Polynomial Degree and Finite Differences

CONDENSED LESSON 7.1 Polynomial Degree and Finite Differences In this lesson, you Learn the terminology associated with polynomials Use the finite differences method to determine the degree of a polynomial Find a polynomial function that models a set of data A polynomial in one variale is any epression that can e written in the form a n n a n1 n1 a 1 1 a 0 where is a variale, the eponents are nonnegative integers, and the coefficients are real numers. A function in the form f() a n n a n1 n1 a 1 1 a 0 is a polynomial function. The degree of a polynomial or polynomial function is the power of the term with the greatest eponent. If the degrees of the terms of a polynomial decrease from left to right, the polynomial is in general form. The polynomials elow are in general form. 1st degree nd degree 3rd degree th degree 3 7 1.8 9 3 11 5 A polynomial with one term, such as 5, is called a monomial. A polynomial with two terms, such as 3 7, is called a inomial. A polynomial with three terms, such as 1.8, is called a trinomial. Polynomials with more than three terms, such as 9 3 11, are usually just called polynomials. For linear functions, when the -values are evenly spaced, the differences in the corresponding y-values are constant. This is not true for polynomial functions of higher degree. However, for nd-degree polynomials, the differences of the differences, called the second differences and areviated D,are constant. For 3rd-degree polynomials, the differences of the second differences, called the third differences and areviated D 3,are constant. This is illustrated in the tales on page 361 of your ook. If you have a set of data with equally spaced -values, you can find the degree of the polynomial function that fits the data (if there is a polynomial function that fits the data) y analyzing the differences in y-values. This technique, called the finite differences method, is illustrated in the eample in your ook. Read that eample carefully. Notice that the finite differences method determines only the degree of the polynomial. To find the eact equation for the polynomial function, you need to find the coefficients y solving a system of equations or using some other method. In the eample, the D values are equal. When you use eperimental data, you may have to settle for differences that are nearly equal. Investigation: Free Fall If you have a motion sensor, collect the (time, height) data as descried in Step 1 in your ook. If not, use these sample data. (The values in the last two columns are calculated in Step.) Discovering Advanced Algera Condensed Lessons CHAPTER 7 93

Lesson 7.1 Polynomial Degree and Finite Differences Complete Steps 6 in your ook. The results given are ased on the sample data. Step The first and second differences, D 1 and D,are shown in the tale at right. For these data, we can stop with the second differences ecause they are nearly constant. Step 3 The three plots are shown elow. 0.5 1.008 (L1, L) (L3, L) (L5, L6) Time (s) Height (m) y D 1 D 0.00.000 0.05 1.988 0.10 1.951 0.15 1.890 0.0 1.80 0.5 1.69 0.30 1.559 0.35 1.00 0.0 1.16 0.01 0.037 0.061 0.086 0.110 0.135 0.159 0.18 0.08 0.05 0.0 0.05 0.0 0.05 0.0 0.05 0.0 [0, 1, 0.5, 0,.5, 0.5] [0, 1, 0.5, 0.5, 0.5, 0.5] [0, 1, 0.5, 0.5, 0.5, 0.5] Step The graph of (L1, L) appears paraolic, suggesting that the correct model may e a nd-degree polynomial function. The graph of (L3, L) shows that the first differences are not constant ecause they decrease in a linear fashion. The graph of (L5, L6) shows that the second differences are nearly constant, so the correct model should e a nd-degree polynomial function. Step 5 A nd-degree polynomial in the form y a c fits the data. Step 6 To write the system, choose three data points. For each point, write an equation y sustituting the time and height values for and y in the equation y a c. The following system is ased on the values (0, ), (0., 1.80), and (0., 1.16). c.000 0.0a 0. c 1.80 0.16a 0. c 1.16 One way to solve this system is y writing the matri equation 0 0 1 a.000 0.0 0. 1 1.80 0.16 0. 1 c 1.16 and solving using an inverse matri. The solution is a.9, 0, and c, so the equation that fits the data is y.9. Read the remainder of the lesson in your ook. 9 CHAPTER 7 Discovering Advanced Algera Condensed Lessons

CONDENSED LESSON 7. Equivalent Quadratic Forms In this lesson, you Learn aout the verte form and factored form of a quadratic equation and the information each form reveals aout the graph Use the zero-product property to find the roots of a factored equation Write a quadratic model for a data set in verte, general, and factored form A nd-degree polynomial function is called a quadratic function. In Lesson 7.1, you learned that the general form of a quadratic function is y a c. In this lesson you will eplore other forms of a quadratic function. You know that every quadratic function is a transformation of y.when a quadratic function is written in the form y k a h or y a h k, you can tell that the verte of the paraola is (h, k) and that the vertical and horizontal scale factors are a and. Conversely, if you know the verte of a paraola and you know (or can find) the scale factors, you can write its equation in one of these forms. The quadratic function y a h k can e rewritten in the form a y ( h) k. The coefficient a comines the two scale factors into one vertical scale factor. In the verte form of a quadratic equation, y a( h) k, this single scale factor is simply denoted a. From this form, you can identify the verte, (h, k), and the vertical scale factor, a. If you know the verte of a paraola and the vertical scale factor, you can write an equation in verte form. The zero-product property states that for all real numers a and, if a 0, then a 0, or 0, or a 0 and 0. For eample, if 3( 7) 0, then 3 0 or 7 0. Therefore, 0 or 7. The solutions to an equation in the form f() 0 are called the roots of the equation, so 0 and 7 are the roots of 3( 7) 0. The -intercepts of a function are also called the zeros of the function (ecause the y-values are 0). The function y 1.( 5.6)( 3.1), given in Eample A in your ook, is said to e in factored form ecause it is written as the product of factors. The zeros of the function are the solutions of the equation 1.( 5.6)( 3.1) 0. Eample A shows how you can use the zero-product property to find the zeros of the function. In general, the factored form of a quadratic function is y a r 1 r. From this form, you can identify the -intercepts (or zeros), r 1 and r, and the vertical scale factor, a. Conversely, if you know the -intercepts of a paraola and know (or can find) the vertical scale factor, then you can write the equation in factored form. Read Eample B carefully. Investigation: Rolling Along Read the Procedure Note and Steps 1 3 in your ook. Make sure you can visualize how the eperiment works. Use these sample data to complete Steps 8, and then compare your results to those elow. (These data have een adjusted for the position of the starting line as descried in Step 3.) Discovering Advanced Algera Condensed Lessons CHAPTER 7 95

Lesson 7. Equivalent Quadratic Forms Time (s) Distance from line (m), y 0. 0.357 0. 0.355 0.6 0.357 0.8 0.18 1.0 0.56 1. 1.0 1. 1.81 1.6.357 1.8.85.0 3.31 Time (s) Distance from line (m), y. 3.570. 3.81.6.08.8.188 3.0.56 3..57 3..193 3.6.06 3.8 3.871.0 3.619 Time (s) Distance from line (m), y. 3.309..938.6.510.8.08 5.0 1.93 5. 0.897 5. 0.61 5.6 0.399 5.8 0.6 6.0 0.19 Step At right is a graph of the data. The data have a paraolic shape, so they can e modeled with a quadratic function. Ignoring the first and last few data points (when the can started and stopped), the second differences, D,are almost constant, at around 0.06, which implies that a quadratic model is appropriate. Step 5 The coordinates of the verte are (3.,.57). Consider (5., 0.897) to e the image of (1, 1). The horizontal and vertical distances of (1, 1) from the verte of y are oth 1. The horizontal distance of (5., 0.897) from the verte, (3.,.57), is, and the vertical distance is 3.36. So, the horizontal and vertical scale factors are and 3.36, respectively. This can e represented as the single vertical scale factor 3. 36 0.8. Therefore, the verte form of a model for the data is y 0.8( 3.).57. Step 6 Sustituting the points (1, 0.56), (3,.56), and (5, 1.93) into the general form, y a c, gives the system a c 0.56 9a 3 c.56 5a 5 c 1.93 The solution to this system is a 0.81, 5.10, and c 3.7, so the general form of the equation is y 0.81 5.10 3.7. Step 7 The -intercepts are aout (0.9, 0) and (5.5, 0). The scale factor, found in Step 5, is 0.8. So, the intercept form of the equation is y 0.8( 0.9)( 5.5). [1, 7, 1, 1, 7, 1] Step 8 In general, you use the verte form when you know either the verte and the scale factor or the verte and one other point you can use to find the scale factor. You use the general form when you know any three points. You use the factored form when you know the -intercepts and at least one other point you can use to find the scale factor. [1, 7, 1, 1, 7, 1] 96 CHAPTER 7 Discovering Advanced Algera Condensed Lessons

CONDENSED LESSON 7.3 Completing the Square In this lesson, you Use the method of completing the square to find the verte of a paraola whose equation is given in general form Solve prolems involving projectile motion Many real-world prolems involve finding the minimum or maimum value of a function. For quadratic functions, the maimum or minimum value occurs at the verte. If you are given an equation in verte form, you can easily find the coordinates of a paraola s verte. It is also fairly straightforward to find the verte if the equation is in factored form. It gets more complicated if the equation is in general form. In this lesson you will learn a technique for converting a quadratic equation from general form to verte form. Projectile motion the rising or falling of ojects under the influence of gravity can e modeled y quadratic functions. The height of a projectile depends on the height from which it is thrown, the upward velocity with which it is thrown, and the effect of gravity pulling down on the oject. The height can e modeled y the function y 1 g v 0 s 0 where is the time in seconds, y is the height (in m or ft), g is the acceleration due to gravity (either 9.8 m/s or 3 ft/s ), v 0 is the initial upward velocity of the oject (in either m/s or ft/s), and s 0 is the initial height of the oject (in m or ft). Read Eample A in your ook. It illustrates how to write a projectile motion equation when you know only the -intercepts (the two times when the height is 0) and how to use the -intercepts to find the coordinates of the verte. It is important to understand that the -coordinate of a paraola s verte is the mean of the -intercepts. This fact allows you to find the verte of a paraola from the factored form of its equation. Look at the diagrams for ( 5) and (a ) in your ook. Make sure you understand the pattern descried after the diagrams. Investigation: Complete the Square Complete the investigation in your ook. When you are finished, compare your answers to those elow. Step 1 a. You must add 9.. 6 6 9 9 ( 3) 9 c. Enter 6 as Y1 and ( 3) 9 as Y, and verify that the tale values or the graphs are the same for oth epressions. Step a. 9. 6 6 9 9 ( 3) 13 c. Enter 6 as Y1 and ( 3) 13 as Y, and verify that the tale values or the graphs are the same for oth epressions. Discovering Advanced Algera Condensed Lessons CHAPTER 7 97

Lesson 7.3 Completing the Square Step 3 a. Focus on 1. To complete a perfect square, you need to add 9. (This gives an epression in the form a a,where a and 7.) You need to sutract 9 to compensate. So, 1 3 1 9 9 3 ( 7) 6. To make a perfect square, you must add,or. You need to sutract to compensate. So, 10 10 10 Step a. 6 1 ( 3) 1 Factor 6. Step 5 3 9 9 1 Complete the square. You add 9, so you must sutract 9. 3 7 Write in the form a( h) k.. a 10 7 a 1 a0 7 Factor a 10. a c a a c Step 6 y a a h a a 1 0 a a5 a a 5 7 Complete the square. You add a 5 a, so you must sutract a 5 a. a 5 a 7 a5 Write in the form a( h) k. a a a a c a a c a Using the results from Step 5, you can write y a c as c. a This is in verte form, y a( h) k, where and k c. So, the verte is, c. a a a Based on your work in the investigation, you now know two ways to find the verte, (h, k), of a quadratic function given in general form, y a c. 1. You can use the process of completing the square to rewrite the equation in verte form, y a( h) k, and then read the verte from the rewritten equation.. You can use the formulas h a and k = c of the verte directly. a to calculate the coordinates You can use either method, ut make sure you are comfortale with completing the square ecause it will come up in your later work. Eample B in your ook demonstrates oth methods. Work through that eample carefully. Eample C applies what you learned in the investigation to solve a projectile motion prolem. Try to solve the prolem on your own efore reading the solution. 98 CHAPTER 7 Discovering Advanced Algera Condensed Lessons

CONDENSED LESSON 7. The Quadratic Formula In this lesson, you Learn how the quadratic formula is derived Use the quadratic formula to solve projectile motion prolems You can use a graph to approimate the -intercepts of a quadratic function. If you can write the equation of the function in factored form, you can find the eact values of its -intercepts. Unfortunately, most quadratic equations cannot easily e converted to factored form. In this lesson you will learn a method that will allow you to find the eact -intercepts of any quadratic function. Read Eample A in your ook carefully, and then read the eample elow. EXAMPLE Find the -intercepts of y 7 1. Solution The -intercepts are the solutions of 7 1 0. See if you can supply the reason for each step in the solution elow. Note that the first four steps involve rewriting the left side in the form a( h) k. 7 1 0 7 1 0 7 9 16 9 8 1 0 7 1 8 0 7 1 8 7 1 1 6 7 1 7 1 7 1 7 1 7 1 The -intercepts are 3.351 and = 0.19. The series of equations after Eample A in your ook shows how you can derive the quadratic formula y following the same steps used in Eample A. The quadratic formula, ac a gives the general solution to a quadratic equation in the form a c 0. Follow along with the steps in the derivation, using a pencil and paper. To make sure you understand the quadratic formula, use it to verify that the solutions of 7 1 7 1 7 1 0 are and. Discovering Advanced Algera Condensed Lessons CHAPTER 7 99

Lesson 7. The Quadratic Formula Investigation: How High Can You Go? Complete the investigation in your ook, and then compare your answers to those elow. Step 1 The equation is y 16 88 3, where y is the height and is the time in seconds. (If you answered this question incorrectly, look ack at the discussion of projectile motion in Lesson 7.3.) Step The equation is 16 88 3. Step 3 In a c 0 form, the equation is 16 88 1 0. For this equation, a 16, 88, and c 1. Sustituting these values into the quadratic formula gives 88 88 (16 )(1 ) 88 8 8 80 600 (16) 3 3 So, 8 8 80 3 0.5 or 8 8 80 3 5.5. So, the all is feet aove the ground 0.5 second after it is hit (on the way up) and 5.5 seconds after it is hit (on the way down). Step The verte is the maimum. The all reaches the maimum height only once. The all reaches other heights once on the way up and once on the way down, ut the maimum point is the height where the all changes directions, so it is reached only once. Step 5 The equation is 1 16 88 3. In a c 0 form, the equation is 16 88 11 0. For this equation, a 16, 88, and c 11. Sustituting these values into the quadratic formula gives 88 88 (16 )(1 1) 88 88 0 (16) 3 3.75 The all reaches a maimum height.75 seconds after it is hit. The fact that there is only one solution ecomes apparent when you realize that the value under the square root sign is 0. Step 6 The equation is 00 16 88 3. In a c 0 form, the equation is 16 88 197 0. For this equation, a 16, 88, and c 197. Sustituting these values into the quadratic formula gives 88 88 (16 88 86 )(19 7) (16) 3 The value under the square root sign is negative. Because the square root of a negative numer is not a real numer, the equation has no real-numer solution. Your work in the investigation shows that when the value under the square root sign, ac, is 0, the equation a c 0 has only one solution, and when the value under the square root sign, ac, is negative, the equation a c 0 has no real-numer solutions. This means that if you are given a quadratic equation in the general form, you can use the value of ac to determine whether the graph will have zero, one, or two -intercepts. Eample B in your ook shows the importance of writing an equation in general form efore you attempt to apply the quadratic formula. Read the eample carefully. 100 CHAPTER 7 Discovering Advanced Algera Condensed Lessons

CONDENSED LESSON 7.5 Comple Numers In this lesson, you Learn that some polynomial equations have solutions that are comple numers Learn how to add, sutract, multiply, and divide comple numers The graph of y.5 has no -intercepts. y If you use the quadratic formula to attempt to find the -intercepts, you get 1 1(1)(.5 1 9 ) (1) 1 9 1 9 The numers and are not real numers ecause they involve the square root of a negative numer. Numers that include the real numers as well as the square roots of negative numers are called comple numers. Defining the set of comple numers makes it possile to solve equations such as.5 0 and 0, which have no solutions in the set of real numers. The square roots of negative numers are epressed using an imaginary unit called i, defined y i 1 or i 1. You can rewrite 9 as 9 1, or 3i. Therefore, the two solutions to the quadratic equation aove can e written as 1 3i and 1 3i, or 1 3 i and 1 3 i. These two solutions are a conjugate pair, meaning that one is in the form a + i and the other is in the form a i. The two numers in a comple pair are comple conjugates. Roots of polynomial equations can e real numers or nonreal comple numers, or there may e some of each. However, as long as the polynomial has real-numer coefficients, any nonreal roots will come in conjugate pairs, such as 3i and 3i or 6 5i and 6 5i. Your ook defines a comple numer as a numer in the form a i, where a and are real numers and i 1. The numer a is called the real part, and the numer is called the imaginary part. The set of comple numers includes all real numers and all imaginary numers. Look at the diagram on page 39 of your ook, which shows the relationship etween these numers, and some other sets of numers you may e familiar with, as well as eamples of numers in each set. Then read the eample in your ook, which shows how to solve the equation 3 0. Discovering Advanced Algera Condensed Lessons CHAPTER 7 101

Lesson 7.5 Comple Numers Investigation: Comple Arithmetic In this investigation you discover the rules for computing with comple numers. Part 1: Addition and Sutraction Adding and sutracting comple numers is similar to comining like terms. Use your calculator to add or sutract the numers in Part 1a d in your ook. Then, make a conjecture aout how to add comple numers without a calculator. Below are the solutions and a possile conjecture. a. 5 i. 5 3i c. 1 9i d. 3 i Possile conjecture: To add two comple numers, add the real parts and add the imaginary parts. In symols, (a i) (c di) (a c) ( d)i. Part : Multiplication Multiplying the comple numers a i and c di is very similar to multiplying the inomials a and c d. You just need to keep in mind that i 1. Multiply the comple numers in Part a d, and epress the answers in the form a i. The answers are elow. a. ( i)(3 5i) 3 5i i 3 i 5i Epand as you would for a product of inomials. 6 10i 1i 0i 6 i 0i 6 i 0(1) Multiply within each term. Comine 10i and 1i. i 1 6 i Comine 6 and 0.. 16 3i c. 1 16i d. 8 16i Part 3: The Comple Conjugates Complete Part 3a d, which involves finding either the sum or product of a comple numer and its conjugate. The answers are given elow. a.. 1 c. 0 d. 3 Possile generalizations: The sum of a numer and its conjugate is a real numer: (a i) (a i) a. The product of a real numer and its conjugate is a real numer: (a i)(a i) a. 10 CHAPTER 7 Discovering Advanced Algera Condensed Lessons

Lesson 7.5 Comple Numers Part : Division To divide two comple numers, write the division prolem as a fraction, conjugate of denominator multiply y conjugate of denominator (to change the denominator to a real numer), and then write the result in the form a i. Divide the numers in Part a d. Here are the answers. a. 7 i 1 i 7 i 1 i 1 i 1 conjugate of denominator Multiply y. i conjugate of denominator 5 9i.5.5i Multiply. The denominator ecomes a real numer. Divide.. 0.56 0.9i c. 0. 0.0i d. 0.6 0.8i Comple numers can e graphed on a comple plane, where the horizontal ais is the real ais and the vertical ais is the imaginary ais. The numer a i is represented y the point with coordinates (a, ). The numers 3 i and i are graphed elow. 5 i 5 Imaginary ais 3 i 5 Real ais 5 Discovering Advanced Algera Condensed Lessons CHAPTER 7 103

CONDENSED LESSON 7.6 Factoring Polynomials In this lesson, you Learn aout cuic functions Use the -intercepts of a polynomial function to help you write the function in factored form The polynomial equations y 6 9 and y ( 3)( 3) are equivalent. The first is in general form, and the second is in factored form. Writing a polynomial equation in factored form is useful for finding the -intercepts, or zeros, of the function. In this lesson you will learn some techniques for writing higher-degree polynomials in factored form. A 3rd-degree polynomial function is called a cuic function. At right is a graph of the cuic function y 3 16 9 36. The -intercepts of the function are, 1.5, and 1.5, so its factored equation must e in the form y a( )( 1.5)( 1.5). To find the value of a, you can sustitute the coordinates of another point on the curve. The y-intercept is (0, 36). Sustituting this point into the equation gives 36 a()(1.5)(1.5). So, a, and the factored form of the equation is y ( )( 1.5)( 1.5) (, 0) 5 y 50 (0, 36) ( 1.5, 0) 50 (1.5, 0) 5 Investigation: The Bo Factory You can make a o from a 16-y-0-unit sheet of paper, y cutting squares of side length from the corners and folding the sides up. Follow the Procedure Note in your ook to construct oes for several different integer values of. Record the dimensions and volume of each o. (If you don t want to construct the oes, try topicture them in your mind.) Complete the investigation, and then compare your results to those elow. Step 1 Here are the results for integer -values from 1 to 6. 0 16 Volume Length Width Height y 1 18 1 1 5 16 1 38 3 1 10 3 0 1 8 38 5 10 6 5 300 6 8 6 19 Step The dimensions of the oes are 0, 16, and. Therefore, the volume function is y (16 )(0 )(). Discovering Advanced Algera Condensed Lessons CHAPTER 7 105

Lesson 7.6 Factoring Polynomials Step 3 The data points lie on the graph of the function. [, 1, 1, 00, 500, 100] Step If you were to epand (16 )(0 )(), the result would e a polynomial, and the highest power of would e 3. Therefore, the function is a 3rd-degree polynomial function. Step 5 The -intercepts of the graph are 0, 8, and 10, so the function is y ( 8)( 10). Step 6 The graphs have the same -intercepts and general shape ut different vertical scale factors. A vertical scale factor of makes them equivalent: y ( 8)( 10). Step 7 If 0, there are no sides to fold up, so a o cannot e formed. For 8, 8-unit-wide strips would e cut off the sides of the sheet. Folding up the sides would mean folding the remaining strip in half, which would not form a o. 8 8 [, 1, 1, 00, 500, 100] 8 Cut off Cut off 8 16 8 Cut off Cut off 8 8 8 0 A value of 10 is impossile ecause it is more than half the length of the shorter side of the sheet. Only a domain of 0 8 makes sense in this situation. By zooming and tracing to find the coordinates of the high point of the graph, you can find that the -value of aout.9 maimizes the volume. Work through the eample in your ook, which asks you to find the factored form of a polynomial function y using the -intercepts of the graph. This method works well when the zeros of a function are integer values. Unfortunately, this is not always the case. Sometimes the zeros of a polynomial are not nice rational or integer values, and sometimes they are not even real numers. With quadratic functions, if you cannot find the zeros y factoring or making a graph, you can always use the quadratic formula. Once you know the zeros, r 1 and r,you can write the polynomial in the form y a r 1 r.read the remainder of the lesson in your ook, and then read the eample on the net page. 106 CHAPTER 7 Discovering Advanced Algera Condensed Lessons

Lesson 7.6 Factoring Polynomials EXAMPLE Write the equation of the function elow in factored form. y 6 (, ) 8 6 8 Solution The factored equation is in the form y a r 1 r,where r 1 and r are the zeros. From the graph, you can see that the only real-numer zero is 3. If the other zero were a nonreal numer, then its conjugate would also e a zero. This would mean there are three zeros, which is not possile. So, 3 must e a doule zero. This means that the function is in the form y a( 3)( 3), or y a( 3).To find the value of a, sustitute (, ): a(1),so a. The factored form of the function is y ( 3). Discovering Advanced Algera Condensed Lessons CHAPTER 7 107

CONDENSED LESSON 7.7 Higher-Degree Polynomials In this lesson, you Descrie the etreme values and end ehavior of polynomial functions Solve a prolem that involves maimizing a polynomial function Write equations for polynomial functions with given intercepts Polynomials with degree 3 or higher are often referred to as higher-degree polynomials. At right is the graph of the polynomial y ( 3),or y 3 6 9. The zero-product property tells you that the zeros are 0 and 3 the values of for which y 0. The -intercepts of the graph confirm this. The graph has other key features in addition to the -intercepts. For eample, the point (1, ) is called a local minimum ecause it is lower than the other points near it. The point (3, 0) is called a local maimum ecause it is higher than the other points near it. You can also descrie the end ehavior of the graph that is, what happens to the graph as increases in the positive and negative directions. For this graph, as increases in the positive direction, y increases in the negative direction. As increases in the negative direction, y increases in the positive direction. The introduction to Lesson 7.7 in your ook gives another eample of a 3rd-degree polynomial and its graph. The graph of a polynomial function with real coefficients has a y-intercept, possily one or more -intercepts, and other features such as local maimums or minimums and end ehavior. The maimums and minimums are called etreme values. y (1, ) 6 6 Investigation: The Largest Triangle Start with an 8.5-y-11 in. sheet of paper. In centimeters, the dimensions are 1.5 y 8 cm. Orient the paper so that the long side is horizontal. Fold the upper left corner so that it touches some point on the ottom edge. Find the area, in cm,ofthe triangle formed in the lower left corner. A A What distance,, along the ottom of the paper produces the triangle with greatest area? To answer this question, first find a formula for the area of the triangle in terms of. Try to do this on your own efore reading on. Here is one way to find the formula: Let h e the height of the triangle. Then the hypotenuse has length 1.5 h. (Why?) h 1.5h Discovering Advanced Algera Condensed Lessons CHAPTER 7 109

Lesson 7.7 Higher-Degree Polynomials Use the Pythagorean Theorem to help you write h in terms of. h (1.5 h), so h 6. 5 3 Now, you can write a formula for the area, y. y 1 6. 5 3 At right is a graph of the area function. If you trace the graph, you ll find that the maimum point is aout (1.,.5). Therefore, the value of that gives the greatest area is aout 1. cm. The maimum area is aout.5 cm. [5, 5, 5, 10, 50, 10] Eample A in your ook shows you how to find the equation for a polynomial with given -intercepts and y-intercept. Read this eample carefully. To test your understanding, find a polynomial function with -intercepts 6,, and 1, and y-intercept 60. (One answer is y 5( 6)( )( 1).) Graphs A D on page 07 of your ook show the possile shapes for the graph of a 3rd-degree polynomial function. Graph A is the graph of the parent function y 3.The graph of any other 3rd-degree polynomial will e a transformation of Graph A. Eample B in your ook shows you how to find a polynomial function with given zeros when some of the zeros are comple. The key to finding the solution is to recall that comple zeros come in conjugate pairs. Read that eample carefully, and then read the eample elow. EXAMPLE Solution Find a th-degree polynomial function with real coefficients and zeros, 3, and 1 i. Comple zeros occur in conjugate pairs, so 1 i must also e a zero. So, one possile function, in factored form, is y ( )( 3)( (1 i))( (1 i)) Multiply the factors to get a polynomial in general form. y ( )( 3)( (1 i))( (1 i)) ( 6)( (1 i) (1 i) (1 i)(1 i)) ( 6)( i i ) ( 6)( ) 3 3 6 1 1 3 3 10 1 Check the solution y making a graph. (You will see only the real zeros.) Note that an nth-degree polynomial function always has n zeros. However, some of the zeros may e nonreal numers, the function may not have n -intercepts. 110 CHAPTER 7 Discovering Advanced Algera Condensed Lessons

CONDENSED LESSON 7.8 More Aout Finding Solutions In this lesson, you Use long division to find the roots of a higher-degree polynomial Use the Rational Root Theorem to find all the possile rational roots of a polynomial Use synthetic division to divide a polynomial y a linear factor You can find the zeros of a quadratic function y factoring or y using the quadratic formula. You can sometimes use a graph to find the zeros of higherdegree polynomials, ut this method may give only an approimation of real zeros and won t work at all to find nonreal zeros. In this lesson you will learn a method for finding the eact zeros, oth real and nonreal, of many higher-degree polynomials. Eample A in your ook shows that if you know some of the zeros of a polynomial function, you can sometimes use long division to find the other roots. Follow along with this eample, using a pencil and paper. Make sure you understand each step. To confirm that a value is a zero of a polynomial function, you sustitute it into the equation to confirm that the function value is zero. This process uses the Factor Theorem, which states that ( r) is a factor of a polynomial function P() if and only if P(r) 0. When you divide polynomials, e sure to write oth the divisor and the dividend so that the terms are in order of decreasing degree. If a degree is missing, insert a term with coefficient 0 as a placeholder. For eample, to divide 13 36 y 9, rewrite 13 36 as 0 3 13 0 36 and rewrite 9 as 0 9. The division prolem elow shows that 13 36 9. 0 9 0 3 13 0 36 0 3 9 0 36 0 36 0 In Eample A, you found some of the zeros y looking at the graph. If the -intercepts of a graph are not integers, identifying the zeros can e difficult. The Rational Root Theorem tells you which rational numers might e zeros. It states that if the polynomial equation P() 0 has rational roots, then they are of the form p,where q p is a factor of the constant term and q is a factor of the leading coefficient. Note that this theorem helps you find only rational roots. Eample B shows how the theorem is used. Work through that eample, and then read the eample on the net page. Discovering Advanced Algera Condensed Lessons CHAPTER 7 111

Lesson 7.8 More Aout Finding Solutions EXAMPLE Find the roots of 7 3 3 56 0. Solution The graph of the function y = 7 3 3 56 appears at right. None of the -intercepts are integers. The Rational Root Theorem tells you that any rational root will e a factor of divided y a factor of 7. The factors of are 1,, 3,, 6, 8, 1, and. The factors of 7 are 1 and 7. You know there are no integer roots, so you need to consider only 1 7, 7, 3 7, 7, 6 7, 8 7, 1 7, and 7. The graph indicates that one of the roots is etween 3 and. None of these possiilities are in that interval. Another root is a little less than 1.This could e 3 7.Try sustituting 3 7 into the polynomial. 7 3 7 3 3 3 7 56 3 7 7 9 7 9 0 [3, 3, 1, 60, 60, 10] So, 3 7 is a root, which means 3 7 is a factor. Use division to divide out this factor. 7 56 3 7 7 3 3 56 7 3 3 56 56 0 So, 7 3 3 56 0 is equivalent to 3 7 7 56 0. To find the other roots, solve 7 56 0. The solutions are 8. So, the roots are 3, 7 8, and 8. Synthetic division is a shortcut method for dividing a polynomial y a linear factor. Read the remainder of the lesson in your ook to see how to use synthetic 7 division. Below is an eample using synthetic division to find 3 3 56. 3 7 Note that in the eample aove you found this same quotient using long division. Known zero 3_ 7 The result shows that 3 7 7 56. Coefficients of 7 3 3 56 7 3 56 Add Add Add 3 0 1 Bring 3 5 7 down 3_ 7 7 7 0 56 0 7 3 3 56 3 7 3_ 7 0 6 3_ 7 56 7 56, so 7 3 3 56 11 CHAPTER 7 Discovering Advanced Algera Condensed Lessons