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  • 1. Advanced Algebra 1 This chapter aims the following objectives: 1. To generalize Polynomial. 2. To familiarize the operation used in polynomial 3. To restate Synthetic Division 4. To determine the zeros of Polynomial 5. To derive Quadratic Formula
  • 2. Advanced Algebra 2 LESSON 1: POLYNOMIALS Objectives To define polynomials. To be aware of the terms of a polynomial. To distinguish the degree of a polynomial. Hi Everybody! I want to introduce to you my friend, Polynomials. He is one variable of an algebraic expression of the form anxn + an-1xn-1 + an-2xn-2 + … + a0, where n is a non-negative integer, and an, an-1, an-2, …,a0 are constants, and an ≠ 0. You want to know him better? Here is an example of a polynomial. x2 + 4xy + 4y2
  • 3. Advanced Algebra 3 After you have learned what polynomial is, you should also know his different terms and degrees. I want you to meet Monomial. He is a polynomial consisting of only one term. Another is Binomial. He is a polynomial consisting of two terms. And the last but not the least is a Trinomial. He is a polynomial consisting of three terms. Now you meet them, I want you to know more about them because it will help you a lot. Here are the examples. Hi there. I ‘am monomial 5x; -2ab; 11a3; 28 And I ‘am binomial 3x2 + y; 8x2y2 – 4; ; a2 + 2 And we are the trinomials. 6x2 + 2xy + y2; m + n – p; x2 + 5x + 6 Yes! This is it. You can now easily determine the terms of a polynomials. You’re now ready to take my challenges. But wait you’re in the half of our study we’re not yet done. You must know what the degree of polynomial is.
  • 4. Advanced Algebra 4 Remember that in determining the degree of a polynomial you must get the highest power of its terms after it has been simplified. 5 is the degree of polynomial because on the term 6x2y x is in the highest power, then x is in the 2nd degree and on the term 12x3y2, x is in the 3rd degree so we add same variables exponent then we will get 5 as the degree of the polynomial. 6x2y + 12x3y2 + 8 And now, this is the moment of truth you’re now ready to solve and answer all my challenges regarding our lesson which is polynomials. I’m sure that you’ll get excellent points if you really understand our lesson.
  • 5. Advanced Algebra 5 NAME: RATING: A. Determine whether the following is a polynomial or not. 1. 3x2 – 5x + 6 ______ 2. A2b2 – 4ab + 8 ______ 8. q + c + c2 ______ 3. x2 – y 2 ______ 9. ______ 4. a3 + b3 + c3 ______ 10. 5x + ______ 5. ______ 11. ______ 6. ______ 12. ______ 7. x1/3 + 5 ______ B. Classify each polynomial according to the number of its terms. 1. 8a + 2 ______ 6. 5x2y – 12xy + y2 ______ 2. -5xy + x + xy ______ 7. xyz ______ 3. +x ______ 8. m + 2n – 3p ______ 4. a2 + 4a + 16 ______ 9. ______ 5. x + y3 ______ 10. 2a3 – 5a2 – 15 ______ C. Give the degree of each of the following polynomials. 1. 4x ______ 5. 3a4b2 + 4ab3 – 6b7 ______ 2. 4x2 + 4x – 8 ______ 6. a3b4 + a2b2 – 6ab3______ 3. 5xy + 8y2 + 13x2 ______ 7. 9 – 6xy + yz + xyz 4. 6x2 + 4x + ______ ______
  • 6. Advanced Algebra 6 8. v- 9. 8xyz2 - ______ ______ 10. m + 9m5n8 + 3m3n5p7 - 7np10 ______ LESSON 2: FUNDAMENTAL OPERATIONS Objectives To explore the addition with subtraction operation. To familiarize student the use of multiplication and division. To perform the synthetic division method. After we discussed the polynomials terms and degree, I will now introduce to you the different operation in solving polynomial expression. Meet addition and subtraction, and follow their rules in solving polynomials.
  • 7. Advanced Algebra 7 Remember that in adding polynomials, simply add the numerical coefficient of like terms. And in subtracting polynomials, just simply change the second polynomial into its additive inverse and then proceed to addition. Here is an example of adding polynomials. (9x4 – x3 + 5x2 – 8x) + (4x4 – x3 – 7x + 7) =? Solution: (9x4 – x3 + 5x2 – 8x) + (4x4 – x3 – 7x + 7) = 9x4 – x3 + 5x2 – 8x + 4x4 – x3 – 7x + 7 = 9x4+ 4x4– x3– x3+ 5x2 – 8x – 7x + 7 = 13x4 – 2x3 + 5x2 – 15x + 7 And here is the example of subtracting polynomials. (15x2 – 4xy + 10y2) – (9x2 + 4xy + 5y2) =? Solution: (15x2 – 4xy + 10y2) – (9x2 + 4xy + 5y2) = 15x2 – 4xy + 10y2 – 9x2 – 4xy – 5y2 = 15x2– 9x2 – 4xy – 4xy + 10y2 – 5y2 = 6x2 – 8xy + 5y
  • 8. Advanced Algebra 8 NAME: RATING: Perform the indicated operations. 1. (6xy – 2x +4) + (11 – 8xy + 6x) 2. (-8mn2 + 12n) – (m3 – 4m2 + 4n) 3. (3b2 – 2b + 9) + (b2 + 6b – 4) 4. (5x2 – 2y + 4z) – (x2 – 3y2 + z) 5. (a2 + 4a + 2) + (2a + 3) 6. (3x2 – y) – (2x2 + 5y) 7. (-7x4y3 – 21x3y3 + 28x5y4) + (7x2y2 + 6xy + 2x2y) 8. (-x2 + 6x – 2) – (x2 – x + 3) – (x + 1) + (x + 2) 9. (3y2 + 3xy + 10) + (4y3 – 10xy – 15) 10. (-2m2 + mn + 5n2) – (-4m2 – 6mn + 3n2) After we have discussed the two operations which are addition and subtraction, we now proceed to Multiplication and division, the other operation in polynomials. You will meet them later in the middle of our study. We all know that mathematics have 4 major operations in solving an equation. Like in arithmetic calculation, polynomials have also those 4 operations. And we formerly discussed addition and subtraction. Now I will introduce to you multiplication.
  • 9. Advanced Algebra 9 Remember that to get the product of polynomial; you must multiply each term of polynomial by each of the other terms of polynomial. Then combine like terms. Here is the example of multiplication for you to understand what multiplication meant to be. (4x + 3) (2x + 5) Solution: (4x + 3) (2x + 5) = ( 4x)(2x) + (4x)(5) + (3)(2x) + (3)(5) = (8x2 + 20x) + (6x + 15) = 8x2 + (20x + 6x) + 15 = 8x2 + 26x + 15 Rules in Dividing Polynomials 1) In dividing polynomials by monomial, divide each term of a polynomial by monomial. 2) In dividing polynomial by another polynomial: a. Arrange the term in descending power with respect to a variable. b. Get the common factor of the dividend and divisor.
  • 10. Advanced Algebra 10 To make it clear, here is the example that clarifies to your curiosity. We factor the dividend with the common factor of the divisor. Solution: = Then I cancel it out = = (a + 2) And this is it. Now you answer again my challenges for me to know that your ability in learning my lesson is satisfy your knowledge. NAME: RATING: A. Perform the indicated operations. 1. (c2 – 16)(c +1) 2. (11p2 – 66p + 99)(p + 1) 3. (a – b)2(a + b)2
  • 11. Advanced Algebra 11 4. (5x2y + 3xy2 – 7x2y3)(xy) 5. (a2 + 2a + 3)(a – 5) + 6. (x + 1)(x + 2)(x + 3) 7. (x2 + x + 2)(2x2 + 3) + 8. 9. 10. (a2n – 3an + 5)(a + 2)(a2 + 4a + 2) B. Find the Quotient. 1. (a2 – 7a + 10) (a – 5) 2. (x3 – 4x2 – 2 + 5x) (x – 1) 3. (3a4 – 2a + 5) (a2 + 3) 4. (2x3 + 5x2 – x – 1) (x – 1) 5. (2x2 – 5x – 6) (2x – 1) LESSON 3: SYNTHETIC DIVISION To define Synthetic division. Objectives of the procedure in using Synthetic division. To be aware To perform synthetic division in solving polynomial function.
  • 12. Advanced Algebra 12 I know that you have a difficulty in getting the common factor to divide the polynomials. I have here a friend that will help you to make your factoring in its easiest and shortest way. I proudly introduce to you Synthetic division. He is a process of division for polynomials in one variable where the divisor is of the form x – c, and c is any real number. So your difficulty in dividing polynomials will lessen through this method. I think that you will use this in the near future of your study in different branches of mathematics.
  • 13. Advanced Algebra 13 Procedure in using synthetic division 1. Arrange the terms into descending power. 2. Copy the numerical coefficient. (If the descending power of the terms is like this x3 + 2x + 1 then the arrangement should be 1 0 2 1: because th the degree of polynomials is in the 4 . ) 3. Substitute the value of x in the divisor. 4. Then, bring down 1st the numerical coefficient w/c is in his 1st term. 5. Multiply to the value of x then subtract product to the 2nd term of numerical Here is the example for you to apply this method. (x3 + 9x2 + 17x – 19) (x + 4) Solution: -4 1 9 17 -19 -4 -20 12 remainder 1 5 -3 -7 The final answer = x2 + 5x – 3 – So, we finally studied this method. And now, you can use it in your mathematics subject. And of course, after we end the lesson we might have a test regarding this lesson. Are you ready? But you should be ready because I rate you according to this scale: NAME: RATING:
  • 14. Advanced Algebra 14 Get the quotient of the following by using the Synthetic Division Method. 1. (x4 + 4x3 + x2 + x + 17) (x – 2) 2. (x4 + 2x2 + x - 11) (x + 5) 3. (x2 + 4x + 21) (x + 7) 4. (2x4 + x3 – x – 12) (x – 2) 5. (x3 – 4x2 + 5x – 2) (x – 1) 6. (x3 – 7x2 – 4x + 24) (x – 6) 7. (3x3 – 2x2 + 5x +1) (x + 5) 8. x5 – 2x4 + 3x3 – 2x2 + 1) (x – 2) 9. (x2 – x – 20) (x – 5) 10. (x3 – 4x2 – 2 + 5x) (x – 4) LESSON 4: ZEROS OF A POLYNOMIAL FUNCTION Objectives To determine the zeros of a polynomials. To be aware in the different techniques in factoring a polynomials. To perform the use of quadratic formula in solving polynomials.
  • 15. Advanced Algebra 15 It’s a long discussion about the operations used in Polynomials. But now another mathematics word you will be meet. I introduced to you the Zeros of polynomial Function. And from the word itself you will be notice that he is pertaining to zero. Zero of a Polynomial Function is the value of t he variable x, which makes the polynomial function equal to zero. And it’s written in symbolic form: f(x) = 0 x = -2; x2 + 4x + 4 f(x) = (-2)2 + 4(-2) + 4 That’s the simplest explanation of it. And I think you already understand it = 4 + (-8) + 4 or maybe I’ll give an example of that. =8–8 =0
  • 16. Advanced Algebra 16 Now, take my challenges. You should get perfect score, because this lesson is very easy anyone can perfect it. So grab it! NAME: RATING: Which numbers -3, -2, -1, 1, 2 and 3 are zeros of the following polynomial functions? 1. f(x) = x3 + 4x2 + x – 6 2. f(x) = 2x3 – 3x2 – 11x + 6 3. f(x) = x3 +3x 2 – x – 3
  • 17. Advanced Algebra 17 4. f(x) = x4 – 4x3 +6x2 – 4x + 1 5. f(x) = x3 – 2x2 – 5x + 6 6. f(x) = x3 – x2 – 9x + 9 7. f(x) = x3 – 3x2 – 5x + 12 8. f(x) = 5x4 + 2x3 – 3x – 3 9. f(x) = 12x3 – 18x2 + 14x + 17 10. f(x) = x3 – 5x2 + 8x – 3 11. f(x) = x2 – 5x + 7 12. f(x) = 8x4 + 5x3 + 2x2 – 6x – 1 13. f(x) = 3x3 – 8x2 + 7x – 6 LESSON 5: FACTORING TECHNIQUES Objectives To determine the different techniques used in factorization To familiarize the use of such techniques To evaluate equation using factorization techniques. Under Zeros of polynomial function we have the factoring. He is the process of expressing a polynomial as a product of factors.
  • 18. Advanced Algebra 18 TECHNIQUES A. Common Factor of Highest Degree D. Perfect Square Trinomial Ex: a2b + a2 = ab(a + b) Ex: a2 + 2ab + b2 = (a + b)2 E. Difference and Sum of Cubes Ex: x3 + y3 = (x + y)(X2 – xy + y2) F. Grouping B. Difference of Squares Ex: am + bm – an – bn = (am + bm) – (an + bn) = m(a + b) – n(a + b) Ex: x2 – y2 = (x + y)(x – y) = (m – n)(a + b) NAME: RATING: Factor the given expressions. 1. a3 + 2a2 + 3a 2. 4x3 – 2x2 + x 3. 9p3q – 51p2q + 216pq2 4. x (x + 8) – 7(x + 8) G. J. M. Perfect Square Trinomial 5. 3z2 – 2z + 6z – 4 Ex: a2 + 2ab + b2 = (a + b)2 6. 6x2 + x + 2 + 12x H. K. N. Difference and Sum of Cubes 7. a2 + 10ab + 25b2 Ex: x3 + y3 = (x + y)(X2 – xy + y2) I. L. O. Grouping Ex: am + bm – an – bn = (am + bm) – (an + bn) = m(a + b) – n(a + b)
  • 19. Advanced Algebra 19 8. x2 – 10xy + 25y2 9. 4c2 – 12cd + 9d2 10. 64xy2 – 9x3 11. (b2 – 2b + 1) – 100d2 12. m4 – 121n4 13. 4a2 – 28ac + 49c2 14. x2 – 3x + 2 15. 27a3 – 8b3 16. 4(x + y) – 64a4y2 17. x3 + 1 18. 16m4 – 25n6 19. 36y4 – 81x4 20. 24a3 + 27a2 + 64a + 12 LESSON 6: QUADRATIC FORMULA Objectives : To derive the quadratic formula To analyze the quadratic formula To use quadratic formula in solving polynomial function
  • 20. Advanced Algebra 20 So you are now familiarized with the different techniques in factoring a polynomial function. Now we have to introduce the formula in getting a polynomial function zero. The zero of a polynomial function is a real number that replace or substitute the value of x in f(x) and make f(x) = 0. This can be obtained through the last topics we discussed like synthetic and factoring techniques. But that method cannot get the exact value of x because of their remainder. And the appropriate method to use is get the exact value of x is the Quadratic formula. Its symbol form is X= Always remember that Quadratic formula is use only in Quadratic form or in the 2nd degree of terms. It always use in ax2 + bx + c = 0 Example:
  • 21. Advanced Algebra 21 F(x) = x2 – 2x – 2 Solution: X= a = 1; b = -2; c = -2 = = = = =1 This is the moment of truth, we done in this formula. So as you go beyond opening this workbook you encounter many challenges that help to develop your skills in Mathematics. And this is another challenge. Answer it carefully, because the one who careless in answering the more mistake he got.
  • 22. Advanced Algebra 22 NAME: RATING: A. Determine whether the given number is a zero of the given polynomial function. 1. f(x) = x3 + 4x – 5; 1 2. f(x) = 3x2 – 2x – 8; 3 3. f(x) = x4 + 3x2 – 2x – 2; 2 4. f(x) = 3x4 – 3x3 – 20x2 + 18; 3 5. f(x) = x4 + x3 – 3x2 – 4x – 4; 2 6. f(x) = x4 – 5x3 + 8x2 + 15x – 2; 3 7. f(x) = 9x3 + 6x2 + 4x + 2; - 8. f(x) = x3 + 2x2 – 25x – 50; 5 B. Find the remaining zeros of the polynomial functions given one zero. Use Quadratic formula if the function is in quadratic form. 1. f(x) = x3 + x2 – x – 1; x1 = 1 2. f(x) = x3 – 3x2 – x + 3; x1 = 1 3. f(x) = x3 + 4x2 + x – 6; x1 = 1 4. f(x) = 3x3 – 4x2 – 13x – 6; x1 = -1
  • 23. Advanced Algebra 23 5. f(x) = 2x3 – 3x2 - - 3x + 2; x1 = -1 6. f(x) = x3 – 4x2 – 7x + 10; x1 = 5 7. f(x) = x4 + x3 – 4x – 4; x1 = -1 8. f(x) = x4 + 5x3 + 5x2 – 5x – 6; x1 = -3 9. f(x) = x4 – 5x3 + 5x2 +5x – 6; x1 = 1 10. f(x) = x4 + 3x3 – x2 + 11x – 4; x1 -4 11. f(x) = x3 + 12x2 + 41x + 42; x1 = -2 12. f(x) = x3 + 2x2 – 2x – 4; x1 = -2 A. Encircle the letter of the correct answer. 1. What is the graph of a linear function? a. ellipse c. circle b. line d. parabola 2. What is the degree of the polynomial -6x4 + x5 – 3x7 + 2x – 1? a. 5 c. 7 b. 4 d. 6 3. How many turning points does the graph of a polynomial function of degree n > 1 have? a. 2n c. n b. n + 1 d. n – 1 4. Given P(x) = 5x3 – 3x2 + x – 7, what is P(2)?
  • 24. Advanced Algebra 24 a. -30 c. 30 b. -23 d. 23 5. What is the remainder if the divisor x – c is a factor of a polynomial function P(x)? a. infinite c. positive b. zero d. negative 6. How many real zeros do P(x) = x4 + 2x3 – 7x2 – 8x - 12 have? a. 3 c. 4 b. 5 d. 7 7. In the polynomial 7x5 + 4x3 – 5x6 + x2 – 3x + 6, the coefficient of the term of highest degree is_____. a. 7 c. -5 b. 1 d. -3 8. If P(x) = x3 + x2 – 5x + 3 is divided by D(x) = x – 1, the quotient Q(x) is_____. a. x2 + 2x – 3 c. 4x3 – 3x2 + 2x – 3 b. x2 + 2x + 3 d. 4x3 – 3x2 – 2x – 3 9. What is the remainder in number 8? a. -6 c. 1 b. 0 d. 6 10. Which of the following describes the graph of f(x) = 2x + 3? a. a line that rises to the right c. a parabola that opens upward b. a line that falls to the right d. a parabola that opens downward B. Solve each of the following: 1. What is the remainder when 2x4 – 5x3 + 3x2 – 4x + 3 are divided by x – 2?
  • 25. Advanced Algebra 25 2. Determine if -2 is a zero of f(x) = x3 + 2x2 – x + 9. 3. If f(x) = x3 – 5x2 + 20x – 16 and f(2) = 0, what can be said about x – 2? 4. What are the zeros of f(x) = x3 – 2x2 – x + 6? 5. If one of the factors of x3 + 5x2 – 3x – 5 is x + 1, what are the other factors? 6. Find all the rational zeros of f(x) = 3x3 + 5x2 – 3x – 5. 7. If x – 1 is a factor of x3 – 2x2 – kx + 6, what is the value of k? C. Challenge 1. If P(x) = 3x4 – x3 + 5x – 7, find P(-2) and P(2). 2. Determine the value of m so that G(-3) = -1 for G(x) = 2x3 – 7x2 + 5x + m. 3. Use synthetic division to find the quotient when P(x) = 2x4 – 3x3 – 12x– 32 is divided by x – 3. 4. What is the remainder if P(x) = 2x2 – 4x -5 is divided by x – 3? 5. Divide P(x) = x3 + 2x2 – 7x – 4 by x – c. Then write the given function in the form P(x) = (x – c) Q(x) + R where Q(x) is a quotient and R is the remainder.
  • 26. Advanced Algebra 26 6. For what values of p and q are x + 1 and x – 2 factors of x3 + px2 +2x + q? 7. Find the polynomial function whose zeros are 2, -2 and 3. 8. If -3 is a zero of the polynomial function P(x) = x3 + 3x2 – 2x – 6, find the other zeros. 9. Find all the zeros of the function P(x) = x3 + 4x2 +x – 6. 10. Find a such that x – 3 will be a factor of x3 – 4x2 +ax – 9. 11. Find the rational zeros of P(x) = 2x4 – 3x3 + 2x2 – x – 3. 12. Which value(s) of k will the polynomial function P(a) = a2 – ka + k + 8 have exactly one real zero?
  • 27. Advanced Algebra 27 This chapter aims the following objectives: 1. To elaborate rational expression 2. To familiarize the operation used in rational expression 3. To synthesize variation 4. To generalize Rational Function
  • 28. Advanced Algebra 28 LESSON 1: RATIONAL EXPRESSION Objectives To define Rational Expression To evaluate rational expression To familiarize in simplifying Rational Expression And finally I will introduce to you RATIONAL EXPRESSION. It is a fraction containing polynomials on the numerator & denominator. This is my numerator having polynomial Hi! I am a Rational Expression!! This is my denom inator Analyze these examples: All of the above examples are rational expressions. Now who is that lowest term? Are you familiar with him? Ok! He is a fraction having numerator & denominator with no common factor aside from one. Remember that RATIONAL EXPRESSION should be in lowest term. In reducing rational expression to lowest term, first, factor the numerator & the denominator, and then divide it by the common factor.
  • 29. Advanced Algebra 29 I am the lowest = term And c is the common factor so I divide the expression by it. R Remember that b0 & c 0. Let us try the following examples: 1. 2. 3. Solution: We get the common factor w/c is. 1. = Then we canceled out or divide the common factor. = = Now, you get me, I am the lowest term! Remember! Here is another one for you to familiarize with reducing rational expression to lowest term. Now, you get again the common factor w/c is x + y. 2. = Then we canceled out or divide the common factor.
  • 30. Advanced Algebra 30 = = You finally get again the lowest term!! Some more? And you will do with your own after this. Common factor. 3. = Cancel out or divide = common factor. Now, you get me, I = am the lowest term! Remember! Now you practice in these 3 examples, you’re ready to take challenge and test yourself how you understand the lesson. NAME: RATING: GET THE LOWEST TERM.
  • 31. Advanced Algebra 31 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
  • 32. Advanced Algebra 32 14. 15. 16. 17. 18. 19. 20. LESSON 2: OPERATION ON RATIONAL EXPRESSION Objectives To define the rule of addition and subtraction To determine the rule on multiplication To be aware on the rule of division To perform the different operations on rational expression.
  • 33. Advanced Algebra 33 The operations used in rational expression similar to ordinary arithmetic operations. Let me introduce to you the first operation. Hi! We are addition and subtraction. And either we have the same procedure or rules in solving problem, you get the sum / difference in two rule which are the common and the no common denominator. Remember that rational expressions with the same denominator can be added or subtracted by simply adding or subtracting their numerator and copy the denominator. Here we are the sum and the difference with common = denominator c. R Remember that c 0. Now you know how to get our sum/difference because we have a common denominator then after you know us, we will introduce to you how to add and subtract rational expressions with different denominator; that isby finding the LCD.
  • 34. Advanced Algebra 34 To find the LCD factor completely the denominators and the product of each unique prime factor raised to the highest power. Then apply the procedures in getting the sum/difference. = = Now you get my sum/difference with This is my LCD! unlike denominator… Let’s try these examples and apply the procedure in our lesson. 1. + - 2. + - 3. - + Solution: If you notice in number 1, we will use the common denominator. 1. + - = We will copy the common denominator w/c is 4a2. = Then we get the sum of the numerator = = Then we simplify by getting the common factor and dividing it w/c is 2. Then you get the sum. You need one more? Here is another one. Get again the 2. + - = common denominator Then add / subtract the same term
  • 35. Advanced Algebra 35 = You get now the sum / difference = You will be challenged more because the next example is a rational expressions having unlike denominators. 3. – + = You will get the LCD of their denominators which is a3. Then I’ll divide the LCD by their denominators and multiply to its numerator. = Then you will combine similar terms. = Now you get their sum / difference
  • 36. Advanced Algebra 36 NAME: RATING: Start to find my sum / difference: 1. – + 2. + – 3. – 4. – 5. + 6. + 7. – – 8. –
  • 37. Advanced Algebra 37 9. – + 10. + – 11. – + 1 12. – 13. + 14. + – 15. – 16. – 17. – 18. + – 19. + – +
  • 38. Advanced Algebra 38 After we, introduced to you addition and subtraction, we now move on to the next level or next operation that will be used in solving rational expression. We all know that the next operation is the hardest among the two operations but now, we made t simpler for you to understand what multiplication and division are all about. In solving rational expressions using multiplication and division, you must remember that there are rules and procedures to be followed. Steps in multiplying rational expression:  Multiply the expression from numerator by numerator and denominator by denominator.  Get the common factor then, cancel out or divide it to eliminate.  Simplify the remaining expression. a x b = ab c d cd
  • 39. Advanced Algebra 39 Steps in dividing rational expression:  Get the reciprocal of the divisor then follow the steps in multiplication. a ÷ b= a x d c d c b ad cb Here are the examples to make it clear for you. 1. 2. x2 + 6x + 9 x2 + 3x – 10 ÷x + 3 x+5 Solution: Multiply numerator by numerator 2 2 2 2 1. 2a d 9b c . (2a d) (9b c) then denominator by denominator 3bc x 16ad2 = (3bc) (16ad2) = (2ad) (3bc) (ab) Find the common factor then divide (3bc) (2ad) (8d) or cancel it = ab 8d Then you get the product of the expression Get the reciprocal of the denominator
  • 40. Advanced Algebra 40 2. x2 + 3x – 10 x + 5 x2 + 3x – 10 x+3 x2 + 6x + 9 ÷ x + 3 = x2 + 6x + 9x x+5 = (x2 + 3x – 10) (x + 3) Then multiply numerator by (x2 + 6x + 9) (x + 5) numerator and denominator by denominator (x + 5) (x – 2) (x + 3) Get the common factor and = (x + 3) (x + 3) (x + 5) divide or cancel it = x–2 You have now the product of x+3 given expression Now, you are fully charge with the examples I have given. So to know if you really understand what I’ am talking about you must answer my challenges so that I can measure your ability to learn.
  • 41. Advanced Algebra 41 Find the product or quotient of the following. 1. x3 + 6x2 + 5x – 12 x x3 – 4 x–1 x2 + 2x – 3 2. a+3 a2 – 9 a2 + a – 12 ÷ a2 + 7a + 12 3. b2 – 25 2b – 10 ÷ (b + 5)2 4b + 20 4. 2x2 – 6x 9x + 81 x2 + 18x + 81 x x2 – 9 5. 2x2 + 5x + 2 x 2x2 + x – 1 4x2 – 1 x2 + x - 2 6. 6x3 – 6x2 ÷ 3x2 – 15x + 12 x4 + 5x3 2x2 + 2x – 40 x x
  • 42. Advanced Algebra 42 7. x2y x (y + 2) 3x (x – 1) y (x – 1) 8. 8x2y x+1 x x+1 6xy2 9. 9m + 6n 12m + 8n m2 n 2 ÷ 5m2 10. (x + 3) (2x – 1)÷ (-x – 3) (2x + 1) x (x + 4) x LESSON 3: Variation Objectives To define variation To familiarize with the types of variation To perform the different equation form in variation To apply variation in Word problem.
  • 43. Advanced Algebra 43 Nowadays, topics in mathematics are really used, without knowing that math is involved. Particularly in the application of science that made a big role in our integral life. Using such proportionality in speed or motion, we all know that math is involved. This lesson will be used in science and in word problems application. Variation is a relationship between variables that is defined using power function. It is in the form f(x) = kxn where k & n are real number. DIRECT VARIATION Variable y is directly proportional to variable x, if the quotient of y divided by x is constant. Written in the form y = kx where k is called proportionality constant. A variable y is said to be directly proportional to the nth power of variable x if y=kxn Where n > 0 INDIRECT (INVERSE) VARIATION Variable y varies indirectly or inversely as the variable x if the product of x & y is constant written as y=k/x or y=kx-1 Where k is not equal to 0 Variable y is said to be inversely proportional to the nth power of he variable x if y=k/xn or y=kx-n Where n >0
  • 44. Advanced Algebra 44 Remember that two variables w/c increase or decrease in the same ratio is said to be direct variation. And if one of the variables increases while the other decreases in the same ratio is called indirect variation. JOINT VARIATION Variable z varies jointly as the variable x & y if z varies directly as x when y is held constant and varies directly as y when x is held constant written as z=kxy Where k is not equal to 0 Variable z said to be jointly proportional to the nth power of the variable x and the nth power of the variable y if z=kyn xm Where n & m > 0 Let’s proceed to our example. Example Direct Variation The pressure of a given volume of gas varies directly as the temperature. Find the constant value if pressure of certain volume is 250 cm3 when its temperature is 170 C.
  • 45. Advanced Algebra 45 Solution: P = kT formulate an equation using direct variation 250 cm3 = k 50C substitute the given value to the equation form K = 250 cm3/50 C divide pressure to temperature to get the constant value. K= 5 cm3/C final answer Indirect Variation If y varies inversely as x, and the constant of variation is k = , what is y when x = 10? xy = k formulate an equation using indirect variation 10y = substitute the given value to the equation form y= × divide k to y to get the constant value = final answer Joint Variation If w varies jointly as x and y if w = 18 when x = 2 and y = 3, find the value of w if x= 4 and y = 5. W = kxy formulate equation using joint variation 18 = k (2) (3) substitute the first given value to look for the constant k = 18 / 6 divide product of x and y to w k=3 this is the constant value w = kxy formulate again equation to find w w = (3) (4) (5) substitute the second given value to look for w w = (12) (5) multiply k,x and y w = 60 you get the value of w. After you studied variation, you must answer the following challenges for you to measure your capability in variation.
  • 46. Advanced Algebra 46 NAME: RATING: Evaluate the following. 1. Let y vary jointly as x and the cube root of z, and inversely as the square of w. What is the effect on y if x is increases by 20%, z is doubled and y is doubled? 2. The surface area of a sphere varies directly as the square of the radius. If the surface area is 36π in2 when the radius is 3 in, what is the surface area of a sphere with a radius of 5 in?
  • 47. Advanced Algebra 47 3. When driver of a running vehicle applies the brakes abruptly, the resulting skid marks has a length (y) that varies directly as the square of the speed (x). A little boy crossing the street was struck by a car, leaving skid marks 15 cm long. If the police know that at a speed of 50 kph, the skid marks would be 12 meters long, how fast was the car running before the driver applied the break? 4. If y varies directly as x and if y is 15 when x = 5, find the value of y if x = 7. 5. If w varies jointly as x and y if w = 15 when x = 2 and y = 3, find the value of w if x= 3 and y = 4. 6. If an athlete could jump 23.5 ft when his takeoff velocity is 9.3 m/s, how far could he jump be if his takeoff velocity is only 9.0 m/s assuming that the jump is proportional to the square of the takeoff velocity. 7. If z varies inversely as x and if z = 6 when x = 4, find the value of z when x = 3. 8. In an remote barangay with a population of only 300, the rate of growth of an epidemic (i.e., the rate of change of the number o f infected person), R is found to be jointly proportional to the number of people infected, x and the number of people who are not infected, (300 – x). Given that the epidemic is growing at the rate of 5 people per day when there are 75 people infected, how fast will the epidemic growing if the number of people infected in the barangay is doubled? 9. Suppose that the maximum number of bacteria that can be supported by particular environment is M and the rate of bacterial growth (y) is jointly proportional to the number of bacteria present (x) and the difference between M and the number of bacteria present. Write the defining equation for y as a function of x and give the domain. 10. The gravitational attraction (F) between two bodies varies jointly as their masses (m1) and (m2) and inversely as the square of the distance(d) between them. What is the effect on the
  • 48. Advanced Algebra 48 gravitational attraction between two bodies if the masses are each halved and the distance between them is doubled? LESSON 4: Rational Function Objectives To define Rational Function To distinguish the different types of Asymptotes To perform operation used in finding Asymptotes A rational function f has the form
  • 49. Advanced Algebra 49 where g (x) and h (x) are polynomial functions and h(x) ≠ 0. Asymptote is a straight line at singularity which graph of function tends to approach but never touches. Asymptote Vertical Asymptote  vertical lines which correspond to the zeroes of the denominator of a rational function  consist of vertical lines of the form x=a where "a" is any value of x resulting in division by zero Horizontal Asymptote  horizontal line including x-axis to which graph of function comes closer and closer but never touches
  • 50. Advanced Algebra 50  consists of a horizontal line of the form y=b where "b" is the value of f(x) as x approaches positive or negative infinity (a) Horizontal Asymptote  If m < n, there is no horizontal asymptote.  If m = n, then y = an/bm is the horizontal asymptote.  f m > n, then y = 0 is the horizontal asymptote. (b) Vertical Asymptotes  Every zero of the denominator q(x) determines a vertical asymptote. If r1, r2, . . . , rk are zeros of q(x), then the lines x = r1, x = r2, . . . , x = rk are all vertical asymptotes of r(x). Proceed to our example: 1. Find the vertical asymptote of x2 + 2x – 8 = 0. 2. Find the horizontal asymptote of 5x3 -1 r(x) = ----------------
  • 51. Advanced Algebra 51 x2 + 3x + 2 Solution: 1. Use any kind of factorization to find the of x x2 + 2x – 8 = 0 Synthetic Division (optional) 2 1 2 -8 2 8 1 4 0 So the value of x are 2 and -4 (x-2) (x+4) = 0 And the value of vertical asymptote is the factor of the equation which is the value of x 2. Only the denominator will be use x2 + 3x + 2 = 0 Use again any factorization to look for x Synthetic Division (optional) -2 1 3 2 -2 -2 1 1 0 So the value of x are -2 and -1 (x+1) (x+2) = 0 And the value of horizontal asymptote is the value of x in which the denominator of a function becomes 0. This is it. Now you are in the end of the lesson in my workbook. I admire for the patient you‘ve done in reading and analyzing the topics I presented to you. So hold you’re tight because I really admire if you answer all the challenges I prepared to you. Be careful for the last challenge because if you careless answer it the more mistake you’ve got.
  • 52. Advanced Algebra 52 Find the vertical and horizontal asymptotes of the following if any: 1. f(x) = 1/(x + 2) 2. f(x) = (5x + 3) / (x – 2) 3. f(x) = (x + 2) / (x + 3) 4. f(x) = x / (x – 4) 5. f(x) = (3x2 – 15x + 12) / (2x2 + 2x – 40) 6. f(x) = (10x – 1) / (2x2 + 1) 7. f(x) = (x2 + 1) / (1 – x2) 8. f(x) = (9x2 + 81) / (x2 + 18x + 81) 9. f(x) = (3x3 – 27x2 + 9x – 81) / (x2 + 4)
  • 53. Advanced Algebra 53 10. f(x) = (x2 + 10x + 25) / (x3 + 3x2 + 9x + 27) A. Simplify the following: 1. (5x2 – 15x) / 10x2 2. (3t4 – 9t3) / 6t2 3. (2t2 +5t – 3) / (2t2 + 7t + 3) 4. (x3 + x2 – x – 1) / (x3 – x2 – x + 1) 5. (t4 + c4) / [(c + t)2 (c2 + t2)] B. Find the sum/difference: 1. [(a + b) / (a – b)] + [(a – b) / (a + b)] – [(b – a) / (a – b)] + (b – a) / (a + b)]
  • 54. Advanced Algebra 54 2. (x – y)2 – (x + y)2 3. (4t2 – 4t + 1)-1 + (4t2 – 1)-1 C. Find the product/quotient: 1. [(r2 + 4rs + 3s2) / (r2 + 5rs + 6s2)] × (r + 2s)-1 ÷ [(r + s) / (r2 + 4rs + 4s2)] 2. [(x2 + 3ax) / (3a – x)] ÷ [(x2 – 4ax + 3a2) / (a2 – x2)] 3. [(u2v) / (u+ v)] × [(u2 + 2uv + v2) / (uv2 – u2v)] D. Solve the following: 1. If r is inversely proportional to s and r = 5 when s = 4, find s when r= 7. 2. If y varies directly as x, and the constant of variation is 3, what is y when x = 7. 3. If a varies jointly as c and d, and a = 20 when c = 2 and d = 4,find d when a = 25 and c = 8. E. Find the vertical and horizontal asymptotes of the following if any: 1. f(x) = 3x / (x +2) 2. f(x) = (x2 + 1) / x2 3. f(x) = 10x2 / (x2 + 1) f(x) = (1 – x2) / x2
  • 55. Advanced Algebra 55