MATH 2 part 1

Mathematics II PART 1

Module 1 System of Linear Equations and Inequalities What this module is about This module is about systems of linear equations and inequalities. As yougo over the exercises, you will develop skills in solving systems of linearequations in two variables and ability to apply these in solving problems. Treatthe lessons with fun and take time to go back if you think you are at a loss. What you are expected to learnThis module is designed for you to:1. define a system of linear equations in two variables, and2. solve systems of linear equation in two variables and identify those whose graphs: • intersect • coincide • are parallel.How much do you know A. Tell whether the graphs of the pair of linear equations intersect,coincide or are parallel.1. 4x – y = -1 3. x – y = 2 8x – 2y = -2 2x – 2y = 42. x + y = 3 4. y = 3x + 3 x – 4y = -2 y = 3x – 2B. Answer the following:5. When do the graphs of the system of linear equations intersect?

6. When do the graphs of the systems of linear equations coincide? 7. When do the graphs of the systems of linear equations parallel? C. Graph the following systems and identify the kinds of linear systems intwo variables. 8. 2x – 2y = 4 x–y=2 9. 3x – y = 3 3x – y = 6 10.x + y = 6 x–y=4 What you will do LESSON 1 Define a System of Linear Equations in Two Variables Two equations in two variables taken together to find a solution is asystem of linear equations in two variables. The graph of the system are straightlines. The solution of this system is the set of ordered pairs that are solutions ofboth equations. One way of solving a system is to graph each equation on the same set ofaxes. The graphed lines represent the solution sets of the individual equations.The intersection of the graphs is the solution of the system.Examples: Consider the following figures: 5. 2x + 2y = 6 y = 4x – 2 There is only one solution. The solution of the system is (1, 2 2

2. y – 2x = 1 y = 2x – 2 There is no solution. The graphs do not intersect. They are parallel. 3. x – y = 1 2y = 2x – 2 There are infinite number of solutions. The graphs coincide. Some solutions are (1,0), (2,1), (3, 2). Consider the given examples, with the illustrations, the three kinds ofsystem of linear equations are: 1. Consistent and independent 2x + 2y = 6 y = 4x – 2 The graphs intersect at one point. The solution therefore of the system is the point of intersection. 2. Inconsistent y – 2x = 1 y = 2x – 2 The graphs of the two equations do not intersect. They are parallel. There is no solution. 3. Consistent and dependent 3

x–y=1 2y = 2x – 2You can see that the two graphs coincide which means they have infiniteor many solutions. Without graphing you can determine the kinds of system of linearequations by the coefficients of each linear equation of the form: a1x + b1y = c1 The coefficients of x is a1 and y is b1 a2x + b2y = c2 The coefficients x is a2 and y is b2 It is more convenient for you to determine the kind of system if eachequation is of the form like the above.Examples:1. x + y = 3 The coefficient of x is 1 and y is 1and c is 3 x – 4y = -2 The coefficient of x is 1 and y is –4 and c is -2 Since the coefficients of x. and y and the value of c in the two equationsare not equal, the lines will intersect and will have one solution. Hence, this typeof system is independent and inconsistent.2. x + 2y = 3 The coefficient of x is 1 and y is 2 and c is 3 x + 2y = -2 The coefficient of x is 1 and y is 2 and c is –2 Since the coefficients of x and y are equal and the values of c are notequal in the two equations, the lines are parallel and will have no solution.Hence, the system is inconsistent.3. x – y = 2 The coefficient of x is 1 and y is –1 and c is 2. 2x – 2y = 4 The coefficient of x is 2 and y is –2 and c is 4. Notice that if you multiply x – y = 2 by 2 it will be equal to 2x – 2y = 4. Wecan conclude that the two lines will coincide and will have an infinite number ofsolutions. Hence, this type of system is dependent and consistent.Try this outA. Complete the following statements.1. ___________ is a system of equations with no solution.2. ___________ is a system of equations one solution.3. ___________ is a system of equations with many solutions. 4

4. If the system of linear equations is consistent and independent, then their graphs ___________. 5. If the system of linear equations is independent, then their graphs _______. 6. If the system of linear equations is dependent and consistent, then their graphs _________. 7. When the relation between variables involves at least two equations, they are called __________. 8. 2x – y = 4 and x – y = 5 is an example of _________ system. 9. 2x – y = 4 and 6x – 3y = 18 is an example of _________ system. 10. 2x – y = 2 and 4x – 2y = 4 is an example of _________ system.B. WHO’S WHO By inspection, match the kind of system of linear equations in Column Bthe given equations in Column A. The letters of the answer will spell out thename of the first woman mathematician. Column A Column B P Dependent 1. x – 5y = -3 H Independent 2x + y = 5 Y Inconsistent T Independent 2. 2x – 3y = 1 I Dependent 2x – 3y = 2 A Inconsistent 3. 4x – y = -1 8x – 2y = -2 4. y = 3x + 2 y = 3x – 2 5. 2x + 3y = 6 2x + y = -2 6. 2x + 3y = 6 6x + 9y = 18 7. 2x – y = 1 6x – 3y = 12Answer: ___ ___ ___ ___ ___ ___ ___ 1234567 5

C. Math Discovery Name this Scientist/MathematicianThe following is a quote from a famous scientist and mathematician. “If I have seen farther than others, it is because I have stood on theshoulder of giants.” Discover the name of the scientist – mathematician by identifying whetherthe graphs of the following system of linear equations intersect, coincide orparallel. Shade the box containing the correct answer and the letters in theunshaded boxes will spell out the name. Choose your answer fromthe boxes:First Row:1. x – 2y = 2 2. x + 3y = 3 x+y=5 -x + y = 5Second Row:3. y = 3x – 1 4. x + 4y = 3 6x – 2y = -6 x + 4y = -2Third Row:5. 4x – 2y = 6 6. 3x – y = 4 y = 2x – 3 9x – 3y = 12Answer: ____ ____ ____ ____ ____ _____ LESSON 2 Solution of System of Linear Equations A solution for a linear system of equations in two variables is an orderedpair of real numbers (x, y) that satisfies both equations in the system. A systemmay have only one solution, many or infinite solutions or no solution at all. 6

One way as stated earlier is by graphing. Consider the linear system: x – 2y = -1 2x + y = 8 The pair (3, 2) is a solution because after substituting 3 for x and 2 for y inthe two equations of the system, you have the two true statements.x – 2y = -1 and 2x + y = 83 – 2(2) = -1 2(3) + 2 = 8 8=8 -1 = -1Both equations are satisfied by (3, 2). Since a solution to a system of equations represents a point on both lines,one approach in finding the solution for a system is to graph each equation onthe same set of coordinate axes and then you identify the point of intersection.Example 1:Solve the system by graphing: 2x + y = 4 x–y=5 First solve for y and tabulate the values of x and y: Domain = set of realnumbers.First Equation: Second Equation:2x + y = 4 x–y=5 y = -2x + 4 y=x-5 Now, solve for y by substituting values for x in the equation y = -2x + 4 andy = x – 5. Use x = {-1, 0, 1, 2, 3, 4} First Equation Second Equation y = -2x + 4 y=x-5If x = -1 y = -2 (-1) + 4 y = -1 - 5 =2+4 =-6 =6If x = 0 7

y = -2(0) + 4 y=0–5=0+4 = -5=4If x = 1y = -2(1) + 4 y=1–5 = -2 + 4 =-4 =2If x = 2y = -2(2) + 4 y=2–5y = -4 + 4 y=-3y=0If x = 3y = -2(3) + 4 y=3–5 = -6 + 4 y = -2 = -2If x = 4y = -2(4) + 4 y=4–5 =-8+4 y = -1 = -4Let’s tabulate the values for easier viewing:X -1 0 1 2 3 4 x -1 0 1 2 3 4y 6 4 2 0 -2 -4 y -6 -5 -4 -3 -2 -1 Do the equations have the same coordinates? Yes, you are right. At x =3, the two tables yield the same values for y which is equal to -2. But what doesthis mean? Let’s find out! From the tabulations, plot the solution points on the same coordinate axesand then draw a straight line through these points. As we have studied earlierthe two systems are independent and therefore their graphs will intersect at onepoint. How is it related to the same values of y at a given x in the table? And,what is that point. Find out by graphing. 8

Solution: The two lines intersect at only one point. The system therefore has one solution. The point of intersection is at (3, -2). This is the only point lying on both lines. Therefore (3, -2) is the only ordered pair satisfying both equations. Hence, the solution for the system. As we have studied earlier, a system of linear equations with two variableshaving the same numerical coefficients for x and y will have no solution. Thesaid system is inconsistent and their graphs will not intersect.Example 2:Solve the system by graphing: 2x – y = 4 6x – 3y = 18 First solve for y and tabulate the values of x and y: Domain = set of realnumbers.First Equation: Second Equation:2x - y = 4 6x – 3y = 18 (dividing by 3) y = 2x - 4 y = 2x - 6 Now, solve for y by substituting values for x in the equation y = 2x - 4 andy = 2x – 6. Use x = {-1, 0, 1, 2, 3, 4} First Equation Second Equation y = 2x - 4 y = 2x - 6If x = -1 y = 2 (-1) - 4 y = 2(-1) - 6 = -2 - 4 = -2 - 6 = -6 = -8 9

If x = 0 y = 2(0) – 6 y = -2(0) - 4 = -6 =0-4 = -4If x = 1y = 2(1) - 4 y = 2(1) – 6 = 2-4 =2-6 = -2 = -4If x = 2y = 2(2) - 4 y = 2(2) – 6 =4-4 =4-6 =0 = -2If x = 3y = 2(3) - 4 y = 2(3) – 6 = 6-4 y=6-6 =2 =0If x = 4y = 2(4) - 4 y = 2(4) – 6 = 8-4 =8-6 =4 =2Now tabulate the values for easier viewing:x -1 0 1 2 3 4 x -1 0 1 2 3 4y -6 -4 -2 0 2 4 y -8 -6 -4 -2 0 2 Do the equations have the same coordinates? You are right, there isnone. But what does this mean? Let’s find out! From the tabulations, plot the solution points on the same coordinate axesand then draw a straight line through these points. 10

Solution: The lines are distinct and parallel. There is no point at which they intersect. Therefore, the system has no solution.Example 3:Solve the system by graphing: 2x – y = 2 4x – 2y = 4 First solve for y and tabulate the values of x and y: Domain = set of realnumbers.First Equation: Second Equation:2x - y = 2 4x – 2y = 4 (dividing by 2) y = 2x - 2 y = 2x - 2 Notice that when the second equation is divided by 2, it is equal to the firstequation. Similarly, when the first equation is multiplied by 2, it is equal to thesecond equation.What does this mean? This only means that the two equations will have graphs that wouldoccupy the same set of points.Let us continue the same procedure used in the two previous examples. Now, solve for y by substituting values for x in the equation y = 2x – 2.Use x = {-1, 0, 1, 2, 3, 4} Since the first equation is equal to the second equation, we can solve forthe value of y using either of the two equations. 11

The Equation y = 2x - 2If x = -1 y = 2 (-1) - 2 = -2 - 2 = -4If x = 0 y = -2(0) - 2 =0-2 = -2If x = 1 y = 2(1) - 2 = 2-2 =0If x = 2 y = 2(2) - 2 =4-2 =2If x = 3 y = 2(3) - 2 = 6-2 =4If x = 4 y = 2(4) - 2 = 8-2 =6Now tabulate the values for easier viewing: x -1 0 1 2 3 4 y -4 -2 0 2 4 6 12

Do the two equations have the same coordinates? You are right, the twoequations have the same values for y for each value of x. But what does thismean? Let’s find out! From the tabulations, plot the solution points on the same coordinate axesand then draw a straight line through these points. Solution: You should see that the two equations have the same graph. They coincide, so they have an infinite number of solutions.Try this out A. Each exercise names the graphs of two equations. Give the solution for the system containing the two equations. 1. a and d 2. b and c 3. c and d 4. b and d 5. a and c 6. a and b 7. e and d 8. e and a 9. b and e 10.c and e 13

B. Find the solution of the following systems of linear equations by graphing: 1. 2x – y = 7 5x + 2y = 4 2. y – 3x = 0 –2y – 3y = 0 3. y = -x + 6 y=5 14

4. x + y = 6 x–y=85. 3x + y = -4 2x + y = 106. x + y = 4 2x + 2y = 8 15

7. x + y = 8 2x + 2y = 168. 3x + y = 9 x – 2y = -49. 3x + y = 1 6x + 2y = 2 16

10. x + y = 6 x–y=2C. Solve the following system of linear equations and match with the graphs below:.1. y = -x + 3 6. x = 2y – 3 y = 2x + 3 y=x+62. 2x – y = 9 7. x + y = 6 5x + y = 14 2x – y = 33. 2x + 5y = 0 8. 2x – y = 12 3x + 4y = -7 x+y=34. 3x – y = 1 9. x + y = 3 2x + y = 4 2x – y = 95. 2x – y = -10 10. 3x + y = 6 x+y=1 2x – y = 4AB 17

CDEFGH 18

IJ LET’S SUMMARIZE1. Systems of linear equations are two or more equations taken together.2. Kinds of systems of linear equations: a. Consistent and independent – the graphs of the two equation intersect at only one point. That point of intersection is the solution of the system of equations. b. Inconsistent – the graphs of the two equations do not intersect. They are parallel. The system has no solution. c. Consistent and dependent – the graphs of the two equations coincide. The system have infinite or many solutions.3. A solution for a linear system of equations in two variables is an ordered pair of real numbers (x, y) that satisfies both equations in the system.What have you learnedA. Identify the graphs of the following pairs of linear equations in twovariables if they intersect, coincide or are parallel.1. x + y = 4 2. 3x + y = 1 2x + sy = 8 2y = 3 – 6x 19

3. 3x + y = 9 4. x + y = 6 x – 2y = 4 x–y=2B. Complete the statement:5. ___________ a system of linear equations with only one solution.6. ___________ a system of linear equations with no solution.7. ___________ a system of linear equations with infinite number ofsolutions.C. Identify the kinds of system of the following and solve by graphing.8. 3x – y = 2 x+y=69. x + 2y = 4 x–y=110. 6x – 3y = 12 y = 2x – 4Answer KeyHow much do you know B.A. 5. consistent and independent 6. inconsstent 1. coincide 7. consistent and independent 2. parallel 3. intersect 4. intersectC.8. Infinite solutions Dependent system 20

9. No solution Inconsistent 10. Solution: (5, 1) Consistent and independentTry this outLesson 2A. 1. Inconsistent 2. Independent and consistent 3. Dependent 4. Intersect 5. Parallel 6. Coincide 7. System of Linear Equations 8. Independent and consistent 9. Inconsistent 10. DependentB. 1. H Independent 21

2. Y Inconsistent 3. P Dependent 4. A Inconsistent 5. T Independent 6. I Independent 7. A Inconsistent Answer: H Y P A T I A 1 234567C.Answer: PASCALLesson 3 6. (3, -1)A. 7. no solution 8. no solution 1. (3, -1) 9. (3, -1) 2. (0, 30 10. no solution 3. (-6, -4) 4. (3, 1) 6. 5. (-3, -1)B.1. 22

2. 7.3. 8.4. 9. 23

5. 10.C. 1. C 2. H 3. E 4. F 5. J 6. I 7. B 8. A 9. G 10. DHow much have you learnedA. 1. Coincide 2. Intersect 3. Coincide 4. parallelB. 5. When the system is consistent and independent 6. When the system is consistent and dependent 7. When the system is inconsistentC. 8. Solution: (2, 4) Independent and consistent system. 24

9. Solution: (2, 1) Consistent and independent system10. Infinite solutions Dependent system 25

Module 1 Rational Algebraic Expressions What this module is about This module is about rational algebraic expressions, or simply rational expressions.As you go over the exercises, you will develop skills in: finding the domain of the givenrational expressions, finding the values that will make a rational expressions meaningless,finding numerical values for the given values of x, and simplifying rational algebraicexpressions. Treat the lessons with fun and take time to go back and review if you thinkyou are at a loss. What you are expected to learnThis module is designed for you to: 1. Define and illustrate rational expressions 2. Find the domain of a rational expression 3. Find the values that will make rational expressions meaningless 4. Find the numerical values of rational expressions for the given values of the variable/s 5. Recall how to factor polynomials 6. Simplify rational algebraic expressions How much do you know1. In _3x__, the value of the variable that must be excluded is2x – 1a. 2 b. –2 c. 1 d. – 1 2 22. The domain of the variable x in the rational expression in #1 isa. {x Є R  x ≠ 2} b. {x Є R  x ≠ –2} c. {x Є R  x ≠ 1 } d. {x Є R  x ≠ – 1 } 223. Find the value of the variable that must be excluded in m2 – 4m + 4. m2 + m – 6

4. Give the domain of the variable m in the rational expression m2 – 4m + 4. m2 + m – 65. Find any value for variable p which 3x – 24 is meaningless. 5p – 256. Find the numerical value of a rational expression – 5x + 1 when x = 2. 2x7. Find the numerical value of a rational expression 8x2y _ when x = –1 and y = 3. –3x + 2y8. Rename – 8y6z5x4 in lowest term. 6y7z3x59. Factor a2 + ac – ab – bc by grouping and then rename into lowest term. a2 – b210. Simplify r2 – r – 20 . r2 + r – 30 What you will do Lesson 1Definition of Rational Expression and Finding the Domain of its Variable Algebraic expressions containing variables that are written in fractional form of theform { p p, q are polynomials, the polynomial q ≠ 0} are called rational expressions. qExamples:a + b , y 6 ( same as y 6 ), 8k3– 3k2 + 2k + 6 , 49a2bc , _1 and 13. a–b 1 4k2 + 5k 21b5 w5 Domain of Rational Expressions: The denominators of the above expressions should not be equal to zero becauseany value assigned to a variable that results in a denominator of zero will make theexpression meaningless and must be excluded from the domain of the variable.For 7a, exclude b = 0. The domain of the expression is the set of b real numbers { b Є R  b ≠ 0} except 0. 2

For 12q – 24, exclude q = – 6. The domain of the variable q is the set of q+6 real numbers { q Є R  q ≠ – 6} except – 6.For m2 – 5 , exclude n = 2 and n = 3. n2 – 5n + 6 The factors of n2 – 5n + 6 is (n – 3)(n – 2) and a value of n = 2 and n = 3 will give azero denominator. Thus, the domain of the variable n is the set of real numbers except 2or 3, or both. In set notation, {n Є R  n ≠ 2, 3}.If n is equal to both 2 and 3, m2 – 5 m2 – 5 m2 – 5 m2 – 5_ m2 – 5 is meaningless.n2 – 5n + 6 (n – 3)(n – 2) (3 – 3)(2 – 2) (0)(0) 0If n is equal to 3, m2 – 5 m2 – 5 m2 – 5 m2 – 5_ m2 – 5 is meaningless.n2 – 5n + 6 (0 ) (n – 2) 0 (n – 3)(n – 2) (3 – 3)(n – 2)If n is equal to 2, m2 – 5 m2 – 5 m2 – 5 m2 – 5_ m2 – 5 is meaningless.n2 – 5n + 6 (n – 3)(n – 2) (n – 3)(2 – 2) (n – 3)(0) 0Examples For each algebraic expression, state the values of the variable that must beexcluded and give the domain of the variable.1. 13m_ m+8Exclude the values for which m + 8 = 0. m +8=0m +8–8=0–8 m=–8 Therefore, m cannot be equal to – 8 and the Domain = {m Є R m ≠ – 8}2. 36x2 54xyExclude the values for which 54xy = 0.Either x and y cannot be equal to 0 and the Domain = {x and y Є R x, y ≠ 0}. 3

3. 2a – 3__ a2 – a – 12Exclude the values for which a2 – a – 12 = 0.Factor the polynomial and set each factor equal to 0. a2 – a – 12 = 0(a – 4)(a + 3) = 0a–4=0 a+3=0 a = 4 or a = – 3Therefore, a cannot be equal to 4 or – 3 and the Domain = {a Є R  a ≠ 4 or a ≠ – 3}4. t – 5 t = – 7 will make the denominator equal 0 t+7 the expression is meaningless for t = – 7.5. ___9b2 ___ To find the number/s that make the denominator 0, b2 – 5b + 6 solve the equation b2 – 5b + 6 = 0. b2 – 5b + 6 = 0 (b – 2)(b – 3) = 0 b – 2 = 0 or b – 3 = 0 b = 2 or b=3Therefore the rational expression is meaningless for 2 and for 3.6. 2p + 9 This denominator can never be equal to 0, so there are no values p2 + 1 for which the rational expression is meaningless.Try this outA. For each algebraic expression, state the values of the variable that must be excluded.1. _13b_ 6. _x (y – 1)_ 39b2 x2(y + 2)2. _21y2z_ 7. _ 25 – p2 _ 49yz3 p2 + p – 303. _ m + 5 _ 8. _c2 – 8c +12_ 7(m – 5) c2 – 6c + 94. _ -5w2xy2 _ 9. _ m2 – 5m + 6 _ x(w2 – 9) m3 – 12m2 + 36m 4

5. _y2 - 4_ 10. _(s + 2)(s – 2)_ y2 – 16 (s2 – 4)( s + 2)B. Give the domain of the variable for each of the following algebraic expressions.1. _13b_ 6. _x (y – 1)_ 39b2 x2(y + 2)2. _21y2z_ 7. _ 25 – p2 _ 49yz3 p2 + p – 303. _ m + 5 _ 8. _c2 – 8c +12_ 7(m – 5) c2 – 6c + 94. _ -5w2xy2 _ 9. _ m2 – 5m + 6 _ x(w2 – 9) m3 – 12m2 + 36m5. _y2 - 4_ 10. _(s + 2)(s – 2)_ y2 – 16 (s2 – 4)( s + 2)C. Find any values for which the following rational expressions are meaningless.1. _3_ 6. _ 8g + 6 _ 5x g2 – 8g + 152. _20m_ 7. _ c + 12 _ m–4 c2 – c – 123. _ 7r_ 8. _ 7k + 21 _ r+9 3k2 – k – 104. _ b2__ 9. _ 8y _ 3b – 2 y2 + 165. _a + 9_ 10. _ 12d _ a2 + 16 d2 – 100 Lesson 2 Finding the Numerical Values of Rational Expressions for the Given Values of the Variable To find the numerical values of rational expressions get the values of x from areplacement set. The set of numbers from which replacements for a variable may bechosen is called a replacement set. 5

Example:1. Find the numerical value of 3x + 6 from the given replacement set x = {1, 2, 3}. 2x – 4 If x = 1, replace x with 1.3x + 6 3(1) + 6 3 + 6 9 _ 92x – 4 2(1) – 4 2 – 4 –2 2If x = 2, replace x with 2.3x + 6 3(2) + 6 6 + 6 12 The rational expression is meaningless.2x – 4 2(2) – 4 4 – 4 0If x = 3, replace x with 3.3x + 6 3(3) + 6 9 + 6 15 or 7 12x – 4 2(3) – 4 6 – 4 2 22. Find the numerical value of ____5x2 __ when x = {-1, 0, 2, 4}. 6 – 3x – x2 If x = –1, replace x with –1.__ 5x2 ____5(–1)2 __5(1) __ 56 – 3x – x2 6 – 3(–1) – (–1)2 6 + 3 – (1) 8If x = 0, replace x with 0.____5x2 ____5(0)2 006 – 3x – x2 6 – 3(0) – (0)2 6If x = 2, replace x with 2.____5x2 ____5(2)2 __5(4) __ _20_ – 56 – 3x – x2 6 – 3(2) – (2)2 6–6–4 –4If x = 4, replace x with 4.___5x2 ___5(4)2 __5(16) _ __80__ _80_ 40 or -3_76 – 3x – x2 6 – 3(4) – (4)2 6 – 12 – 16 6 – 28 – 22 -11 11Try this outA. Find the numerical value of each expression when x= {-2, 0, 2}. 6

1. _4x – 5 6. _ x + 8 _ 6x x2 – 4x + 22. _ 3x _ 7. _ 2x + 1 _ – 4x + 1 x2 – 7x +33. 7x – 2x2 8. __– 2x2 _ _ 8x 8 + x – x24. _ x2__ 9. _ 2x + 5 _ 3x2 – 12 x2 + 3x –105. _(– 8x) 2 10. _ 3x – 7 _ 3x + 9 2x2 – 3x – 2B. Find the numerical value of each expression when a = 3, b = – 2 and c = 4.1. _2a – 3b 6. _ a2 – 9 _ 6c b2 – 2b + 12. _7b + 7c _ 7. _ abc + 1 _ – 7a + 1 a2 – 7a +123. 9a – 5b2 8. __– 2ab2c3 _ _ 8c 2c + c – c24. _ ab2__ 9. _ 2ab + c2 __ 3b2 – c2 a2 + 3a –105. _(3ab2) 2 10. _ 3b3 – 7 _ 2c + 3b3 2c2 – 3c – abC. Given the replacement set x = {Set of positive integers}, find the value that will make thefollowing rational expressions meaningless.1. _2x2 – 3 6. x2 + 8x + 16_ – 2x2 – 8 x2 – 8x + 72. _ x – 9x2 _ 7. _10x2 + 1 _ – 4x + 12 x2 – 4x +33. x3 – 5x2 8. __x2 + x – 6 _ 8x3 – 1 25 – x24. _ x5__ 9. _ x2 + 5x + 4 _ 3x2 – 12 x2 + 18x + 81 7

5. _(2x) 2 10. _ x4 – 64 __ 3x + 9 2x3 – 3x2 – 2x Lesson 3 Recalling How to Factor Polynomials Before you proceed to simplifying rational expressions, you must review first someconcepts that will make your task easier.Let’s have another look at factoring polynomials and get connected!Greatest Common Factors of Polynomials A polynomial, specifically monomial, is written in factored form when it is expressedas the product of prime numbers and variables where no variables has an exponentgreater than 1.Examples1. Factor 45a2b4.45a2b4 = 5 • 9 • a • a • b • b • b • b The GCF of two or more monomials is the product of their common factors, wheneach monomial is expressed as a product of prime factors.2. Find the GCF of 12a3bc4 and 30ab2c3.12a3bc4 = 2 • 2 • 3 • a • a • a • b • c • c • c • c30ab2c3 = 2 • 3•a • b • c • c • c • 5•bGCF= 2 • 3 • a • b • c • c • c = 6abc3The GCF of 12a3bc4 and 30ab2c3 is 6abc3.Factoring Using the Distributive Property3. Use the distributive property to factor 24x3yz2 + 36xy2z3. First find the GCF for 24x3yz2 + 36xy2z3. Note that 12 is the largest number that willdivide evenly into 24 and 36, the numerical coefficients of the polynomials. Also, xyz2 isthe largest factor of x3yz2 and xy2z3.24x3yz2 + 36xy2z3 = 12xyz2(2x2) + 12xyz2(3yz) = 12xyz2 (2x2 + 3yz) 8

This process is called factoring out the greatest common factor (GCF).4. Factor 24a4b2– 6a3b3 + 18a2b4. The GCF of the numerical part is 6. To find the variable part, write each variablethe least number of times it appears in any term. Thus a2 is the least power of a thatappears, while b2 is the least power of b that appears. Using this, you can see that 6a2b2 isthe greatest common factor, and so you have24a4b2– 6a3b3 + 18a2b4 = 6a2b2(4a2) – 6a2b2(ab) + 6a2b2 (3b2) = 6a2b2(4a2 – ab + 3b2) The factored form of 24a4b2– 6a3b3 + 18a2b4 is 6a2b2(4a2 – ab + 3b2).5. Factor –m3 + 7m2 – 11m. There are two ways to factor this polynomial, both of which are equally acceptable.You can use m as the greatest common factor, so you have–m3 + 7m2 – 11m = m (–m2) + m (7m)+ m (–11) = m (–m2 + 7m – 11). Alternatively, we can use –m as greatest common factor and write–m3 + 7m2 – 11m = –m (m2) + –m (–7m)+ –m (11) = –m (m2 –7m +11). Sometimes, in a particular problem, there will be reason to prefer one of theseforms over the other, but both are equally correct.6. Find the greatest common factor of –16(p + 5)5 – 48(p + 5)4 + 8(p + 5)3 and factor. Here the greatest common factor is 8(p + 5) 3, and you have –16(p + 5)5 – 48(p + 5)4 + 8(p + 5)3 = 8(p + 5)3 [ –2(p + 5) 2 – 6(p + 5) + 1] We can also use here –8(p + 5)3 as the GCF.–16(p + 5)5 – 48(p + 5)4 + 8(p + 5)3 = –8(p + 5)3 [ 2(p + 5) 2 + 6(p + 5) – 1 ]Factoring by Grouping Polynomials with four or more terms, like 8x2y –5x – 24xy + 15, can be factored bygrouping terms of polynomials. One method is to group the terms into binomials that caneach be factored using the distributive property.8x2y –5x – 24xy + 15 = (8x2y –5x) + (– 24xy + 15) = x (8xy – 5) + (–3) (8xy – 5) 9

Then use the distributive property again with a binomial as the common factor.Notice that (8xy – 5) is the common factor, hence (x – 3) (8xy – 5)7. Factor 3pw – 21w + 5p – 35.3pw – 21w + 5p – 35 = (3pw – 21w) + (5p – 35) = 3w (p – 7) + 5 (p – 7) = (3w + 5) (p – 7) Sometimes you can group the terms in more than one way when factoring apolynomial. Like 3pw – 21w + 5p – 35, we could have factor it in this way.3pw – 21w + 5p – 35 = 3pw + 5p – 21w – 35 = p (3w + 5) – 7(3w + 5) = (p – 7) (3w + 5)8. Factor 6x2 – 6xy + 3xz – 3zy in two different ways.1st Method: 6x2 – 6xy + 3xz – 3zy = (6x2 – 6xy) + (3xz – 3zy) = 6x(x – y) + 3z( x – y) = (6x + 3z) (x – y) = 3(2x + z) (x – y)2nd Method: 6x2 – 6xy + 3xz – 3zy = (6x2 + 3xz) – (3zy + 6xy) = 3x(2x + z) – 3y(z + 2x) = (3x – 3y) (2x + z) = 3(x – y) (2x + z)9. Factor 15a – 3ab + 4b – 20. Factoring out –1 will make 5 – b15a – 3ab + 4b – 20 = (15a – 3ab) + (4b – 20) equal to b – 5. = 3a ( 5 – b) + 4 (b – 5) = –3a ( b – 5) + 4 (b – 5) = (–3a + 4) (b – 5)Factoring Quadratic TrinomialsIf two binomials are multiplied, each binomial is a factor of the product. Consider the binomials 2a + 3 and 3a + 9. You can use the FOIL (First OutsideInside Last terms) method to find their product.F OI L(2a + 3)(3a + 9) = (2a)(3a) + (2a)(9) + (3)(3a) + (3)(9)= 6a2 + 18a + 9a + 27= 6a2 + (18 + 9)a + 27= 6a2 + 27a + 27The binomials 2a + 3 and 3a + 9 are factors of 6a2 + 27a + 27. . 10

When using the FOIL method, take note of the product of the coefficients of the firstand last terms, 18 and 9. Notice that it is the same as the product of the two terms, 18 and9, whose sum is the coefficient of the middle term. You can use this pattern to factorquadratic trinomials, such as 6x2 + 23x + 20.10. Factor 6x2 + 23x + 20.6x2 + 23x + 20  The product of 6 and 20 is 120.6x2 + ( __ + __ )x + 20  You need to find two integers whose product is 120 and whose sum is 23. How about using the guess-and-check strategy to find these numbers.Factors of 120 Sum of Factors1, 120 1 + 120 = 121 no2, 60 2 + 60 = 62 no3, 40 3 + 40 = 43 no4, 30 4 + 30 = 34 no5, 24 5 + 24 = 29 no6, 20 6 + 20 = 26 no8,15 8 + 15 = 23 yes6x2 + (8+15)x + 20  select the factors 8 and 156x2 + 8x +15x + 20(6x2 + 8x) + (15x + 20) group terms that have common monomial factor factor2x(3x + 4) + 5(3x + 4) use the distributive property(2x + 5)(3x + 4)Therefore, 6x2 + 23x + 20 = (2x + 5)(3x + 4)11. Factor 6x2 + 7x – 20. Follow the given procedure.a. Identity the first, the middle and the last terms First term: 6x2 Middle term: + 7x Last Term: -20b. Find the possible factors for 6x2, the first term and – 20, the last term whose cross- products will give you a sum of +7x, the middle term.Possible factors: 2. (3x + 10)(2x – 2) 3. (3x – 4)(2x + 5)1. (6x – 4)(x + 5) +20x -8x - 6x -4x +14x +15x +30x + 7x +26x 11

c. The only factors that gives the middle term is +7x is (3x – 4)(2x + 5)d. Therefore the factors of 6x2 + 7x – 20 are (3x – 4) and (2x + 5).Factoring the Difference of Two Squares You have learned in Elementary Algebra that (a + b)(a – b) = a2 – b2. This product,called the difference of two squares can be used in factoring. 100a2 – 49b2 can be expressed as the difference of two squares by writing (10a) 2 –(7b) 2, which can be factored as (10a + 7b)(10a – 7b).12. Factor 81k2 – 25d2. ____ ____81k2 – 25d2 = (9k) 2 – (5d) 2 √81k2 = 9k and √25d2 = 5d = (9k – 5d)(9k + 5d)Factoring Perfect Square Trinomial Since (a + b) 2 = a2 + 2ab + b2, the trinomial a2 + 2ab + b2 is the square of thebinomial a + b. Likewise (a – b) 2 = a2 – 2ab + b2, the trinomial a2 – 2ab + b2 is the squareof the binomial a – b. For this reason a2 ± 2ab + b is called a perfect square trinomial.We can use this pattern to factor any perfect square trinomial.13. Factor 144a2 – 120ab + 25b2.From the given pattern (a – b) 2 = a2 –2ab + b2, you have144a2 – 120ab + 25b2 = (12a – 5b)214. Factor 16p2 + 80pq + 100q2.From the given pattern (a + b) 2 = a2 + 2ab + b2, you have16p2 + 80pq + 100q2 = (4p + 10q)2Factoring the Difference of Two Cubes The difference of two cubes a3 – b3 can be factored by writinga3 – b3 = (a – b)( a2 + ab + b2).15. Factor n3 – 8. ___ __ ³√ n3 = n and ³√ 8 = 2n3 – 8 = (n – 2)[n2 + (n)(2) + (2)2] = (n – 2) (n2 + 2n + 4)16. Factor 27b3 – 8m3.27b3 – 8m3 = (3b – 2m)[(3b)2 + (3b)(2m) + (2m)2] = (3b – 2m)( 9b2 + 6bm + 4m2) 12

Factoring the Sum of Two Cubes The sum of two cubes a3 + b3 can be factored by writinga3 + b3 = (a + b)( a2 – ab + b2).17. Factor w3 + 64.w3 + 64 = (w + 4) [w2 – (w)(4) + (4)2] = (w + 4) ( w2 – 4w + 16 )18. Factor 27y3 + 1000z3.27y3 + 1000z3 = (3y + 10z) [ (3y)2 – (3y)(10z) + (10z)2] = (3y + 10z) ( 9y2 – 30yz + 100z2)Let’s summarize the methods of special factorizationDifference of Two Squares a2 – b2 = (a + b)(a – b)Perfect Square Trinomials a2 + 2ab + b2 = (a + b) 2 a2 – 2ab + b2 = (a – b) 2Difference of Two Cubes a3 – b3 = (a – b)(a2 + ab + b2)Sum of Two Cubes a3 + b3 = (a + b)(a2 – ab + b2)Try this outA. Factor the following polynomials using common monomial factor. 1. 8t2 + 16t 2. 21a5 + 14a3 3. 5x4z3 + 25x3z2 – 50x2z 4. 65m9 – 35m5 5. – 121p4q2r + 66p2r4B. Factor completely. 1. 6(2 – m)3 – 12(2 – m)5 2. 7(x – y)3 + 21(x – y)5 – 14(x – y)7 3. 5(3n + 4)2 + 15(3n + 4)3 + 25(3n + 4)4 4. 15(x – 4y)5 – 60(x – 4y)7 + 120(x – 4y)9 5. – 18(2x2 – 5y)10 + 72(2x2 – 5y)4 – 9(2x2 – 5y)8C. Factor by grouping. 1. 2ax + 6xc + ba + 3bc 2. 3ax – 6bx + 8b – 4a 3. 2ab + 2am – b – m 4. 6mx – 4m + 3rx – 2r 5. 5a2 – 4ab + 12b3 – 15ab2 13

D. Factor the following trinomials. 1. 100x2 – 90x + 20 2. 6x3 + 12x2 – 90x 3. –3x4 – 6x3 + 72x2 4. 13y3 + 39y2 – 52y 5. –4x3 – 24x2 + 64xE. Factor completely. 1. 4x2 – 25y2 2. 100n2 – 64m2 3. 25h2 – 20hy + 4y2 4. x3y + 6x2y2 + 9xy3 5. 49a2 + 28ab + 4b2 6. 9e2 + 24ep + 16p2 7. 8k3 – 1 8. 125g3 – 64a3b3 9. 27r9 + 64p3 10. 1000y3 + 343h6 Lesson 4Simplify Rational Expressions The fundamental property of rational expressions permits us to write a rationalexpression in lowest terms, in which numerator and denominator have no common factorother than 1.Examples 1. Write in lowest terms.a. 30 b. _14a4b3 _ 72 2a2b2Begin by factoring. _14a4b3 _ __2 • 7 • a • a • a • a • b • b • b30 __2 • 3 • 572 2 • 2 • 2 • 3 • 3 2a2b2 2•a•a•b•bGroup any factors common to the numerator and denominator.30 __5 • (2 • 3) _14a4b3 _ _7 • a • a • b • (2 • a • a • b • b)_72 2 • 2 • 3 • (2 • 3) 2a2b2 (2 • a • a • b • b) Use the fundamental property.30 __5 • (2 • 3) _14a4b3 _ __7 • a • a • b • (2 • a • a • b • b)_72 2 • 2 • 3 • (2 • 3) 2a2b2 (2 • a • a • b • b) 14

30 __5____ _5_ _14a4b3 _ __7 • a • a • b 7a2b72 2 • 2 • 3 12 2a2b2 12. Write the rational expression 3x – 12 in lowest terms. 5x – 20Begin by factoring both numerator and denominator.3x – 12 3 (x – 4) 35x – 20 5 (x – 4) 53. Write the rational expression x2 + 2x – 8 in lowest terms. 2x2 – x – 6Begin by factoring both numerator and denominator.x2 + 2x – 8 _(x + 4)(x – 2)_ _x + 4_2x2 – x – 6 (2x + 3)(x – 2) 2x + 34. Write a – b in lowest terms. b–a At first glance, there’s no way in which we can factor a – b and b – a to get acommon factor. However, b – a = –1(– b + a) = –1(a – b).With these factors, use the fundamental property to simplify the rational expressions.a – b _1(a – b) _ _1_ –1  the quotient of two expressions that are exactlyb – a –1(a – b) –1 opposite in sign is –1.5. Express 8m2 + 6m – 9 in lowest terms. 16m2 – 98m2 + 6m – 9 (2m + 3)(4m – 3) 2m + 3 16m2 – 9 (4m + 3)(4m – 3) 4m + 36. Write p3 + r3 in lowest terms. p2 – q2p3 + r3 (p + q)(p2 – pq + q2)p2 – q2 (p + q) (p – q) p2 – pq + q2 p–q 15

Try this outA. Write each expression in lowest terms.1. _12k2 6. _3y2 – 3y _ 6k 2(y – 1)2. 12a2b5 _ 7. _9p + 12 _ –48a3b2 21p + 283. 6(b + 5)_ 8. _k2 – t2_ 18(b + 5) k+t4. _12x2 – 9_ 9. _11s – 22s2 _ 3 6 –12s5. _2m – 6 _ 10. _ x + 5 _ 5m – 15 x2 + 3x – 10B. Write each expression in lowest terms.1. _4a2 – b2 6. _p2 – q2 _ (2a + b) q–p2. _m2 – 4m + 4 7. _ b2 – 4b_ m2 + m – 6 4b - b23. a2 – a – 6_ 8. _w2 – 2w – 15 _ a2 + a – 12 w2 + 7w + 124. _x2_+ 3x – 4 9. _c2 + c – 30 _ x2 – 1 c2 – 6c + 55. _8n 2 + 6n – 9_ 10. _d4 – r4 _ 16n2 – 9 d+rC. Simplify the following rational expressions and express your answer in lowest terms.1. _4a2 – 20a 6. _g3 + c3 _ a2 – 4a – 5 g2 – c22. _ m2 – 4___ 7. _ a2 + ac – ab – bc _ m2 + 4m + 4 a2 – b23. 12 – 7x + x2 _ 8. _m2 – mp + mn – np _ (x – 3)(4 + x) m2 + 2mn + n2 16

4. _ n2_– p2 9. _xy – yw + xz – zw _ p–n xy + yw + zx + zw5. _r 2 – r – 20 _ 10. ac + ad – bc – bd _ r2 + r – 30 ab + ac – b2 – bc Let’s Summarize Rational expression is the quotient of two polynomials with denominator not equal tozero. Rational expression is any element of the set { p  p, q are polynomials, the qpolynomial q ≠ 0}. Any value/s assigned to a variable that results in a denominator of zero will make arational expression meaningless. To find the numerical values of rational expressions you will get the values of x fromthe replacement set. The set of numbers from which replacements for a variable may bechosen is called a replacement set. Factoring is the process of writing an indicated sum as a product of factors. Factoring a polynomial or finding the factored form of a polynomial means to find itscompletely factored form. To find the variable part of the polynomial you are factoring, write each variable theleast number of times it appears in any terms of the polynomial. The fundamental property of rational expressions permits us to write a rationalexpression in lowest terms, in which numerator and denominator have no common factorother than 1. The quotient of two expressions that are exactly opposite in sign is –1. What have you learned1. In _6x2 + x__ , the value of the variable that must be excluded are ______. 4x2 – 162. Give the domain of the variable x in the rational expression __a2 + 12__. a2 + 7a + 12 17

3. Find the value of the variable that must be excluded in (3 – m)(4 – m). m2 – m – 124. Give the domain of the variable m in the rational expression _ x2 – y2 . x2 – 2x – 245. Find any value for variable p which 2x2 – 2 is meaningless. 4x2 – 646. Find the numerical value of a rational expression 2x2 – 5x + 3 when x = – 3. 3x2 – 5x + 27. Find the numerical value of a rational expression 7a2b3 _ when a = –2 and b = –1. –3a2 + 4b8. Give the lowest term of - m7n4p8 _ . -18m12n2p69. Simplify _y4 – 13y2 + 36_ and then rename into lowest term. y2 + 5y + 610. Simplify _x4 – 16 _ . x4– 8x2 + 16 18

Key to correctionHow much do you know1. 2x – 1 = 0 6. – 5(2) + 1 -9 -2 1 x = 1 letter c 2(2) 4 4 2 7. ___8(–1) 2 (3)__ 8 (1)(3) 24 2 22. c (–3)( –1) + 2(3) 3+6 9 33. m2 + m – 6 = 0 8. – 4z2 (m – 2)(m + 3) = 0 3xy m = 2, m = – 34. { m Є R m ≠ 2, 3} 9. (a2 + ac) + (–ab –bc) a(a + c) – b(a + c) (a – b)(a + b) (a – b)(a + b)5. 5p – 25 = 0 5(p – 5) = 0 10. (r – 5) (r + 4) r + 4 (a – b)(a + c) = a + c p=5 (r – 5) (r + 6) r + 6 (a – b)(a + b) a + bTry this outLesson 1A. 6. x = 0; y + 2 = 01. b = 0 y=–22. y = 0; z = 0 7. (p + 6)(p – 5) = 0 p=–6;p=53. m – 5 =0 8. (c – 3)(c – 3) = 0 m=5 c=34. w2 – 9 = 0 9. m(m – 6)(m – 6) = 0 w2 = 9 m = 0; m = 6 w2 = 9 w = 3; w = –3 ; x = 05. y2 – 4 = 0 y2 – 16 = 0 10. (s – 2)(s + 2)(s + 2) = 0 y2 = 4 y2 = 16 s = 2; s = – 2y2 = 4 y 2 = 16 y = 2; y = –2; y = 4; y = –4 19

B. 6. Domain = {x, y Є R x ≠ 0; y ≠ -2}1. Domain = {b Є R b ≠ 0}2. Domain = {y, z Є R y, z ≠ 0} 7. Domain = {p Є R p ≠ -6, 5}3. Domain = {m Є R m ≠ 5} 8. Domain = {c Є R c ≠ 3}4. Domain = {x, w Є R x ≠ 0; x ≠ 3, -3} 9. Domain = {m Є R m ≠ 0, 6}5. Domain = {y Є R y ≠ 4, -4} 10. Domain = {s Є R s ≠ 2, -2}C. 6. (g – 5)(g + 3) = 01. x = 0 g = 5; g = -32. m = 4 7. (c – 4)(c + 3) = 0 c = 4; c = -33. r = -9 8. (3k + 5)(k – 2) = 0 3k = -5; k = 24. 3b = 2 k -5 b2 3 3 9. y2 + 16 = 0 y2 = -16 No value.5. a2 + 16 = 0 10. d2 – 100 = 0 a2 = -16 d2 = 100It is impossible to extract d = 10; d = -10square root of negative number. No valueLesson 2A. x=-2 x=0 x=21 4(-2) – 5 -8 - 5 -13 11 4(0) – 5 – 5_ 4(2) – 5 8 – 5 3 1 6(-2) -12 -12 12 6(0) 0 6(2) 12 12 4 meaningless2 __3(-2)__ -6 - 2 __3(0)__ 0 0 __3(2)__ 6 - 6 -4(-2) + 1 9 3 -4(0) + 1 1 -4(2) + 1 -7 7 x=-2 x=0 x=23 7(-2) – 2(-2) 2 -22 1 3 7(0) – 2(0) 2 0 7(2) – 2(2) 2 6 3 8(-2) -16 8 8(0) 0 8(2) 16 8 meaningless4 __(-2) 2__ __4__ 4 __(0) 2__ _0_ 0 __ (2) 2_ __4__ 4 3(-2)2 -12 12 -12 0 3(0)2 -12 -12 3(2)2 -12 12 -12 0 meaningless meaningless 20

5 [ -8(-2)] 2 (16) 2 85 1 [ -8(0)] 2 (0) 0 [ -8(2)] 2 (-16) 2 17 1 3(2) + 9 6 + 9 153(-2) + 9 -6 + 9 3 3(0) + 9 96 ___-2 + 8____ 6 3 ___0 + 8____ 8 4 ___2 + 8____ 10 -5 (-2)2 – 4(-2) +2 10 5 (0)2 – 4(0) +2 2 (2)2 – 4(2) +2 -27 __2(-2) + 1_ -3 - 1 __2(0) + 1_ 1 __2(2) + 1_ 5 -5 (-2)2 -7(-2)+3 21 7 (0)2 -7(0)+3 3 (2)2 -7(2)+3 - 7 78 ___-2(-2) 2___ -8 – 4 ___-2(0) 2__ 0 0 ___-2(2) 2__ -8 -1 18 +(-2) –(-2) 2 2 8 +(0) –(0) 2 8 8 +(2) –(2) 2 6 39 __2(-2) + 5__ -4+5 -1 __2(0) + 5_ 5 -1 __2(2) + 5__ 4 + 5 9(-2)2+3(-2) -10 -12 12 (0)2+3(0) -10 -10 2 (2)2+3(2) -10 10 -10 0 meaningless10 __3(-2) – 7___ -6-7 -13 _3(0) – 7___ - 7 3 1 __3(2) – 7__ 6 – 72(-2)2 - 3(-2) -2 12 12 2(0)2- 3(0) -2 - 2 2 2(2)2- 3(2) -2 0 meaninglessB. 12_ 1 1 931. _2(3) – 3(-2)__ __6 + 6__ (-2) 2 – 2(-2) + 1 4 + 4 + 12. _7(-2) + 7(4) _-14 + 28_ _14_ -_7_– 7(3) + 1 -21 + 1 -20 103. 9(3) – 5(-2)2 27 – 5(4) 27 – 20 _7_ 8(4) 32 32 324. _ 3(-2)2___ __3(4)___ 12__ 12 -3 3(-2)2 – (4)2 3(4) – 16 12 – 16 -45. _[3(3)(-2)2] 2 __(36)2_ 1296 -81 2(4) + 3(-2)3 8 + 3(-8) –166. (3) 2 – 9 9 – 9 _0 0 6(4) 24 247. _ 3(-2)(4) + 1 __-24 + 1__ - 23 meaningless (3)2 – 7(3) +12 9 – 21 + 12 08. _– 2(3)(-2)2(4)3 _ _–6(4)(64)_ –1 536 3842(4) + 4 – (4)2 8 + 4 – 16 –49. __2(3)(-2) + (4)2 –12 + 16_ 4 1 (3)2 + 3(3) – 10 9 + 9 –10 8 2 21

10. _ _ 3(-2)3 – 7 ___ _ ___3(-8) – 7__ –24 – 7 – 31 -1_5_2(4)2 – 3(4) – (3)(-2) 2(16) – 12 + 6 26 26 26C. 6. x2 – 8x + 7 = 01. -2x2 – 8 = 0 No real number value. (x – 7)(x – 1) = 0 x = 7; x = 1 -2x2 = 8 x2 = 8 -2 x2 = -42. -4x + 12 = 0 7. x2 – 7x + 3 = 0 -4x = -12 ( x – 3)(x – 1) = 0 x=3 x = 3; x = 13. 8x3 – 1 = 0 8. - x2 + 25 = 0 8x3 = 1 - x2 = – 25 x2 = 25 3 8x3 = 3 1 x=5 2x = 1 x = 1 Not an integer x = – 5 not a positive integer 24. 3x2 – 12 = 0 9. x2 + 18x + 81 = 0 3x2 = 12 (x + 9)(x + 9) = 0 x2 = 4 x = – 9 not a positive integer x=2 x = – 2 not a positive integer5. 3x + 9 = 0 10. 2x3 – 3x2 – 2x = 0 3x = – 9 x(2x2 – 3x – 2) = 0 x = – 3 not a positive integer x (2x + 1)(x – 2)=0 x=0;x=2Lesson 3 x = – ½ not a positive integerA. 1. 8t(t + 2) 2. 7a3(3a2 + 2) 3. 5x2z(x2z2 + 5xz – 10) 4. 5m5 (13m4 – 7) 5. 11p2r (– 11p2q2 + 6r3)B. 1. 6(2 – m)3 – 12(2 – m)5 = 6(2 – m)3 [ 1 – 2(2 – m)2] 2. 7(x – y)3 + 21(x – y)5 – 14(x – y)7 = 7(x – y)3[1 + 3(x – y)2 – 2(x – y)4] 3. 5(3n + 4)2 + 15(3n + 4)3 + 25(3n + 4)4 = 5(3n + 4)2 [ 1+ 3(3n + 4) + 5(3n + 4)2] 4. 15(x – 4y)5 – 60(x – 4y)7 + 120(x – 4y)9 = 15(x – 4y)5[1 – 4(x – 4y)2 + 8(x – 4y)4] 5. –18(2x2 – 5y)10 + 72(2x2 – 5y)4 – 9(2x2 – 5y)8 = –9(2x2 – 5y)4 [ 2(2x2 – 5y)6 – 8 + (2x2 – 5y)4] 22

C. 1. 2ax + 6xc + ba + 3bc = (2ax + 6xc) + (ba + 3bc) = 2x(a + 3c) + b(a + 3c) = (2x + b)(a + 3c)2. 3ax – 6bx + 8b – 4a = (3ax – 6bx) + (8b – 4a) = 3x(a – 2b) + 4(2b – a) = 3x(a – 2b) – 4 (a – 2b) = (3x – 4)(a – 2b)3. 2ab + 2am – b – m = (2ab + 2am) – (b + m) = 2a(b + m) – (b + m) = (2a – 1)(b + m)4. 6mx – 4m + 3rx – 2r = (6mx – 4m) + (3rx – 2r) = 2m(3x – 2) + r(3x – 2) = (2m + r)(3x – 2) 5. 5a2 – 4ab + 12b3 – 15ab2 = (5a2 –15ab2) + (12b3 – 4ab) = 5a(a – 3b2) + 4b(3b2 – a) = 5a(a – 3b2) – 4b(a – 3b2) = (5a – 4b)(a – 3b2)D. 1. 100x2 – 90x + 20 = (10x – 4)(10x – 5) 2. 6x3 + 12x2 – 90x = x (6x2 + 12x – 90) = x (3x – 9)(2x + 10) 3. –3x4 – 6x3 + 72x2 = –3x2 (x2 + 2x2 – 24) = –3x2 (x + 6)(x – 4) 4. 13y3 + 39y2 – 52y = 13y(y2 + 3y – 4) = 13y (y + 4)(y – 1) 5. –4x3 – 24x2 + 64x = –4x (x2 + 6x – 16) = – 4x (x + 8)(x – 2)E. 1. 4x2 – 25y2 = (2x – 5y)(2x + 5y) 2. 100n2 – 64m2 = (10n – 8m)(10n + 8m) 3. 25h2 – 20hy + 4y2 = (5h – 2y)2 4. x3y + 6x2y2 + 9xy3 = xy (x2 + 6xy + 9y2) = xy (x + 3y)2 5. 49a2 + 28ab + 4b2 = (7a + 2b)2 6. 9e2 + 24ep + 16p2 = (3e + 4p)2 7. 8k3 – 1= (2k – 1)(4k2 + 2k + 1) 8. 125g3 – 64a3b3 = (5g – 4ab)(25g2 + 20abg + 16a2b2) 9. 27r9 + 64p3 = (3r3 + 4p)(9r6 – 12pr3 + 16p2) 10. 1000y3 + 343h6 = (10y + 7h2)(100y2 – 70h2y + 49h4)Lesson 4A. Write each expression in lowest terms.1. _12k2 6k(2k) 2k 6. _3y2 – 3y 3y ( y – 1) 3y 6k 6k 2(y – 1) 2 (y – 1) 2 23

2. 12a2b5 _ _12a2b2(b3)_ b3 7. _9p + 12 _ 3(3p + 4) 3 –48a3b2 2a2b2 (–4a) -4a 21p + 28 7(3p + 4) 73. 6(b + 5)_ 1 8. _k2 – t2_ ( k+ t)( k – t) k – t 18(b + 5) 3 k + t ( k + t)4. _12x2 – 9 3 (4x2 – 3) 4x2 – 3 9. _11s – 22s2 11s(1 – 2s)_ 11s 33 6 –12s 6(1 – 2s) 65. _2m – 6 2(m – 3) 2 5m – 15 5(m – 3) 5 10. _ x + 5 _ x + 5___ 1__ x2 + 3x – 10 (x – 2)(x + 5) x – 2B. Write each expression in lowest terms.1. _4a2 – b2 (2a + b)(2a – b) 2a – b(2a + b) (2a + b)2. _m2 – 4m + 4 (m – 2)(m – 2) m–2 m2 + m – 6 (m – 2)(m + 3) m+33. a2 – a – 6_ (a – 3)(a + 2) a + 2 a2 + a – 12 (a – 3)(a + 4) a + 44. _x2_+ 3x – 4 (x + 4)(x – 1) x + 4x2 – 1 (x + 1)(x – 1) x + 15. _8n 2 + 6n – 9_ (4n – 3)(2n + 3) 2n + 316n2 – 9 (4n – 3)(4n + 3) 4n + 36. _p2 – q2 (p +q)(p – q) –(p + q) = – p – q q–p –1(p – q)7. _ b2 – 4b_ b(b – 4) b(b – 4) –14b - b2 b(4 – b) –b(b – 4)8. _w2 – 2w – 15 (w – 5)(w + 3) w – 5 w2 + 7w + 12 (w + 4)(w + 3) w + 49. _c2 + c – 30 (c + 6)(c – 5) c + 6 c2 – 6c + 5 (c – 1)(c – 5) c – 110. _d4 – r4 (d2 – r2)(d2 + r2) (d – r)(d + r)(d2 + r2) (d – r)(d2 + r2) d+r d+r d+r 24