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BIC 10303 ALGEBRA

BIC 10303 ALGEBRA. By: Dr Norfaradilla Wahid. Chapter 3: Polynomial. ALGEBRA. BIC 10303. CHAPTER OVERVIEW. POLYNOMIAL 3.1 Basic concept 3.2 Algebra operation for polynomial 3.3 Synthetic divider 3.4 Remainder theorem and factor theorem 3.5 Quadratic equation

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BIC 10303 ALGEBRA

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  1. BIC 10303ALGEBRA By: DrNorfaradilla Wahid Chapter 3: Polynomial

  2. ALGEBRA BIC 10303

  3. CHAPTER OVERVIEW POLYNOMIAL 3.1 Basic concept 3.2Algebra operation for polynomial 3.3 Synthetic divider 3.4 Remainder theorem and factor theorem 3.5 Quadratic equation 3.6 Inequality and partial fraction

  4. Basic Concept 3.1

  5. 3.1 Basic Concept • A polynomial is an algebraic expression in which the exponents of the variables are nonnegative integers and there are no variables in the denominator • An expression that can be written in the form • a + bx + cx2 + dx3 +ex4 + …. • Things with Surds (e.g. x + 4x +1 ) and reciprocals (e.g. 1/x + x) are not polynomials • The degree is the highest index • e.g 4x5 + 13x3 + 27x is of degree 5

  6. They can be added: 2x2 - 5x - 3 + 2x4 + 4x2 + 2x (3x2 - 5x - 3) (2x4 + 4x2) Or multiplied: 3x2 -5x -3 Set-up a multiplication table 2x4 +4x2 Polynomials can be combined to give new polynomials = 2x4 + 6x2 -3x -3 6x6 -10x5 -6x4 +12x4 -20x3 -12x2 +12x4 6x6 -10x5 -20x3 -12x2 -6x4 Gather like terms = 6x6 - 10x5 + 6x4 - 20x3 -12x2

  7. f(x) x Function Notation • An polynomials function can be written as • f(x) = a + bx + cx2 + dx3 +ex4 + …. • f(x) means ‘function of x’ • instead of y = …. • e.g f(x) = 4x5 + 13x3 + 27x • f(3) means …. • “the value of the function when x=3” • e.g. for f(x) = 4x5 + 13x3 + 27x • f(3) = 4 x 35 + 13 x 33 + 27 x 3 • f(3) = 972 + 351 + 81 = 1404

  8. Example • Identify which expressions are polynomials. If an expression is not a polynomial, explain why. • (a) • (b) • (c) • (d)

  9. Some polynomials have special names depending on the number of terms contained in the polynomial. • A monomial is a polynomial containing one term. A term may have more than one factor. • A binomial is a polynomial containing two terms. • A trinomial is a polynomial containing three terms

  10. Polynomials in one variable • The degree of a polynomial in one variable is the largest exponent of that variable. • A constant has no variable. It is a 0 degree polynomial. • This is a 1st degree polynomial. 1st degree polynomials are linear. • This is a 2nd degree polynomial. 2nd degree polynomials are quadratic. • This is a 3rd degree polynomial. 3rd degree polynomials are cubic.

  11. Example Classify the polynomials by degree and number of terms. Classify by number of terms Classify by degree Polynomial Degree a. b. c. d. Zero Constant Monomial First Linear Binomial Second Quadratic Binomial Trinomial Third Cubic

  12. Identify the Degree of Terms • Exponents of variable factors must be integers greater than 0. • For a term that is constant, the degree is 0. • For a term that has only one variable factor, the degree is the same as the exponent of the variable. • For a term that has more than one variable factor, the degree is the sum of the exponents of the variables. • A linear term has degree 1 • A quadratic term has degree 2 • A cubic term has degree 3 • Try this: • Identify the degree of each term in the polynomial

  13. Identify the Degree of Polynomials • Identify the degree of each term of the polynomial. • Compare the degrees of each term of the polynomial and select the greatest degree as the degree of the polynomial. • Try this: • Identify the degree of the polynomial • (a) has a degree of 2 • (b) has a degree of 3 and is a cubic poly • (c) has a degree of 0 and is a constant • (d) has a degree of 1 and is a linear polynomial

  14. Algebra operation for polynomial 3.2

  15. 3.2 Basic Operations with Polynomials • (1) Add and Subtract Polynomials • Combine the coefficients of the like terms using the rules for adding or subtracting signed numbers • (2) Multiply Polynomials • Multiply the coefficients using the rules for signed numbers. • Multiply the letter factors using the laws of exponents for factors with like base • Distribute if multiplying a polynomial by a monomial • (3) Divide Polynomials • Divide the coefficients using the rules for signed numbers. • Divide the variable factors using the laws of exponents for factors with like base

  16. Take a break • What does simplify mean? • - instruction to ‘simplify the expression’ are vague but often used. • To simplify an expression means to write the expression using fewer terms or reduced or with lower coefficients or exponents. • Examine the expression and see which laws or operations allow you to rewrite the expression in a simpler form.

  17. Exercises • Simplify the algebraic expressions by combining like terms. • (a) (e) • (b) (f) • (c) (g) • (d)

  18. Exercises (multiply polynomials)

  19. Exercise (Divide polynomials) • Express the answers with positive exponents

  20. Divide polynomials (cont..) • There are two procedures for dividing polynomials: 1. Long division 2. Synthetic division • These procedures are especially valuable in factoring and finding zeros of polynomial functions. • Break-even points in financial applications. • Intervals where polynomial functions are positive, negative.

  21. 1. Long division Divide P(x) = x3 – 7x – 6 by x – 4 P(x) = x3 – 7x – 6 is a dividend; Q(x) = x – 4 is a divisor 1. The divisor goes on the outside of the box. The dividend goes on the inside of the box. 2. Divide 1st term of dividend by first term of divisor to get first term of the quotient. Line up the first term of the quotient with the term of the dividend that has the same degree. 3. Take the term found in step 2 and multiply it times the divisor. 4. Subtract this from the line above. 5. Repeat until done (until the degree of the "new" dividend is less than the degree of the divisor). x – 4 x3 –0x2 - 7x – 6 x2 x – 4 x3 –0x2 - 7x – 6 x2 x – 4 x3 –0x2 - 7x – 6 x3 –4x2 4x2 - 7x

  22. 1. Long division (cont..) Divide P(x) = x3 – 7x – 6 by x – 4 P(x) = D(x)Q(x) + R dividend = divisor x quotient + remainder x2 + 4x + 9 x – 4 x3 –0x2 - 7x – 6 x3 –4x2 4x2 - 7x 4x2 – 16x 9x – 6 9x –36 30 P(x) = D(x)Q(x) + R x3 – 7x – 6= (x - 4) (x2 + 4x + 9) + 30 22

  23. Take a break • Divide using long division and use the result to factor the polynomial completely.

  24. Synthetic Divider 3.3

  25. 2. Synthetic Division Synthetic Division is a special case of dividing by a linear factor binomial (x - k) or (jx - k) if x is variable, and j = const, k = const. Divider is a constant , which turns a binomial divisor into zero: for a divisor (x – k) use the divider k; for a divisor (jx – k) use the divider k/j. Divide x4 - 3x + 5 by (x - 4), (x- 4) is a divisor, then divider is 4.

  26. 2. Synthetic Division (cont..) Divide x4 - 3x + 5 by (x - 4) 1.Write dividend in descending powers and insert 0's for any missing terms x4 + 0x3 + 0x2 - 3x + 5 2. Find the number which turns the divisor into 0: x- 4 = 0, if x = 4 3. Write the coefficients of the dividend to the right: 4| 1 0 0 -3 5 4. Bring down the leading coefficient to the bottom row 4| 1 0 0 -3 5 1 5. Multiply c by the value just written on the bottom row 4| 1 0 0 -3 5 4 16 64 244 1 4 16 61 249 Note; Vertical pattern : add terms Diagonal pattern : Multiply by k

  27. 2. Synthetic Division (cont..) 4| 1 0 0 -3 5 4 16 64 244 1 4 16 61 249 The numbers in the last row make up your coefficients of the quotient as well as the remainder: quotient = x3 + 4x2 + 16x + 61 remainder = 249 x4 - 3x + 5 = (x - 4)*(x3 + 4x2 + 16x + 61 ) + 249 P(x) = D(x)Q(x) + R dividend = divisor x quotient + remainder

  28. Exercise • Use synthetic division to divide

  29. Remainder Theorem and Factor Theorem 3.4

  30. 3.4 Remainder Theorem and Factor Theorem 1. Remainder Theorem • The remainder obtained in the synthetic division process has an important interpretation, as described below • The Remainder Theorem tells you that synthetic division can be used to evaluate a polynomial function. • That is, to evaluate a polynomial function f(x) when x = k, divide f(x) by x – k. The remainder will be f(k)

  31. 3 5 –4 3 Example: Use the Remainder Theorem to evaluate the following function at Using synthetic division, you obtain the following Because the remainder is r = 36, you can conclude that f(3) = 36. This means that (3,36) is a point on the graph of f(x). You can check this by substituting x=3 into original solution. 15 33 5 11 36

  32. Example: Check! Notice that the value obtained when evaluating the function at f(3) and the value of the remainder when dividing the polynomial by x – 3 are the same. f(x) = 5x2 – 4x + 3 f(3) = 5(3)2 – 4(3) + 3 f(3) = 5 ∙ 9 – 12 + 3 f(3) = 45 – 12 + 3 f(3) = 36

  33. Example: • Use Remainder Theorem and synthetic division to find the function value

  34. 2. Factor Theorem • Lets look at this example again

  35. 2. Factor Theorem • Another important theorem is the Factor Theorem. This theorem states that you can test to see whether a polynomial has (x – k) as a factor by evaluating the polynomial at x = k. If the result is 0, (x – k) is a factor. • If the remainder is zero, that means f(k) = 0. This also means • that the divisor resulting in a remainder of zero is a factor of • the polynomial.

  36. Example: Show that (x – 3) is factors of f(x) = x3 + 4x2 – 15x – 18. Then find the remaining factors of f(x) Using synthetic division, you obtain the following Then function f(x) can be written as f(x) = (x – 3)( x2 + 7x + 6 ) f(x) = (x – 3)(x + 6)(x + 1) Therefore, (x – 3) is factors of f(x) = x3 + 4x2 – 15x – 18. Remaining factors of f(x) are (x + 6)(x + 1) 3 1 4 –15 –18 3 21 18 1 7 6 0 Since the remainder is zero, f(3) = 0, show that (x – 3) is a factor of x3+ 4x2 – 15x – 18. Factoring x2+ 7x +6 x2 + 7x +6 = (x + 6)(x + 1)

  37. The Factor Theorem (x – 3)(x + 6)(x + 1). Compare the factors of the polynomials to the zeros as seen on the graph of x3 + 4x2 – 15x – 18.

  38. Example: Given a polynomial and one of its factors, find the remaining factors of the polynomial. x3 – 11x2 + 14x + 80 x – 8 Ans: (x – 8)(x – 5)(x + 2)

  39. 3. The Rational Zero Test • The Rational Zero Test • Suppose P(x) = anxn + an-1xn-1 + an-2xn-2+ …+ a2x2 + a1x+ a0 • is a polynomial with integer coefficients, and • Rational zero = p/q is a rational zero of P(x). • Then p is a factor of constant term, q is a factor of leading coefficient.

  40. Find the rational zeros of f(x) =1x3- 3x2 - 2x + 6 The constant term, p = 6 and leading coefficient, q = 1 p = {1, 2, 3, 6} q = {1} p/q = {1, 2, 3, 6} f(-1) = (-1)3 - 3(-1)2 -2(-1) + 6 = 4 f(1) = (1)3 - 3(1)2 -2(1) + 6 = 2 f(-2) = (-2)3 - 3(-2)2 -2(-2) + 6 = -10 f(2) = (2)3 - 3(2)2 -2(2) + 6 = -2 f(-3) = (-3)3 - 3(-3)2 -2(-3) + 6 = -42 f(3) = (3)3 - 3(3)2 -2(3) + 6 = 0 List all rational numbers whose numerators are factors of the constant terms and whose denominators are factors of the leading coefficient. List all the possible rational zeros Use trial-and-error method to determine which , if any, are actual zeros of the polynomial. (x - 3) is a factor, gives that x = 3 is a rational zero f(x) =x3 - 3x2 - 2x + 6 = (x - 3)(…………….…)

  41. x3 - 3x2 - 2x + 6 1 -3 -2 6 3 0 -6 1 0 -2 0 x2- 2 3 Cont. Find the rational zeros of f(x) =x3 - 3x2 - 2x + 6 ? f(x) = x3 - 3x2 - 2x + 6 = (x - 3)(…………….…) Now find the remaining factor Applying synthetic division f(x) = x3 - 3x2 - 2x + 6 = (x - 3)(x2 - 2) f(x) = (x - 3)( x - 2)(x + 2) Rewrite the polynomial by including the factor Factor of x2 – 2 produce x = - 2 and x = 2 We conclude that the rational zeros of f are x = 3 , x = - 2 and x = 2

  42. Example: Find the rational zeros of the following f(x) = x4 - 6x2 - 27

  43. Inequality and partial fraction 3.6

  44. Linear equation/inequalities(re-cap) • Linear equality • ax +c = 0 • Linear inequalities • A linear inequality in x is any inequality (with a≠0) ax +c < 0, ax +c > 0, ax +c ≤ 0, or ax +c ≥ 0 • E.g: 6x – 27 < 9

  45. Rational Equation • A rational expression is a fraction in which the numerator and/or the denominator are polynomials. • i.e., • denominator ≠ 0 • A rational function/equation is made up from rational expression • i.e.,

  46. 3.5 Partial Fraction Polynomial - Intro To write a sum or difference of fractional(rational) expressions as a single fraction, we bring them to a common denominator. easy

  47. 3.5 Partial Fraction Polynomial • However, for some applications of algebra to calculus, we must reverse this process. • We must express a fraction such as 3x/(2x2 – x – 1) as the sum of the simpler fractions 1/(x – 1) and 1/(2x + 1)

  48. Partial Fraction • These simpler fractions are called partial fractions. • In this section, we learn how to find them. • Let r be the rational functionwhere the degree of P is less than the degree of Q.

  49. Quadratic Equation 3.5 Partial Fraction involving Quadratic Equation

  50. Partial Fraction • Every polynomial with real coefficients can be factored completely into linear and irreducible quadratic factors. • That is, factors of the form ax +b and ax2 + bx +c where a, b, and c are real numbers.

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