Solving and Graphing Polynomials

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1 UNIT 9 Solving and Graphing Polynomials You can see laminar and turbulent fl ow in a fountain. Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

2 When water comes out of a pipe, it usually looks very smooth. This is called laminar flow. After time, the water becomes less smooth as its flow becomes more turbulent. Engineers use polynomials to model both types of flow. Big Ideas You can apply the rules of arithmetic to comple algebraic epressions as well as simple numeric epressions. A function is a correspondence between two sets, the domain and the range, that assigns to each member of the domain eactly one member of the range. Functions can model many events in the physical world. If you can create a mathematical model for a situation, you can use the model to solve other problems that you might not be able to solve otherwise. Algebraic equations can capture key relationships among quantities in the world. Unit Topics Foundations for Unit 9 Polynomial Long Division Synthetic Division The Polynomial Remainder Theorem Factors and Rational Roots Graphing Polynomials Factoring Polynomials Completely Applications: Polynomials SOLVING AND GRAPHING POLYNOMIALS 31 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

3 Foundations for Unit 9 Before you study factoring and graphing polynomial functions, you should know how to do the following: Perform long division with numbers. Solve a quadratic equation by factoring. Graph power functions. Long Division It is important to know the terms used in division. Definitions dividend divisor quotient remainder the number that is being divided the number by which the dividend is being divided the result of a division a number that is left over after a division In long division, the dividend is written under a division symbol. When performing long division, you repeat the process of dividing, multiplying, and subtracting. Eample 1 Divide 3 ) Step 1 The divisor, 3, divides into 56 two times, so write in the quotient. Multiply by 3; write the product directly below 56. Subtract. Bring down the net number in the dividend. Step Because 3 divides into 108 four times, write 4 as the net number in the quotient. Multiply 4 by 3 and write the product below 108. Subtract. Bring down the net number in the dividend. Step 3 Because 3 divides into 167 seven times, write 7 as the net number in the quotient. Multiply 7 by 3 and write the product below 167. Subtract. There are no more numbers to bring down. The remainder is 6. The quotient can be written 47 R6 or You can check your answer to a long division problem by multiplying. Quotient Divisor + Remainder = Dividend Check In Eample 1, = 5687, so 47 R6 is correct. TIP If a number divides evenly into another number, the remainder is 0. quotient divisor ) dividend 47 R6 3 ) UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

4 Unit 9 Foundations Problem Set A Divide and check ) ) ) ) ) ) 197 Solving Quadratic Equations You can use the zero product property to solve quadratic equations that are easily factorable. Zero Product Property For any real numbers a and b, if ab = 0, then a = 0 or b = 0. Eample Solve each equation. A 15 = 0 B = 0 ( + 3)( 5) = 0 (3 + 1)( + 4) = = 0 or 5 = = 0 or + 4 = 0 = 3 = 5 = 3 1 = 4 The solution set is { 3, 5}. The solution set is { 4, 3 1 }. Problem Set B Solve = 0 8. r 5r + 6 = = h h 8 = a + 11a + 30 = = = d + 39d + 8 = = 0 FOUNDATIONS FOR UNIT 9 33 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

5 Unit 9 Foundations Graphing Power Functions To graph a power function of the form f () = a n + b, plot a few ordered pairs and use what you know about the basic shape of the graph as well as the end behavior of the function. The end behavior of a power function relies on the values of both n and a. The End Behavior of f ( ) = a n + b a is positive a is negative f () f () n is even As, f () As, f () As, f () As, f () f () f () n is odd As, f () As, f () As, f () As, f () Eample 3 Graph f () = and describe the graph s end behavior. Make a table and plot ordered pairs. y f () Since a = and n = 3, the values of f () will continue to decrease as decreases, and continue to increase as increases. 4 (1, 3) (0, 1) ( 1, 1) 4 Problem Set C Graph and describe the graph s end behavior. 16. f () = f () = g () = g () = f () = h() = f () = 1 3. h() = g () = UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

6 Polynomial Long Division If you know how to use long division with numbers, then you can use the same strategy to divide polynomials. You can set up a division problem with polynomials the same way you do with numbers. Place the dividend inside the division symbol and the divisor to the left of it. The quotient goes on top of the division symbol. Numbers 5 15 ) = 5 Polynomials 3 ) = Dividing a Second-Degree Polynomial by a First-Degree Binomial As you do when you divide numbers, repeat the process of divide, multiply, and subtract until no more divisions can be made. Eample 1 Divide. A ( 6 7) ( + 3) Solution Step ) 6 7 ( + 3) 9 7 Divide: =. Multiply: ( + 3) = + 3. Subtract: ( 6) ( + 3) = 9. Bring down 7. Step 9 Divide: 9 = ) 6 7 ( + 3) 9 7 ( 9 7) Multiply: 9 ( + 3) = Subtract: ( 9 7) ( 9 7) = 0. So 6 7 divided by + 3 is 9. POLYNOMIAL LONG DIVISION 35 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

7 B by 1. Solution Step 1 1 ) ( ) 14 4 Step ) ( ) 14 4 (14 7) 3 Divide =. Multiply: ( 1) =. Subtract: ( + 13) ( ) = 14. Bring down 4. Divide: 14 = 7. Multiply: 7 ( 1) = Subtract: (14 4) (14 7) = 3. So divided by 1 is + 7, with remainder 3. Representing and Checking a Quotient NAMES OF EXPRESSIONS IN A POLYNOMIAL DIVISION Divisor ) ( ) 14 4 (14 7) 3 Quotient Dividend Remainder 3 Different Ways to Epress Related Multiplication and Division Facts With Numbers With Polynomials 13 5 =, R3 ( ) ( 1) = + 7, R = = Because multiplication and division are inverse operations, you can check your answer to a division problem by multiplying. Using Multiplication to Check a Quotient Quotient Divisor + Remainder = Dividend 36 UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

8 Eample A Determine if is the quotient of ( 6 18) ( 5). Solution ( + 4)( 5) Quotient Divisor + Remainder = Dividend Multiply = 6 18 Simplify. So is the quotient of ( 6 18) ( 5). B Determine if 3 5 is the quotient of (6 15) ( + 3). Solution (3 5)( + 3) 6 15 Quotient Divisor = Dividend Multiply Simplify. So 3 5 is not the quotient of (6 15) ( + 3). Dividing a Fourth-Degree Polynomial by a First-Degree Binomial Eample 3 Divide by + 1. Solution Rewrite the dividend so that its terms are in order of decreasing degree, and write 0 3 for the missing 3 term = Now divide ) ( ) ( 3 ) 9 4 (9 + 9) ( 13 13) 14 TIP When you multiply a term in the quotient by the divisor, align the terms of the product with their like terms in the dividend. Notice that in Eample 3, the product ( ) is written so that 4 is directly below 4 and 3 is directly below 0 3. The quotient is The remainder is 14. So ( ) ( + 1) = POLYNOMIAL LONG DIVISION 37 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

9 Determining Whether One Polynomial Is a Factor of Another Polynomial Eample 4 Use division to determine whether a is a factor of a 3 7a + 6. Solution a + a 3 a ) a 3 + 0a 7a + 6 (a 3 a ) a 7a (a 4a) 3a + 6 ( 3a + 6) 0 The remainder is 0, so a 3 7a + 6 = (a )(a + a 3), and a is a factor of a 3 7a + 6. THINK ABOUT IT When the remainder is 0, both the divisor and the quotient are factors of the dividend. To check your answer, multiply (a )(a + a 3). Verify that the product is a 3 7a + 6. Problem Set Divide. 1. ( 3) ( + 1). (t 5t 14) (t + ) 3. ( ) ( + 1) 4. ( y + 5y 7) ( y 1) 5. (r + 4r 1) (r + 6) 6. ( 100) ( + 10) 7. (3q + 1q + 15) (3q + 3) 8. ( ) ( + 1) 9. ( ) (4 1) 10. (n + 13n 7) (n + 7) 11. (3g 4 1g 3 + g 4) ( g 4) 1. ( ) ( + 3) 13. (h 3 1) (h 1) 14. ( p 4 4p 1) ( p + ) 15. (b 4 + 7b + 5) (b + 3) Use multiplication to check the quotient. 16. Determine whether is the quotient of ( ) ( + 1). 17. Determine whether 9y + 5 is the quotient of (9y + 4y 5) ( y + 1). 18. Determine whether 3c + 1 is the quotient of (3c 4c + 1) (c 1) Determine whether n + 6 n 5 is the quotient of (n + n 35) (n 5). Use division to determine whether the binomial is a factor of the trinomial. 0. Determine whether 3r 7 is a factor of 6r 11r Determine whether + 1 is a factor of Determine whether 3a + is a factor of 9a + 1a Determine whether p 1 is a factor of 6p p 1. Solve. * 4. Challenge What value of k makes 3 a factor of k 3? * 5. Challenge The volume of a rectangular prism is The length of the prism is + and the width is 3. Find the height of the prism. 38 UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

10 Synthetic Division As long as you re careful, long division always works, but it takes time; another method you can use to divide polynomials is synthetic division. For synthetic division to work, the divisor must be a binomial in the form k, where k is a constant. Using Synthetic Division to Divide Polynomials Suppose you want to divide by. The process of finding the quotient using both long division and synthetic division is shown below. Long Division Synthetic Division + 7 ) ( 4) (7 14) Synthetic division is considered a shorthand method for long division, but remember it only works when the divisor is in the form k. Eample 1 Divide by + 3. Solution The divisor is + 3. To get the form k, write + 3 as ( 3), so k = 3. THINK ABOUT IT When you use synthetic division, notice that the coefficients of the dividend appear in the top row and the coefficients of the quotient appear with the remainder in the bottom row. Step Put the value for k in the bo and put the coefficients to the right of it. Step Step Step Bring down the first coefficient, 3, then multiply by 3 and put the result in the net column. Add 11 and 9, then multiply the sum,, by 3 and put the result in the net column. Add and 6, then multiply the sum, 8, by 3 and put the product in the net column. SYNTHETIC DIVISION 39 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

11 Step Step R0 Add 4 and 4. The sum, 0, is your remainder. Now, write the quotient. Begin writing the quotient with a power of that is one less than the greatest power of in the dividend. The quotient has no remainder, so ( ) ( + 3) = Eample Divide by 5. Solution The divisor is 5, so k = 5. Step Put the value for k in the bo and put the coefficients to the right of it. Step Step Step Step Step R Bring down the first coefficient, 1, then multiply by 5 and put the result in the net column. Add 6 and 5, then multiply the sum, 1, by 5 and put the result in the net column. Add 1 and 5, then multiply the sum, 6, by 5 and put the product in the net column. Add 3 and 30. The sum,, is your remainder. Now, write the quotient. Begin writing the quotient with a power of that is one less than the greatest power of in the dividend. The quotient has remainder, so ( ) ( 5) = Eample 3 Divide by + 4. Solution The divisor is + 4. To get the form k, write + 4 = ( 4) so k = 4. The dividend is missing an term, so use 0 for the coefficient. Step Put the value for k in the bo and put the coefficients to the right of it. Step Bring down the first coefficient,, then multiply by 4 and put the result in the net column. 330 UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

12 Step Step Step Step Step R0 Add 9 and 8, then multiply the sum, 1, by 4 and put the result in the net column. Add 0 and 4, then multiply the sum, 4, by 4 and put the product in the net column. Add 7 and 16, then multiply the sum, 9, by 4 and put the product in the net column. Add 36 and 36. The sum, 0, is your remainder. Now, write the quotient. Begin writing the quotient with a power of that is one less than the greatest power of in the dividend. The quotient has no remainder, so ( ) ( + 4) = Dividing Polynomials Using Synthetic Division when the Given Divisor Is Not in the Form k Eample 4 Divide 4y 3 5y 10 by y 3. Solution To use synthetic division, you must have a divisor in the form k. Rewrite the division problem by factoring from the divisor and the dividend. 4y 3 5y 10 y 3 = ( y 3 5 y 5 ) ( y 3 ) = y3 5 y 5 y 3 The dividend is now y y 5 and the divisor is y, so k =. 3 The dividend is missing a y term, so use 0 for the coefficient. Step Put the value for k in the bo and put the coefficients to the right of it. Step Bring down the first coefficient,, then multiply by 3 and put the result in the net column. Step Add 0 and 3, then multiply the sum, 3, by 3 and put the result in the net column. SYNTHETIC DIVISION 331 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

13 Step 4 Step 5 Step y + 3y + R Add 5 and 9, then multiply the sum,, by 3 and put the product in the net column. Add 5 and 3. The sum,, is your remainder. Now, write the quotient. Begin writing the quotient with a power of that is one less than the greatest power of in the dividend. The quotient has a remainder of, so (4y 3 5y 10) (y 3) = y + 3y + y 3 or, y + 3y + 4 y 3. REMEMBER To simplify a compound fraction, multiply the numerator and denominator by their least common denominator: y 3 = ( y 3 ) = 4 y 3. Problem Set Divide using synthetic division. 1. (4a + 11a + 6) (a + ) 14. ( ) ( + 1). ( ) ( + 5) 15. (b + 10b + ) (b + ) 3. (v 9v 35) (v 7) 16. (8y 18y 1) (y 4) 4. (3y 7y 1) (y ) 17. ( 4 4) ( 8) 5. (w 4w 45) (w + 5) 18. (9r 33r 4) (3r 4) 6. ( ) ( + 6) 19. (5f + 10f 6) (5f + ) 7. ( ) ( 9) 0. ( ) (3 1) 8. (t + 6t 0) (t + 9) 1. (9h + 1h + 4) (3h + 6) 9. (1a 3 + 7a) (a + 4). (9 3 56) (3 + 7) 10. ( ) ( + 3) 3. ( ) (3 + 6) 11. (q 4 + 6q 3 7q + 0) (q + 4) 4. (0g 4 4g 10) (4g 4) 1. ( ) ( ) 5. (15n 4 + n 3 n 8n + 3) (5n 3) 13. (m 4 + m 3 + 6m 9) (m 1) 6. ( ) ( + 5) Solve. * 7. Challenge Use synthetic division to divide by + 1. * 8. Challenge The area of a triangle is given by Find an epression for the base of the triangle if the height is given by UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

14 The Polynomial Remainder Theorem You can use the polynomial remainder theorem to evaluate a polynomial function. THE POLYNOMIAL REMAINDER THEOREM If a polynomial p () is divided by a, then the remainder is p (a). p () = q() ( a) + p (a) dividend = quotient divisor + remainder To prove this theorem, recall that you can check a quotient by using multiplication: Dividend = Quotient Divisor + Remainder Let p() represent the dividend, q() represent the quotient, and r () represent the remainder. If a is the divisor, then the formula above can be written as follows: p () = q() ( a) + r () Now substitute a for. p (a) = q(a) (a a) + r (a) Substitution p (a) = q(a) 0 + r (a) a a = 0 p (a) = r (a) Zero Property of Multiplication Therefore, the remainder is the value of the polynomial evaluated for a. Evaluating a Polynomial Sometimes synthetic division is an easier and quicker method to evaluate a polynomial, especially when a calculator is not readily available. Eample 1 Given f () = , find f ( ). THE POLYNOMIAL REMAINDER THEOREM 333 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

15 Solution Method 1 Method Use the remainder theorem. Substitute for in f (). Divide by +. f () = f ( ) = ( ) 4 3 ( ) 3 + ( ) = 16 3 ( 8) = The remainder is 43. = 43 So f ( ) = 43. Using the Remainder Theorem to Graph a Polynomial Eample Graph each function by using the remainder theorem. A f () = 4 1 Solution Use synthetic division repeatedly to find and plot ordered pairs. Organize your work in a table. List the coefficients of the dividend at the top. Perform the y steps mentally and write only the numbers that appear in the ( 1, 4) bottom row. Plot each pair (a, f (a)) B g () = Solution Use synthetic division repeatedly to find and plot ordered pairs (0, 1) (4, 1) 3 (1, 4) (3, 4) 6 (, 5) (5, 4) (1, 6) 6 (0, 4) (3, 4) 3 (, 4) ( 1, 8) y (4, 1) 334 UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

16 Application: Projectile Motion Eample 3 A baseball is hit straight up into the air with an initial velocity of 0 meters per second. The height of the baseball after t seconds is modeled by the polynomial function h(t) = 9.8t + 0t + 1, where h(t) is the height in meters (neglecting air resistance). What is the height of the baseball after 1.5 seconds? Solution The height of the baseball in meters after t seconds is h(t). So, to find the height of the baseball after 1.5 seconds, find h(1.5). Method 1 Method Use the remainder theorem. Substitute 1.5 for t in h(t). Divide 9.8t + 0t + 1 by t 1.5, using synthetic division. h(t) = 9.8t + 0t h(1.5) = = = 8.95 The remainder is 8.95, so h(1.5) = The height of the baseball after 1.5 seconds is 8.95 meters. TIP You can use one method to solve the problem and the other method to check your answer. Problem Set Use the polynomial remainder theorem to evaluate the polynomial for the given value. 1. f () = ; f (1). a(r) = 8r 3 r + 4r + 6; a() 3. f () = ; f ( 3) 4. h(t) = t + 3t 9; h(3) 5. f () = 3 + 5; f ( 5) 6. g(v) = v 7v 19; g(7) 7. f () = ; f (6) 8. r(z) = z 4 9z 3; r(3) 9. f () = ; f ( ) 10. P( y) = y 4 + 3y 3 7y 7y + 1; P(4) For each problem, do the following: A. Use the polynomial remainder theorem to find at least four ordered pairs for each polynomial. B. Use the points to graph the polynomial. 11. f () = f () = b(n) = n 3 + 3n 4n S( y) = 4y + 3y f () = f (a) = a 3 3a f () = f () = g() = f () = h(t) = t 3 + 8t + 10t Challenge *. k(t) = t 4 8t + 3 * 3. f () = THE POLYNOMIAL REMAINDER THEOREM 335 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

17 Solve. 4. A tennis ball is thrown straight up into the air with an initial velocity of 0 feet per second. The height of the tennis ball after t seconds is modeled by the polynomial function h(t) = 16t + 0t + 5, where h(t) is the height in feet (ignoring air resistance). What is the height of the tennis ball after 0.5 seconds? 5. The temperature of a city for January can be modeled by the polynomial function T() = , where is the day of the month. What is the temperature of the city on January 15? 6. The polynomial function, f () = , models the amount of fish caught and sold by commercial fisherman in thousands of metric tons, where is the month of the year. What is the amount of fish caught and sold for the month of June? 7. The distance (km) from an earthquake s epicenter that a seismic wave can be felt, minutes after the earthquake occurs, can be modeled by the polynomial function f () = What is the distance of a seismic wave 8 minutes after the earthquake occurs? * 8. Challenge A soccer ball is kicked into the air with an initial upward velocity of 5 meters per second and a horizontal velocity of 15 meters per second. The height of the soccer ball after t seconds is modeled by the polynomial function h(t) = 9.8t + 5t, where h(t) is the height in meters (ignoring air resistance). The distance traveled by the soccer ball after t seconds is modeled by the function d(t) = 15t. What is the height of the soccer ball if it has traveled a horizontal distance of 37.5 meters? 336 UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

18 Factors and Rational Roots Synthetic division can help you factor polynomials. THE FACTOR THEOREM For any polynomial p(), the binomial a is a factor of p() if and only if p (a) = 0. The remainder theorem is used to prove the factor theorem. For the proof, use p () for the dividend, q() for the quotient, and r () for the remainder. Therefore, p () = q() ( a) + r (). Because the theorem uses if and only if, there are two cases. Case 1: Assume p (a) = 0. Then by the remainder theorem, r (a) = 0. Because the quotient does not have a remainder, a is a factor of the dividend. Case : Assume a is a factor. Then the remainder r (a) equals 0. By the remainder theorem, r (a) = p (a), and 0 = p (a) by substitution. Using the Factor Theorem So what does the factor theorem mean? It means that if you evaluate a polynomial for some value of and the result is zero, then you can use that value to factor the polynomial. Eample 1 Determine whether + 3 is a factor of f () = Solution Since + 3 = ( 3), a = 3. So find f ( 3). f ( 3) = ( 3) 4 + ( 3) 3 11( 3) + ( 3) 6 = 81 + ( 7) = = 0 Since f ( 3) = 0, + 3 is a factor of f () = FACTORS AND RATIONAL ROOTS 337 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

19 For any polynomial function p (), a solution of the equation p () = 0 is called a zero or root of p (). In Eample 1, you could say that 3 is a zero of p (). Eample Factor the polynomial f () = , given that = 5 is a zero. Solution Step 1 If = 5 is a zero, then 5 is a factor. Use synthetic division to divide by Step Write the polynomial as a product of 5 and the quotient from Step 1. Then factor the quotient = ( 5)( ) = ( 5)( + 3)( + ) So f () = = ( 5)( + 3)( + ). Eample 3 Find all the roots of = 0. Solution By the factor theorem, if a is a factor of f (), then f (a) = 0. Eample shows that = ( 5)( + 3)( + ). ( 5) is a factor of f () = , so f (5) = 0. ( + 3) is a factor of f () = , so f ( 3) = 0. ( + ) is a factor of f () = , so f ( ) = 0. Therefore 5, 3, and are the roots of = 0. REMEMBER A number that satisfies an equation is called a root, or solution, of the equation. Finding Possible Rational Roots It s easy to find all the roots if you are given a head start, but what if you aren t given any of the roots? How can you find all the roots on your own? The rational root theorem can give you a set of values to try. THE RATIONAL ROOT THEOREM If p q is in simplest form and is a rational root of the polynomial equation a n n + a n 1 n a 1 + a 0 = 0 with integer coefficients, then p must be a factor of a 0 and q must be a factor of a n. REMEMBER A rational number is a real number that can be written as a quotient of two integers. 338 UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

20 Eample 4 Find the possible rational roots of each equation. A = 0 p Solution If q is a rational root of the polynomial , then p is a factor of the constant term, 5, and q is a factor of the leading coefficient,. List the factors of 5 and the factors of. The factors of 5 are ±1 and ±5. The factors of are ±1 and ±. The possible roots are p q are ± 1 1, ± 1, ± 1 5, and ± 5. Simplifying gives the following possible rational roots of = 0: ±1, ± 1, ±5, and ± 5. B = 0 p Solution If q is a rational root of the polynomial , then p is a factor of the constant term, 9, and q is a factor of the leading coefficient, 3. List the factors of 9 and the factors of 3. The factors of 9 are ±1, ±3, and ±9. The factors of 3 are ±1 and ±3. The possible roots p q are ± 1 1, ± 3 1, ± 1 3, ± 3 3, ± 9 1, and ± 9 3. Simplifying gives the following possible rational roots of = 0: ±1, ± 1 3, ±3, and ±9. Using the Rational Root Theorem to Factor a Polynomial Equation Eample 5 Factor completely. A = 0 Solution Identify the possible rational roots: ±1, ±, ±4, and ±8. Test for possible roots Because 1 is a root, ( + 1) is a factor. The other factor is 8, which factors into ( 4)( + ). The factored form is ( + 1)( 4)( + ) = 0. TIP You may need to test several roots before finding one that makes the remainder 0. Test easier numbers such as 1 and 1 before testing larger numbers and fractions. FACTORS AND RATIONAL ROOTS 339 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

21 B = 0 Solution Identify the possible rational roots: ±1, ±3, ±5, and ±15. Test for possible roots Because 3 is the first possible root with a remainder of 0, ( 3) is a factor. The other factor is 5. Solve 5 = 0 to f ind the other roots. 5 = 0 = 5 = ± 5 The factored form is ( 3) ( + 5 )( 5 ) = 0. C = 0 Solution The possible rational roots are ± 3 1, ± 3 4, ±1, ± 3, ±, ± 3 8, ±3, ±4, ±6, ±8, ±1, and ± Testing the possible roots by synthetic division reveals that 3 is a root. So ( 3 ) is a factor. The other factor is , which factors into 3( + 4)( 3). The factored form is 3 ( 3 ) ( + 4)( 3) = 0, which can also be written as (3 )( + 4)( 3) = 0. TIP Organize your work in a table. List the coefficients of the polynomial at the top. Perform the steps mentally and write only the numbers that appear in the bottom row. Problem Set Determine whether the given binomial epression is a factor of f () f () = f () = f () = f () = f () = f () = f () = f () = UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

22 Factor the polynomial equation with the given zero = 0; = is a zero = 0; = 5 is a zero = 0; = 3 is a zero = 0; = is a zero Find the roots of the polynomial equation = ( 1)( + )( ) = = ( + 4)( + 1)( + 1) = = ( )( + )( 3) = = ( + 5)( + 9)( 1) = 0 Find the possible rational roots of the polynomial equation = = = = = = 0 Factor completely = = = = = = 0 * 9. Challenge = 0 Solve. * 30. Challenge The rectangular prism shown has a volume of 6 cubic inches. A. Write an equation for the volume of the prism. B. Find the possible rational roots of the equation. + 1 FACTORS AND RATIONAL ROOTS 341 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

23 Graphing Polynomials Because any root of f () = 0 is also an -intercept of the graph of f (), you can determine the roots of a polynomial equation from its graph. Using a Graph to Determine the Roots of a Polynomial Equation Eample 1 A Find the roots of = 0. Solution The graph of f () = appears to intersect the -ais at (3, 0) and ( 7, 0), which means that 3 and 7 seem to be roots, both of which are consistent with the possible rational roots according to the rational root theorem. Check Find f (3) and f ( 7) to check your answer. f (3) = f ( 7) = ( 7) + 4 ( 7) 1 = = = 0 = 0 Since f (3) = 0 and f ( 7) = 0, 3 and 7 are the roots of = 0. B Find the roots of = 0. Solution The graph of f () = appears to intersect the -ais at ( 1, 0), (, 0), and (3, 0), which means that 1,, and 3 seem to be roots, all of which are consistent with the possible rational roots according to the rational root theorem. Check You can check your answer by factoring. Use any one of the roots obtained from the graph and turn it into a factor: ( 1) = + 1. Use synthetic division to divide Use the quotient to write the polynomial as a product of factors: = ( + 1)( 5 + 6) = ( + 1)( )( 3) The factors show that the roots are 1,, and 3. f () f () UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

24 Graphing a Polynomial Function by Using Intercepts You can draw the graph of a function by finding its -intercepts and incorporating what you know about the end behavior of power functions. Eample Draw the graph of f () = Solution Step 1 Use the rational root theorem to find the possible rational roots of f () = 0. The leading coefficient is 1 and the constant term is 10, so the possible rational roots are ±1, ±, ±5, and ±10. Step Use synthetic division to test the possible roots Since 1 is a root, 1 is a factor. You can write the polynomial as ( 1)( ). Now use synthetic division to find a factor of the cubic polynomial The leading coefficient is 1 and the constant term is 10, so the possible rational roots are ±1, ±, ±5, and ± Since 1 is a root, + 1 is a factor. Now you can write the polynomial as ( 1)( + 1)( 3 10). Step 3 You could continue to use synthetic division to find the factors of the quadratic polynomial, but factoring it into two binomials is quicker. The function in factored form is f () = ( 1)( + 1)( 5)( + ). The -intercepts, then, are 1, 1, 5, and. Plot the -intercepts. Step 4 Graph the polynomial. Because the function is a polynomial with even degree and a positive leading coefficient, the graph approaches positive infinity as approaches both negative and positive infinity. Use synthetic division to find additional points. TIP Note that there are an infinite number of polynomials with these four roots. These polynomials all have the form f () = a( 1)( + 1)( 5)( + ). When you find another point, you can determine a by substituting for and f () in the equation f () 40 0 (0, 10) (, 0) ( 1, 0) (1, 0) (5, 0) ( 1.5, 4.065) 40 (, 36) (4, 90) GRAPHING POLYNOMIALS 343 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

25 Graphing a Polynomial Function by Using Roots Eample 3 Draw the graph of p() = Solution Step 1 Use the rational root theorem to find the possible rational roots of p() = 0. The leading coefficient is 1 and the constant term is, so the possible rational roots are ±1 and ±. Step Test the possible rational roots until you find a root of the polynomial Since is a root, + is a factor. You can write the polynomial as ( + )( 3 + 1). Step 3 Use the quadratic formula to solve 3 ± ( 3) = 0 and find the remaining two = irrational roots. = 3 ± 1 5 So the roots of the polynomial are, 3 + 5, and 3 5. Step 4 Graph the polynomial. First, plot the -intercepts, which are, , and The function is cubic, so it is S-shaped. Because the function is a polynomial with odd degree and has a positive leading coefficient, the graph approaches negative infinity as approaches negative infinity, and the graph approaches positive infinity as approaches positive infinity. Find and plot a few additional points p() ( 1, 5) (, 0) p () 6 4 (0, ) (0.38, 0) (.6, 0) (1, 3) ( 3, 19) UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

26 Problem Set Find the roots of the polynomial function. 1. f () = + 30 y 4. g() = y f () = + 4 y 5. h() = y g() = y 6. p() = y GRAPHING POLYNOMIALS 345 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

27 Describe the end behavior of the function. 7. h() = f () = p () = f () = f () = g () = h() = f () = Graph the polynomial function by using its intercepts. 15. p () = f () = g () = f () = h() = h() = f () = 3 +. g () = f () = p () = f () = h() = p() = f () = f () = Answer the question. * 30. Challenge Describe the degree and the leading coefficient of the function whose graph is shown below. 6 4 y UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

28 Factoring Polynomials Completely Every polynomial of degree n > 0 can be written as the product of n linear factors. Factoring a Binomial Over the Set of Comple Numbers Eample 1 Write + 5 as the product of two linear factors. Solution The equation + 5 = 0 is equivalent to the equation = 5, which has solutions 5i and 5i. Therefore, + 5 = ( + 5i)( 5i). The polynomial in Eample 1 has no real zeros, but it has two comple zeros. Some polynomials have both real zeros and comple zeros. The Multiplicity of a Root For a polynomial that has a linear factor appearing more than once, the idea of multiplicity is useful. DEFINITION For any polynomial p(), a root a of p() = 0 has multiplicity m if the factor a occurs m times in the factorization of p(). If a is a real root with odd multiplicity, the graph of the function crosses the -ais at = a. If a is a real root with even multiplicity, the graph of the function touches, but does not cross, the -ais at = a. Eample Given that f () = = ( 3)( 3)( 1)( + ): A Find the multiplicity of each root of f () = 0. B Describe the behavior of the graph at each root. Solution The roots are 3, 3, 1, Solution The graph of and. The root 3 occurs two f () = times, so it has a multiplicity of. shows the behavior of the function The roots 1 and occur once, at its zeros. The graph crosses the so each has a multiplicity of 1. -ais at the roots of odd multiplicity: = 1 and =. The graph touches, but does not cross, the -ais at the root of even multiplicity: = 3. THINK ABOUT IT Recall that the set of real numbers is a subset of the set of comple numbers. If a polynomial has both real zeros and comple zeros, then all of its zeros are comple. In fact, all the zeros of a polynomial are comple. 6 (, 0) (3, 0) 4 4 (1, 0) f () FACTORING POLYNOMIALS COMPLETELY 347 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

29 Finding the Number of Roots of a Polynomial Equation THE FUNDAMENTAL THEOREM OF ALGEBRA Every polynomial equation p() = 0 with degree n, n > 0, has at least one comple root. Corollary Every polynomial equation p() = 0 with degree n, n > 0, has n comple roots as long as any root with multiplicity m is counted m times. Eample 3 Find the number of comple roots of each equation. A = 0 B 3 = 0 Solution The equation has degree Solution The equation has degree 5, so there are five comple roots. 1, so there is one comple root. Factoring a Polynomial Over the Set of Comple Numbers HOW TO FACTOR A POLYNOMIAL COMPLETELY OVER THE SET OF COMPLEX NUMBERS Step 1 Use the degree of the polynomial to determine the total number of comple roots you need to find. Step Identify the possible rational roots. Step 3 Find rational roots (using a graph, synthetic division, or other strategies) and use them to factor the polynomial. Step 4 Factor the remaining polynomial. Eample 4 Factor each polynomial completely. Then sketch the graph. A f () = Solution Step 1 The polynomial is of degree 3, so there are three comple roots. Step Use the rational root theorem to identify all possible rational roots. The leading coefficient is 1 and the constant is 8, so the possible rational roots are ±1, ±, ±4, and ±8. Step 3 Of the possible rational roots, only 4 is actually a root So f () = ( 4)(1 + ) = ( 4)( + ). REMEMBER All real numbers are comple numbers of the form a + bi, where b = 0. The real root 4 is also the comple root 4 + 0i. 348 UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

30 Step 4 Find the two remaining roots. Use the quadratic formula to solve + = 0. = ± ( ) = ± 4 = ± i = 1 ± i The roots 1 + i and 1 i are comple. So the complete factorization of f () is f () = ( 4)[ (1 + i)][ (1 i)]. The roots of the equation = 0 are 4, 1 + i, and 1 i. In the graph of a function with both real roots and comple roots, only the real roots are -intercepts. The graph of f () shows that only the real root, 4, is an -intercept. B f () = Solution Step 1 The polynomial is of degree 4, so there are four comple roots. Step By the rational root theorem, the possible rational roots are ±1 and ±. Step 3 Synthetic division shows 1 and 1 as roots So f () = ( + 1)( 1)(1 ). Step 4 Find the two remaining roots. Use the quadratic formula to solve = 0. = ± ( ) 4 1( ) 1 = ± 1 = ± 3 = 1 ± 3 The roots and 1 3 are irrational. The factorization of f () is f () = ( +1)( 1)[ (1 + 3 )][ (1 3 )]. Because all the roots are real, all four roots are -intercepts, as shown in the graph. The irrational roots are approimately 0.73 and (4, 0) TIP f () Use the quotient from the first synthetic division to set up the synthetic division for the net root. This will give you less work, as well as a quotient that is a quadratic. f () ( 1, 0) (1, 0) (.73, 0) ( 0.73, 0) FACTORING POLYNOMIALS COMPLETELY 349 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

31 Problem Set Find the number of comple roots of the polynomial equation = = = = = = = = 0 Find the multiplicity of each root of p() = 0. Describe the behavior of the graph at each root. 9. p() = = ( + 1)( + 1)( 7)( + 3) 10. p() = = ( + )( + )( )( ) 11. p() = = ( 1)( + )( 3)( + 4) 1. p() = = ( 1)( 1)( 5)( 5) Find the number of real and rational roots of the polynomial equation = = = = 0 Factor completely. 17. g () = f () = h () = g () = p () = h () = f () = f () = p () = f () = g () = * 8. Challenge f () = UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

32 Applications: Polynomials You can use polynomials to model and solve many types of real-world problems. Application: Finance Eample 1 The daily closing price of a share of stock during one week can be modeled by f () = , where is the day of the week. What was the stock s closing price on day 4? Solution You need to evaluate the polynomial for = 4. Use synthetic division. The remainder is 7. The stock s closing price on day 4 was $ REMEMBER f(4) is equal to the remainder of f() ( 4). Application: Business Eample The number of customers, in hundreds, entering a theme park on its grand-opening day is given by f () = 3 +, where is the number of hours since the park opened. How long does it take for a total of 1400 customers to enter the park? Solution Because f () is given in hundreds, substitute 14 for f (). 14 = = Subtract 14 from each side to get f () = 0. To solve the equation, you need to factor the polynomial. Use the rational root theorem to find possible roots: ±1, ±, ±7, and ± TIP All numbers in the domain are positive, so there is no need to check negative roots. Synthetic division shows that is a root. The polynomial factors into ( )( ). Use the quadratic formula to factor the quadratic epression. = 4 ± = ± i 3 The factored form of the equation is 0 = ( )[ ( + i 3 )][ ( i 3 )]. Since is the only real root, it is the only solution that makes sense for this problem. It takes hours for 1400 customers to enter the park. APPLICATIONS: POLYNOMIALS 351 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

33 Application: Geometry Area and volume of geometric figures are sometimes represented by polynomials. Eample 3 A The area of a rectangle is Find an epression for the length of the rectangle if the width is Solution Divide by Synthetic division shows that ( ) ( + 10) = 3 +. You can write = (3 + )( + 10). Because A = lw, the length is given by 3 +. B The length of a crate is 3 meters greater than its width, and its height is meters greater than its width. If the volume is 40 cubic meters, what is the height of the bo? Solution Let represent the width. Then + 3 represents the length and + represents the height. Because V = lwh, the volume can be modeled by the equation V = ( + 3)( + ). Once you substitute 40 for V, you have 40 = ( + 3)( + ). 40 = Multiply on the right. 0 = Subtract 40 from both sides. Use the factors of 40 to test for a root of the equation. Synthetic division shows that is a root The equation is 0 = ( )( ). Use the quadratic formula to factor = 7 ± = 7 ± i 31 The only real solution is. The width is meters, and the height is meters greater than the width, or 4 meters. REMEMBER Length must be positive, so try only positive factors of 40 when you test roots. Application: Manufacturing Eample 4 Javier forms a bo without a top by cutting squares of equal size from the corners of a piece of cardboard measuring 10 centimeters by 8 centimeters. After removing the squares, he folds up the sides. Write and graph the function that gives the volume of the bo, where is the side length of a square Javier removes. Use the graph to estimate the greatest possible volume. 35 UNIT 9 SOLVING AND GRAPHING POLYNOMIALS Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

34 Solution A diagram of the cardboard piece is shown at the right. Once Javier folds up the sides the height will be, the length will be 10, and the width will be 8. This makes the volume function V() = (10 )(8 ). Multiply the right side to get V() = To graph the function, use synthetic division to evaluate the polynomial for different values of. Because the side of a square must be greater than 0, choose only positive values of. Use these ordered pairs and what you know about the end behavior of a cubic equation with a positive leading coefficient to complete the graph The side length of a removed square must be less than 4 centimeters because the 10 width is only 8 centimeters. Therefore, the domain for this situation is restricted to values between 0 and 4. In this region, the greatest volume occurs when is about 1.5. When = 1.5, V () = 5.5. The greatest possible volume is about 5.5 cubic centimeters y 8 THINK ABOUT IT When = 0, V () = 0 because this means no squares are being removed, and no bo can be made. A flat sheet of cardboard has no volume. Problem Set Solve. 1. Over the course of a day, the price of a share of stock can be modeled by P() = , where is the number of hours after 9:30 a.m., when the Stock Echange opens. What was the stock s price at noon?. The number of birds living in a tree over the course of a year is given by B() = , where is the number of months. How long does it take for there to be 54 birds living in the tree? 3. The width, in meters, of a rectangular garden is + 5. If the area of the garden is , what is the length? 4. The volume of a crate in meters is V() = The length is given by + 5, while the width is + 3. Find the value of for the crate to have a volume 900 m The width of a cooler is 1 centimeters shorter than its length, and its height is 6 centimeters shorter than its length. If the volume is 5184 cubic centimeters, what is the length of the cooler? APPLICATIONS: POLYNOMIALS 353 Copyright 009, K1 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.

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