1 – 2 + (2)(3)] 3 = 𝟓 2x4 31 EXPONENTS & RADICALS EXAMPLE PROBLEM 8 (ME BOARD 1998) Find the value of x that will satisfy the following expression: 𝑥−2= 𝑥 +2 Solution: 𝑥−2= 𝑥 + 2 - square both sides 𝑥 − 2 2 = ( 𝑥 + 2)2 x -2 = x + 2(2)( 𝑥) + 4 x –x - 4 𝑥 = 4 + 2 - 4 𝑥 = 6 (square both sides) 16x = 36 𝟑𝟔 𝟗 x= = 𝟏𝟔 𝟒 32