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Properties of Exponents

Properties of Exponents. Product of Powers Power of a Power Power of a Product Negative Exponents Zero Exponent Quotient of Powers Power of a quotient . Examples. Polynomial Functions -exponents are whole numbers -coefficients are real numbers.

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Properties of Exponents

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  1. Properties of Exponents • Product of Powers Power of a Power Power of a Product Negative Exponents Zero Exponent Quotient of Powers Power of a quotient

  2. Examples

  3. Polynomial Functions-exponents are whole numbers-coefficients are real numbers -2 is the leading coefficient 4 is the degree ( the highest exponent) -7 is the constant term

  4. The degree of a polynomial function is the exponent of the leading term when it is in standard form • Degreetype • 0 Constant • 1 Linear • 2 Quadratic • 3 Cubic • 4 Quartic

  5. You can evaluate polynomial functions using-direct substitution-synthetic substitution • Evaluate

  6. EVALUATING A FUNCTIONgiven a value for x DIRECT SUBSTITUTION: - replace each x with the given value - evaluate expression, following PEMDAS Example: f(x) = 2x⁴ - 8x² + 5x – 7, for x = 3 2(3)⁴ - 8(3)² + 5(3) - 7 2(81) – 8(9) + 5(3) - 7 162 – 72 + 15 -7 98

  7. SYNTHETIC SUBSTITUTION: - write polynomial expression in standard form (include all degree terms) - write only coefficients (including zeros) -use the given value of x in the process below Example: f(x) = 2x⁴ - 8x² + 5x – 7, for x = 3 2x⁴ + 0x³- 8x² + 5x – 7 2 0 -85-7 X = 3 6 6 18 10 30 35 105 98 2 The solution is the last number written.

  8. The time t ( in seconds)it takes a camera battery to recharge after flashing n times can be modeled by: • Find the recharge time after 100 flashes. 11.3 seconds

  9. End Behavior of Polynomial Functions is determined by the degree (n) and leading coefficient (a) Use your graphing calculator to investigate the end behavior of several polynomial functions. Write a paragraph to explain how the leading coefficient and degree of the function affect the end behavior of these graphs For a>0 and n even For a>0 and n odd For a<0 and n even For a <0 and n odd

  10. END BEHAVIORSWHAT THE GRAPH DOES AT THE ENDS? If the degree is even the ends both go in the same direction -If the leading coefficient is positive they both go up -If the leading coefficient is negative they both go down If the degree is odd the ends go in opposite directions -if the leading coefficient is positive it’s climbing the stairs( going up from left to right) -If the leading coefficient is negative it’s going down the stairs ( going down from left to right)

  11. POLYNOMIAL GRAPHSIT’S A MATTER OF DEGREES MAX # OF ZEROS MAX TURNING POINTS DEGREE/TYPE EXAMPLE END BEHAVIORS 0 /Constant y = 3 Horizontal 0 or infinity 0 1/linear y = -2x + 4 Alternate 1 0 2/quadratic y = x2 + 2x – 1 Same 2 1 3/cubic y = x3 – 3x2 + 2 Alternate 3 2 4/quartic y = x4 – 4x3 – x2 + 12x – 2 Same 4 3 n (odd) Alternate n n - 1 n (even) Same n n - 1

  12. 6.3 OPERATIONS ON POLYNOMIALS ADDITION: Aka: combine like terms EXAMPLE:+ Horizontally: Vertically: SUBTRACTION: add the opposite of the second polynomial EXAMPLE: + +

  13. MULTIPLY EXAMPLE: ( Horizontally: Vertically:

  14. APPLICATIONSOF POLYNOMIAL FUNCTIONS From 1985 through 1995, the gross farm income G, and farm expenses, E (in billions of dollars), in the United States can be modeled by G(t) = and E(t) = Where t is the number of years since 1985. Write a model for the net farm income, N, for those years N(t) = G(t) - E(t) N(t) = () - () N(t) =

  15. APPLICATION: BOOK BUSINESS From 1982 through 1995, the number of softbound books, N (in millions) sold in the United States, and the average price per book, P (in dollars) can be modeled by Where t is the number of years since 1982. Write a model for the total revenue, R received from the sales of softbound books. R(t) = P(t) x N(t) What was the total revenue from softbound books in 1990? Method #1: evaluate R with t = 8 Method #2: graph R and determine R(8) $7020 million ($7.02 billion)

  16. After vacation warm up Simplify Write in standard form Graph Use synthetic substitution to evaluate for x=-2

  17. SUM x DIFFERENCE: (a + b)(a – b) = a² - b² Example: (x + 4)(x – 4) = x² - 16 SPECIAL PRODUCT PATTERNS SQUARE OF A BINOMIAL: (a + b)² = a² + 2ab + b² Example: (x + 4)² = x² + 8x + 16 NOTE: The square of a binomial is always a trinomial. CUBE OF A BINOMIAL: (a + b)³ = a³ + 3a²b + 3ab² + b³ Example: (x + 5)³ = a³ + 3a²b + 3ab² + b³ x³ + 3x²b(5) + 3x(25)² + 125 x³ + 15x² + 75x + 125

  18. FACTORING REVIEW COMMON FACTOR: 6x² + 15x + 27 = 3( ) TRINOMIAL: 2x² -5x – 12 = ( )( ) PERFECT SQUARE TRINOMIAL: x² + 20x + 100 = ( )( ) DIFFERENCE OF TWO SQUARES: x² - 49 = ( )( )

  19. MORE SPECIAL FACTORING PATTERNS SUM OF 2 CUBES: a³ + b³ = (a + b)(a² - ab + b²) Example: x³ + 27 = (x + 3)(x² - x(3) + 9 (x + 3)(x² - 3x + 9) DIFFERENCE OF 2 CUBES: a³ - b³ = (a - b)(a² + ab + b²) Try these: x³ - 125 x³ + 64 27x³ - 8 343x³ + 1000

  20. Warm-up • Factor

  21. ZERO PRODUCT RULE(STILL GOOD!) Solving polynomial equations: 1. Transform equation to make one side zero 2. Factor other side completely 3. Determine values to make each factor zero Example: 2x⁵ + 24x = 14x³ 2x⁵ - 14x³ + 24x = 0 2x(x⁴ - 7x² + 12) = 0 2x(x² - 3)(x² - 4) = 0 2x(x² - 3)(x - 2)(x + 2) = 0 Set each factor to zero: 2x = 0 x² - 3 = 0 x – 2 = 0 x + 2 = 0 x = 0 x = ±√3 x = 2 x = -2

  22. (X + 3)(X² – 3X + 9) = 0 OR X + 3 = 0 X² – 3X + 9 = 0

  23. FACTOR BY GROUPINGUse for polynomials with 4 terms Separate into 2 binomials: + Factor out GCF of each: + 6) Factor out new GCF: TRY THESE: 25 CHECK: (X² + 7)(X + 6) (z² - 16)(z – 2) (5p - 1)(5p + 1)(p – 1) (3m - 2)(3m + 2)(m + 2)

  24. Suppose you have 250 cubic inches of clay with which to make a rectangular prism for a sculpture. If you want the height and width each to be 5 inches less than the length, what should the dimensions of the prism be? Solve by factoring.

  25. You are building a bin to hold cedar mulch for your garden. The bin will hold 162 cubic feet of mulch. The dimensions of the bin are x feet by 5x-9 feet by 5x-6 feet. How tall will the bin be?

  26. In 1980 archeologists at the ruins of Caesara discovered a huge hydraulic concrete block with a volume of 330 cubic yards. The blocks dimensions are x yards high by (13x – 11) yards long by • (13x – 15) yards wide. What are the dimensions of the block?

  27. You are building a bin to hold cedar mulch for your garden. The bin will hold 162 cubic feet of mulch. The dimensions of the bin are x ft. by (5x-6)ft. by (5x-9) ft. How tall will the bin be?

  28. 32040 /15 2 1 3 6 LONG DIVISION - Remember 4th grade? 15 3 2 0 4 0 Write dividend “inside the house” LONG DIVISION REVIEW 3 0 Divide 1st digit(s) in dividend by the divisor; write answer in quotient 2 0 Multiply 1 5 5 4 Subtract 4 5 Bring down next digit 9 0 Repeat process as needed 9 0 0 Answer: 2136

  29. (2x⁴ + 3x³ + 5x – 1) /(x² - 2x + 2) 2x² +7x + 10 LONG DIVISION X² - 2x+ 2 2x⁴ + 3x³ + 0x² + 5x – 1 Write dividend in standard form (include all degrees) POLYNOMIAL DIVISION 2x⁴ - 4x³ + 4x² Divide 1st term in dividend by 1st term in divisor 7x³ - 4x² +5x Multiply 7x³ - 14x² + 14x Subtract 10x² -9x -1 10x² - 20x + 20 Bring down next term 11x - 21 Repeat process as needed 11x – 21 X² - 2x + 2 Answer: 2x² + 7x + 10

  30. TRY THESE: Divide x² + 6x + 8 by x + 4 SOLUTIONS: X + 2 Verify: (x + 4)(x + 2) Divide x² + 3x – 12 by x – 3 X + 6 R 6 or x + 6 Verify: (x – 3)(x + 6) + 6

  31. SYNTHETIC DIVISION: To divide polynomial f(x) by (x – k), - write polynomial expression in standard form (include all degree terms) - write only coefficients (including zeros) - use the given value of k in the process below * - The remainder is the last number written - the other numbers in the answer are the coefficients/constant of the quotient Example: Divide x3- 3x2- 16x– 12 , by ( x – 6) x3- 3x2 - 16x – 12 1 -3 -16-12 K = 6 12 0 18 2 6 3 1 Quotient: 1x2 + 3x + 2

  32. TRY THESE: Divide x² + 6x + 8 by x + 4 SOLUTIONS: X + 2 Verify: (x + 4)(x + 2) Divide x² + 3x – 12 by x – 3 X + 6 R 6 or x + 6 Verify: (x – 3)(x + 6) + 6

  33. RELATED THEOREMS REMAINDER THEOREM: If a polynomial f(x) is divided by x – k, then the remainder, r, equals f(k). Remember synthetic substitution? Example: (x3 + 2x2 – 6x – 9) ⁄ (x – 2) 1 2 -6 -9 K = 2 2 8 4 1 4 2 -5 f(2) = -5

  34. WORKOUT Use the Remainder Theorem to evaluate P(-4) for P(x) = 2x4 + 6x3 – 5x2 - 60 • 2 6 -5 0 -60 • -8 8 -12 48 -4 2 -2 3 -12 -12 P(-4) = -12

  35. SPECIAL CASE Use the Remainder Theorem to evaluate P(-3) for P(x) = 2x3+ 11x2 + 18x + 9 2 11 18 9 - 3 -6 -15 -9 5 3 0 2 Since P(-3) = 0: 1. 3 is a zero of P(x) 2. (x - ¯3) is a factor (x + 3) Quotient: 2x2 + 5x + 3 Note: the quotient is also factorable: 2x2 + 5x + 3 = (2x + 3) (x + 1) Therefore, 2x3 + 11x2 + 18x + 9 = (x + 3) (2x + 3) (x + 1) Try: if one zero is 2

  36. OBSERVATION When dividing f(x) by (x-k), if the remainder is 0, then (x – k) is __ ____________ of f(x). Determine whether each divisor is a factor of each dividend: a) (2x2 – 19x + 24) b) (x3 – 4x2 + 3x + 2) (x + 2) no yes

  37. FACTOR THEOREM: A polynomial f(x) has a factor (x - k) if and only if f(k) = 0. Factor f(x) = 2x3 + 7x2 - 33x – 18 given that f(-6) = 0 2 7 -33 -18 -12 30 18 2 -5 -3 0 -6 f(-6) = 0, so (x + 6) is a factor Quotient: 2x2 – 5x -3 (which is the other factor, and can be factored into (2x + 1) (x – 3) Therefore, 2x³ + 7x² - 33x – 18 = (x + 6)(2x + 1)(x – 3)

  38. TRY THIS: Given one zero of the polynomial function, find the other zeros. F(x) = 15x3 – 119x2 – 10x + 16; 8 Since 8 is a zero, (x – 8) is a factor. Since the quotient is 15x2 + x -2, it is also a factor. Since 15x2+ x -2 can be factored into (5x + 2) (3x - 1). The factors of 15x3 – 119x2 – 10x + 16 are (x – 8) (5x + 2) (3x – 1)

  39. Warm-up • Divide using long division.

  40. The volume of a box is represented by the function The box is (x-4) high and (2x+1) wide. Find the length. • V=lwh

  41. WRITING A FUNCTIONGIVEN THE ZEROS Given: 2 and 4 are the zeros of the function f(x). Write the function f(x) = (x – 2) (x – 4) f(x) = x2 – 6x + 8 Given: 3 and -4 and 1 are the zeros of the function f(x). Write the function f(x) = (x – 3) (x + 4) (x – 1) f(x) = (x2 + x – 12) (x – 1) f(x) = x3 – 13x + 12

  42. Try these: Given the zeros of a function, write the function. • SOLUTIONS: • f (x) = x3 - 6x2 + 5x + 12 • f (x) = x3 – 8x2– 23x + 30 • f (x) = x3 – 7x2 + 2x + 40 • f (x) = x2 – 3x + 2 -1, 3, 4 -3, 1, 10 -2, 4, 5 1, 2

  43. The Rational Zero Theorem • If a polynomial function has integer coefficients then every rational zero of the function has the following form: • P = factor of the constant term • Q factor of the leading coefficient

  44. Find the rational zeros of • List the possible zeros • Test the zeros using synthetic division • Divide out the factor and factor the remaining trinomial to find the other zeros. • (You may use your calculator to guide you)

  45. List all the possible rational zeros of the function.

  46. Find all zeros of the function.

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