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Chapter 5 5.1 Polynomials and Functions A polynomial function is a function of the form f(x) = an x nn + an – 1 x nn – 11 +· · ·+ a 1 x + aa00 Where ann 00 and the exponents are all whole numbers. For this polynomial function, aan is the leading coefficient, coefficient n aa00 is the constant constant term, term and n is the degree. degree A polynomial function is in standard form if its terms are descending order order of of exponents exponents from from left left to to right. right. written in descending You are already familiar with some types of polynomial functions. Here is a summary of common types of polynomial functions. Degree Type Standard Form 0 Constant f (x) = a 0 1 Linear f (x) = a1x + a 0 2 Quadratic f (x) = a 2 x 2 + a 1 x + a 0 3 Cubic f (x) = a 3 x 3 + a 2 x 2 + a 1 x + a 0 4 Quartic f (x) = a4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0 Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f (x) = 1 x 2 – 3x4 – 7 2 SOLUTION The function is a polynomial function. 4 Its standard form is f (x) = – 3x + 1 2 x – 7. 2 It has degree 4, so it is a quartic function. The leading coefficient is – 3. Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f (x) = x 3 + 3 x SOLUTION The function is not a polynomial function because the x term 3 does not have a variable base and an exponent that is a whole number. Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. – f (x) = 6x 2 + 2x 1 + x SOLUTION The function is not a polynomial function because the term 2x –1 has an exponent that is not a whole number. Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f (x) = – 0.5x + x 2 – 2 SOLUTION The function is a polynomial function. Its standard form is f (x) = x – 0.5x – 2 It has degree 2, so it is a quadratic function. The leading coefficient is . 2. Identifying Polynomial Functions Polynomial function? f (x) = 12 x 2 – 3x 4 – 7 f (x) = x 3 + 3x f (x) = 6x2 + 2x– 1 + x f (x) = – 0.5x + x2 – 2 Decide whether the function is a polynomial function. If it is, write the function in standard form and state the degree and leading coefficient. Goal 1: To evaluate polynomial functions Goal 2: To simplify polynomial functions Chapter 5 5.2 Addition and Subtraction of Polynomials Goal 1: To add polynomial functions Goal 2: To subtract polynomial functions The additive inverse of a polynomial. The additive inverse of a polynomial can be found by replacing each term by its additive inverse. The sum of a polynomial and its additive inverse is O. The additive inverse of a polynomial. The additive inverse of a polynomial can be found by replacing each term by its additive inverse. The sum of a polynomial and its additive inverse is O. Thus, to subtract one polynomial from another, we add its additive inverse. Simplify the polynomial HW #5.1-2 Pg 208-209 1-29 Odd, 30-36 Pg 212-213 1-31 Odd, 33-35 HW Quiz HW #5.1-2 Wednesday, May 24, 2017 Pg 208 32 Pg 208 34 Pg 212 31 Pg 212 34 Pg 208 30 Pg 208 32 Pg 212 34 Pg 212 35 Chapter 5 5.3 Multiplication of Polynomials Find the product. Simplify. Based on your answers to parts to the above, write a general formula. Use “2n” to represent a general even integer and let “2n + 1” represent a general odd integer, and use “…” for missing terms. Answers to challenge HW #5.3 Pg 217-218 1-39 Odd, 40-49 HW Quiz HW #5.3 Wednesday, May 24, 2017 Missing Parts 5.4 Factoring Do Examples from Regular book la205bad HW 5.4 Pg 222-223 3-60 Every Third, 61-76 5.5 More Factoring HW Handout Factoring 5.6 Factoring A General Strategy Do bonus problems from Great Factoring Problems WS HW Pg 231 1-37 Odd, 38-47 HW Quiz HW #5.6 Wednesday, May 24, 2017 Row 1, 3, 5 Factor Completely 1. x2 18x 72 2. x4 18x2 81 3. 4 x3 108 4 2 4. x 9 x 81 Row 2, 4, 6 Factor Completely 1. x2 6 x 72 4 2. x 16 3. 4 x3 108 4 4. x 64 5.7 Solving by Factoring HW #5.7 Pg 233 1-42 Left Column, 43-46 Pg 228 89-99 Odd 5.8 Using Polynomial Equations A candy factory needs a box that has a volume of 30 cubic inches. The width should be 2 inches less than the height and the length should be 5 inches greater than the height. What should the dimensions of the box be? For the city park commission, you are designing a marble planter in which to plant flowers. You want the length of the planter to be six times the height and the width to be three times the height. The sides should be one foot thick. Since the planter will be on the sidewalk, it does not need a bottom. What should the outer dimensions of the planter be if it is to hold 4 cubic feet of dirt? 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? HW #5.8a Pg 235-236 1-17 Odd, 18-20 5 15 17 18 7 13 17 20 Test Review Geometry Express the area A of a rectangle as a function of the length x if the length of the rectangle is twice its width. Geometry Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides. 1. Find the degree of a polynomial 2. Find the degree of a monomial 3. Find the function value of a polynomial 4. Standard form of a polynomial 5. Find volume of a box 6. Properties of Exponents 7. Additive Inverse 8. Polynomial arithmetic 9. Factoring 10. Solving Polynomial equations 11. Study all the challenge problems in the book If h and k areconstantsand x 2 kx 6 factors into ( x 6)( x h)findk. Evaluate: (30)2 + 2(30)(22) + (22)2 x 2 7 xy 10 y Find ( x 5 y ) 2 15and ( x 2 y) 5, Factor 1. 2. 3. 4. 5. 6. a9 b9c9d2 - d2 x2n + xn yn – yn-1 + yn-2 72x2n + 120xn + 50 50x4 – 72y6 x2 – w2 –16 + 8w Factor 1. 2. 3. 4. 5. 6. 7. x4 + 9x2 + 81 x4 + x2y2 + 25y4 x4 + 64 7y2a + b – 5ya + b + 3ya + 2b 4x2a – 4xa – 3 a6 – 64b6 9a3 +9a2b – 4ab2 – 4b3 HW #R-4 Pg 241 1-42