1. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Chapter 4 Polynomials

2. 1-2 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Polynomials Terms Types of Polynomials Degree and Coefficients Combining Like Terms Evaluating Polynomials and Applications 4.3

3. 1-3 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Terms A term can be a number, a variable, a product of numbers and/or variables, or a quotient of numbers and/or variables. A term that is a product of constants and/or variables is called a monomial. Examples of monomials: 8, w, 24 x 3 y A polynomial is a monomial or a sum of monomials. Examples of polynomials: 5w + 8,  3x 2 + x + 4, x, 0, 75y 6

4. 1-4 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Identify the terms of the polynomial 7p 5  3p 3 + 3. Solution The terms are 7p 5,  3p 3, and 3. We can see this by rewriting all subtractions as additions of opposites: 7p 5  3p 3 + 3 = 7p 5 + (  3p 3 ) + 3 These are the terms of the polynomial.

5. 1-5 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Types of Polynomials A polynomial that is composed of two terms is called a binomial, whereas those composed of three terms are called trinomials. Polynomials with four or more terms have no special name. The degree of a term of a polynomial is the number of variable factors in that term. MonomialsBinomialsTrinomialsNo Special Name 5x25x2 3x + 43x 2 + 5x + 9 5x 3  6x 2 + 2xy  9 84a 5 + 7bc 7x 7  9z 3 + 5a 4 + 2a 3  a 2 + 7a  2  8a 23 b 3  10x 3  76x 2  4x  ½6x 6  4x 5 + 2x 4  x 3 + 3x  2

6. 1-6 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Determine the degree of each term: a) 9x 5 b) 6y c) 9 Solution a) The degree of 9x 5 is 5. b) The degree of 6y is 1. c) The degree of 9 is 0.

7. 1-7 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Coefficient The part of a term that is a constant factor is the coefficient of that term. The coefficient of 4y is 4.

8. 1-8 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Identify the coefficient of each term in the polynomial. 5x 4  8x 2 y + y  9 Solution The coefficient of 5x 4 is 5. The coefficient of  8x 2 y is  8. The coefficient of y is 1, since y = 1y. The coefficient of  9 is simply  9.

9. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. The leading term of a polynomial is the term of highest degree. Its coefficient is called the leading coefficient and its degree is referred to as the degree of the polynomial. Consider this polynomial 4x 2  9x 3 + 6x 4 + 8x  7. The terms are 4x 2,  9x 3, 6x 4, 8x, and  7. The coefficients are 4,  9, 6, 8 and  7. The degree of each term is 2, 3, 4, 1, and 0. The leading term is 6x 4 and the leading coefficient is 6. The degree of the polynomial is 4.

10. 1-10 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Combine like terms: a) 4y 4  9y 4 b) 7x 5 + 9 + 3x 2 + 6x 2  13  6x 5 c) 9w 5  7w 3 + 11w 5 + 2w 3 Solution a) 4y 4  9y 4 = (4  9)y 4 =  5y 4 b) 7x 5 + 9 + 3x 2 + 6x 2  13  6x 5 = 7x 5  6x 5 + 3x 2 + 6x 2 + 9  13 = x 5 + 9x 2  4 c) 9w 5  7w 3 + 11w 5 + 2w 3 = 9w 5 + 11w 5  7w 3 + 2w 3 = 20w 5  5w 3

11. 1-11 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Evaluate  x 3 + 4x + 7 for x =  3. Solution For x =  3, we have  x 3 + 4x + 7 =  (  3) 3 + 4(  3) + 7 =  (  27) + 4(  3) + 7 = 27 + (  12) + 7 = 22

12. 1-12 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example In a sports league of n teams in which each team plays every other team twice, the total number of games to be played is given by the polynomial n 2  n. A boys’ soccer league has 12 teams. How many games are played if each team plays every other team twice? Solution We evaluate the polynomial for n = 12: n 2  n = 12 2  12 = 144  12 = 132. The league plays 132 games.

13. 1-13 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example The average number of accidents per day involving drivers of age r can be approximated by the polynomial 0.4r 2  40r + 1039. Find the average number of accidents per day involving 25-year-old drivers. Solution 0.4r 2  40r + 1039 = 0.4(25) 2  40(25) + 1039 = 0.4(625)  1000 + 1039 = 250  1000 + 1039 = 289 There are, on average, approximately 289 accidents each day involving 25-year-old drivers.