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 a m a n a m n (a m ) n a mn (ab) m a mb m 1 a m m , a 0 a a 0 1, a 0 am mn a ,a 0 n a m am a m ,b 0 b b (23 ) 5 3 5 Examples 2 (2) 3 (2) 5 r 5 s 2 (7b 3 ) 2 b 5b ( xy 2 ) 2 x 3 y 1 Polynomial Functions -exponents are whole numbers -coefficients are real numbers f ( x) 2 x 5 x 2 x x 7 4 3 2 -2 is the leading coefficient 4 is the degree ( the highest exponent) -7 is the constant term The degree of a polynomial function is the exponent of the leading term when it is in standard form • • • • • • Degree 0 1 2 3 4 type Constant Linear Quadratic Cubic Quartic You can evaluate polynomial functions using -direct substitution -synthetic substitution • Evaluate f ( x) 2 x 4 8 x 2 5 x 7 EVALUATING A FUNCTION given 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 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 -8 5 -7 2 6 6 18 10 30 35 105 98 X=3 The solution is the last number written. 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 END BEHAVIORS WHAT THE GRAPH DOES AT THE ENDS? DEGREE 0 1 2 3 4 LEADING COEFFICIENT POSITIVE NEGATIVE POLYNOMIAL GRAPHS IT’S A MATTER OF DEGREES DEGREE/TYPE EXAMPLE END BEHAVIORS MAX # OF MAX ZEROS TURNING POINTS 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 Same 4 3 n (odd) Alternate n n-1 n (even) Same n n-1 4/quartic y = x4 – 4x3 – x2 + 12x – 2 Degre even Leading Coefficie nt positive End behavior of the function Graph of the function Example: f (x) = x2 Degree even Negative Leading coefficient Example: f (x) = –x2 Degree Odd Positive Leading coefficient Example: f (x) = x3 Degree Odd Negative Leading coefficient Example: f (x) = –x3 • The time t ( in seconds)it takes a camera battery to recharge after flashing n times can be modeled by: t 0.000015n 0.0034n 0.25n 5.3 3 2 • Find the recharge time after 100 flashes. 6.3 OPERATIONS ON POLYNOMIALS ADDITION: Aka: combine like terms EXAMPLE: (3𝑥 3 + 2𝑥 2 − 𝑥 − 7) + (𝑥 3 −10𝑥 2 + 8) Horizontally: SUBTRACTION: add the opposite of the second polynomial EXAMPLE: (3𝑥 3 + 2𝑥 2 − 𝑥 − 7) − (𝑥 3 −10𝑥 2 + 8) (3𝑥 3 + 2𝑥 2 − 𝑥 − 7) + (−𝑥 3 +10𝑥 2 − 8) Vertically: MULTIPLY EXAMPLE: (𝑥 − 3)(3𝑥 2 − 2𝑥 − 4) Horizontally: Vertically: APPLICATIONS OF 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) = −.246𝑡 2 + 7.88𝑡 + 159 and E(t) = .174𝑡 2 + 2.54𝑡 + 131 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) = (−.246𝑡 2 + 7.88𝑡 + 159) - (.174𝑡 2 + 2.54𝑡 + 131) N(t) = −.42𝑡 2 + 5.34𝑡 + 28 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 𝑁(𝑡) = 1.36𝑡 2 + 2.53𝑡 + 1076 𝑎𝑛𝑑 𝑃(𝑡) = .314𝑡 + 3.42 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) 𝑅(𝑡) = .42704𝑡 3 + 5.44562𝑡 2 + 346.5166𝑡 + 3679.92 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) After vacation warm up (32 ) 5 38 Simplify 3 xy 9 x 3 y 4 Write in standard form ( x x 7) (2 x 4 x 3) 3 3 Graph ( x 3)( x 4 x 4) y x3 1 Use synthetic substitution to evaluate y 2x 4x 1 3 for x=-2 2 SPECIAL PRODUCT PATTERNS SUM x DIFFERENCE: Example: SQUARE OF A BINOMIAL: (a + b)(a – b) = a² - b² (x + 4)(x – 4) = x² - 16 (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: Example: (a + b)³ = a³ + 3a²b + 3ab² + b³ (x + 5)³ = a³ + 3a²b + 3ab² + b³ x³ + 3x²b(5) + 3x(25)² + 125 x³ + 15x² + 75x + 125 FACTORING REVIEW COMMON FACTOR: 6x² + 15x + 27 = 3( TRINOMIAL: 2x² -5x – 12 = ( )( ) ) PERFECT SQUARE TRINOMIAL: x² + 20x + 100 = ( DIFFERENCE OF TWO SQUARES: x² - 49 = ( )( )( ) ) 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 Warm-up • Factor y 3 27 8 x 2 10 x 3 x 125 3 5 x 3 40 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 x = 0 x² - 3 = 0 x = ±√3 x–2=0 x=2 x+2 = 0 x = -2 𝑋 3 + 27 = 0 (X + 3)(X² – 3X + 9) = 0 X+3=0 OR X² – 3X + 9 = 0 FACTOR BY GROUPING Use for polynomials with 4 terms 𝑟 3 − 3𝑟 2 + 6𝑟 − 18 Separate into 2 binomials: Factor out GCF of each: 𝑟 3 − 3𝑟 2 + 6𝑟 − 18 𝑟 2 (𝑟 − 3) + 6(𝑟 − 3) Factor out new GCF: TRY THESE: 𝑋 3 + 6𝑋 2 + 7𝑋 + 42 𝑧 3 − 2𝑧 2 − 16𝑧 + 32 25𝑝3 − 25𝑝2 − 𝑝 + 1 9𝑚3 + 18𝑚2 − 4𝑚 − 8 (𝑟 − 3)(𝑟 2 + 6) CHECK: (X² + 7)(X + 6) (z² - 16)(z – 2) (5p - 1)(5p + 1)(p – 1) (3m - 2)(3m + 2)(m + 2) • 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. • 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? • 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? • 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? LONG DIVISION REVIEW 32040 /15 213 6 LONG DIVISION - Remember 4th grade? Write dividend “inside the house” 1st Divide digit(s) in dividend by the divisor; write answer in quotient Multiply Subtract Bring down next digit Repeat process as needed Answer: 2136 15 3 2 0 4 0 3 0 2 0 1 5 54 4 5 9 0 9 0 0 POLYNOMIAL DIVISION (2x⁴ + 3x³ + 5x – 1) /(x² - 2x + 2) 2x² +7x + 10 LONG DIVISION Write dividend in standard form (include all degrees) Divide 1st term in dividend by 1st term in divisor Multiply Subtract 2x⁴ + 3x³ + 0x² + 5x – 1 2x⁴ - 4x³ + 4x² 7x³ - 4x² +5x 7x³ - 14x² + 14x 10x² -9x -1 10x² - 20x + 20 Bring down next term Repeat process as needed Answer: 2x² + 7x + 10 X² - 2x + 2 11x – 21 X² - 2x + 2 11x - 21 • TRY THESE: Divide x² + 6x + 8 by x + 4 SOLUTIONS: X+2 Verify: (x + 4)(x + 2) Divide x² + 3x – 12 by x – 3 6 X + 6 R 6 or x + 6 𝑥−3 Verify: (x – 3)(x + 6) + 6 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 1 6 3 18 2 12 0 K=6 Quotient: 1x2 + 3x + 2 • TRY THESE: Divide x² + 6x + 8 by x + 4 SOLUTIONS: X+2 Verify: (x + 4)(x + 2) Divide x² + 3x – 12 by x – 3 6 X + 6 R 6 or x + 6 𝑥−3 Verify: (x – 3)(x + 6) + 6 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 2 -6 8 -9 4 1 4 2 -5 K=2 f(2) = -5 Use the Remainder Theorem to evaluate P(-4) for P(x) = 2x4 + 6x3 – 5x2 - 60 WORKOUT -4 P(-4) = -12 2 6 -8 -5 8 0 -12 -60 48 2 -2 3 -12 -12 SPECIAL CASE Use the Remainder Theorem to evaluate P(-3) for P(x) = 2x3 + 11x2 + 18x + 9 2 11 18 9 2 -6 5 -15 3 -9 0 -3 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: f ( x) x 3 2 x 2 9 x 18 if one zero is 2 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) ÷ (𝑥 − 8) yes b) (x3 – 4x2 + 3x + 2) ÷ (x + 2) no 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 -6 2 7 -33 -18 -12 30 18 -5 -3 0 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) 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) Warm-up • Divide using long division. (3 x 2 x 5 x 1) (3 x 1) 3 2 (4 x 7 x 8) (2 x 1) 3 • The volume of a box is represented by the function f ( x) 2 x 11x 10 x 8 The box is (x-4) high and (2x+1) wide. Find the length. • V=lwh 3 2 WRITING A FUNCTION GIVEN 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 Try these: Given the zeros of a function, write the function. 1. 2. 3. 4. -1, 3, 4 -3, 1, 10 -2, 4, 5 1, 2 SOLUTIONS: 1. f (x) = x3 - 6x2 + 5x + 12 2. f (x) = x3 – 8x2 – 23x + 30 3. f (x) = x3 – 7x2 + 2x + 40 4. f (x) = x2 – 3x + 2 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 • Find the rational zeros of f ( x) x 3 2 x 2 11x 12 • List the possible zeros 1,2,3,4,6,12 • 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) • • List all the possible rational zeros of the function. f ( x) 2 x 3 7 x 2 7 x 30 f ( x) x 3 4 x 2 x 4 Find all zeros of the function. f ( x) x 4 x 11x 30 3 2 f ( x) 2 x 5 x 2 x 5 3 2 Molten Glass At a factory, molten glass is poured into molds to make paperweights. Each mold is a rectangular prism whose height is 3 inches greater than the length of each side of the square base. A machine pours 20 cubic inches of liquid glass into each mold. What are the dimensions of the mold? United States Exports • For 1980 through 1996, the total exports E (in billions of dollars) of the United States can be modeled by E 0.131t 3 5.033t 2 23.2t 233 • where t is the number of years since 1980. In what year were the total exports about • $312.76 billion? Fundamental Theorem of Algebra • If f(x) is a polynomial of degree n and n is greater than zero, then the equation f(x)=0 has at least one root in the set of complex numbers. • (written by Carl Friedrich Gauss) • When all real and imaginary solutions are counted (with all repeated solutions counted individually), an nth degree polynomial equation has exactly n solutions. Any nth degree polynomial function has exactly n zeros. Turning points of a graph • The graph of every polynomial function of degree n has at most n-1 turning points. If the function has n distinct real zeros then its graph has exactly n-1 turning points. • Polynomial functions have local maximum and local minimum points, these are the turning points. • Quadratic functions have only one maximum or minimum point. Finding Turning Points • Use your calculator to graph f ( x) x 3x 2 3 2 • Identify the x intercepts and the points where the local maximums and minimums occur. Maximizing a Polynomial Model You are designing an open box to be made of a piece of cardboard that is 10 inches by 15 inches. The box will be formed by making the cuts at the corners and folding up the sides so that the flaps are square. You want the box to have the greatest volume possible. How long should you make the cuts? What is the maximum volume? What will the dimensions of the finished box be?
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