Polynomial Degree and Finite Differences

CONDENSED LESSON 7.1 Polynomial Degree and Finite Differences In this lesson you will 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 variable is any epression that can be written in the form a n n a n 1 n 1 a 1 1 a 0 where is a variable, the eponents are nonnegative integers, the coefficients are real numbers, and a n 0. A function in the form f () a n n a n 1 n 1 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 below 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 binomial. 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 abbreviated D, are constant. For 3rd-degree polynomials, the differences of the second differences, called the third differences and abbreviated D 3, are constant. This is illustrated in the tables on page 379 of your book. If you have a set of data with equally spaced -values, you can find the lowest possible degree of a polynomial function that fits the data (if there is a polynomial function that fits the data) by analyzing the differences in y-values. This technique, called the finite differences method, is illustrated in the eample in your book. 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 by 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 described in Step 1 in your book. If not, use these sample data. (The values in the last two columns are calculated in Step.) (continued) Discovering Advanced Algebra Condensed Lessons CHAPTER 7 93 010 Key Curriculum Press

Lesson 7.1 Polynomial Degree and Finite Differences (continued) Complete Steps 6 in your book. The results given are based on the sample data. Step The first and second differences, D 1 and D, are shown in the table at right. For these data, we can stop with the second differences because they are nearly constant. Time (s) Height (m) y 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 0.5 1.008 Step 3 The three plots are shown below. (time, height ) (time, d 1) (time 3, d ) D 1 0.01 0.037 0.061 0.086 0.110 0.135 0.159 0.18 0.08 D 0.05 0.0 0.05 0.0 0.05 0.0 0.05 0.0 Step The graph of (time, height ) appears parabolic, suggesting that the correct model may be a nd-degree polynomial function. The graph of (time, d 1 ) shows that the first differences are not constant because they decrease in a linear fashion. The graph of (time 3, d ) shows that the second differences are nearly constant, so the correct model should be a nd-degree polynomial function. Step 5 A nd-degree polynomial in the form y a b c fits the data. Step 6 To write the system, choose three data points. For each point, write an equation by substituting the time and height values for and y in the equation y a b c. The following system is based on the values (0, ), (0., 1.80), and (0., 1.16). c.000 0.0a 0.b c 1.80 0.16a 0.b c 1.16 One way to solve this system is by writing the matri equation 0 0.0 0.16 0 0. 0. 1 1 1 a b.000 1.80 c 1.16 and solving using an inverse matri. The solution is a.9, b 0, and c, so an equation that fits the data is y.9. Read the remainder of the lesson in your book. 9 CHAPTER 7 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 7. Equivalent Quadratic Forms In this lesson you will learn about the verte form and factored form of a quadratic equation and the information each form reveals about 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 b 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 b a h or y b a h k, you can tell that the verte of the parabola is (h, k) and that the horizontal and vertical scale factors are a and b. Conversely, if you know the verte of a parabola and you know (or can find) the scale factors, you can write its equation in one of these forms. The quadratic function y b a h k can be rewritten in the form y b a ( h) k. The coefficient b a combines 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 parabola and the vertical scale factor, you can write an equation in verte form. Work through Eample A carefully. The zero-product property states that for all real numbers a and b, if ab 0, then a 0, or b 0, or a 0 and b 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 (because the corresponding y-values are 0). The function y 1.( 5.6)( 3.1), given in Eample B in your book, is said to be in factored form because 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 B 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 parabola and know (or can find) the vertical scale factor, then you can write the equation in factored form. Read Eample C carefully. Investigation: Rolling Along Read the Procedure Note and Steps 1 3 in your book. Make sure you can visualize how the eperiment works. Use these sample data to complete Steps 8, and then compare your results to those below. (These data have been adjusted for the position of the starting line as described in Step 3.) (continued) Discovering Advanced Algebra Condensed Lessons CHAPTER 7 95

Lesson 7. Equivalent Quadratic Forms (continued) 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 parabolic shape, so they can be 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 be the image of (1, 1). The horizontal and vertical distances of (1, 1) from the verte of y are both 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 be 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 Substituting the points (1, 0.56), (3,.56), and (5, 1.93) into the general form, y a b c, gives the system a b c 0.56 9a 3b c.56 5a 5b c 1.93 The solution to this system is a 0.81, b 5.09, and c 3.7, so the general form of the equation is y 0.81 5.09 3.7. Step 7 The -intercepts are about (0.9, 0) and (5.5, 0). The scale factor, found in Step 5, is 0.8. So the factored form of the equation is y 0.8( 0.9)( 5.5). 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. 96 CHAPTER 7 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 7.3 Completing the Square In this lesson you will use the method of completing the square to find the verte of a parabola whose equation is given in general form solve problems involving projectile motion Many real-world problems 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 a quadratic equation in verte form, you can easily find the coordinates of the parabola 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 objects under the influence of gravity can be modeled by 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 object. The height can be modeled by 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 object (in either m/s or ft/s), and s 0 is the initial height of the object (in m or ft). Read Eample A in your book. It illustrates how to write a projectile motion equation when you know only the -intercepts and how to use the -intercepts to find the coordinates of the verte. Investigation: Complete the Square Complete the investigation in your book. When you are finished, compare your answers to those below. Step 1 a. 5 5 5 10 5 b. ( 8) is the binomial epression being squared, and 16 6 is the perfect square trinomial. 5 5 5 5 5 c. ( 1) 1 d. a ab b The first term of the trinomial, a, is the square of the first term of the binomial. The second term of the trinomial, ab, is twice the product of the binomial terms. The third 1 1 1 1 term of the trinomial, b, is the square of the last term of the binomial. Step a. You must add 9. b. 6 6 9 9 ( 3) 9 c. Enter 6 as f 1 () and ( 3) 9 as f (), and verify that the table values or the graphs are the same for both epressions. (continued) Discovering Advanced Algebra Condensed Lessons CHAPTER 7 97

Lesson 7.3 Completing the Square (continued) Step 3 a. 9 b. 6 6 9 9 ( 3) 13 c. Enter 6 as f 1 () and ( 3) 13 as f (), and verify that the table values or the graphs are the same for both epressions. Step a. Focus on 1. To complete a perfect square, you need to add 9. You need to subtract 9 to compensate. So 1 3 1 9 9 3 ( 7) 6 b. To make b a perfect square, you must add b _, or b subtract to compensate. So b 10 b b b 10 b b b. You need to 10 Step 5 a. 6 1 ( 3) 1 Factor 6. 3 9 9 1 Complete the square. You add 9_ must subtract 9_. 3 7 Write in the form a ( h) k. b. a 10 7 a 10 a 7 Factor a 10. a 10 a 5 a a 5 a 7 Complete the square. You add a 5 must subtract a 5 a 5 a 7 5 a Write in the form a ( h) k. Step 6 The -coordinate is b a. Substitute b a for in y a b c to find the y-coordinate, which is c b a. a., so you a, so you Read Eample B carefully. Based on your work in the investigation and Eample B, you now know two ways to find the verte, (h, k), of a quadratic function given in general form, y a b c. 1. You can use the process of completing the square to rewrite the equation in verte form, y a( h) k. The verte is (h, k).. You can use the formulas h b a and k c b a to calculate the coordinates of the verte directly. You can use either method, but make sure you are comfortable with completing the square because it will come up in your later work. Eample C applies what you learned in the investigation to solve a projectile motion problem. Try to solve the problem on your own before reading the solution. 98 CHAPTER 7 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 7. The Quadratic Formula In this lesson you will learn how the quadratic formula is derived use the quadratic formula to solve projectile motion problems 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. However, most quadratic equations cannot easily be 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 book carefully, and then read the eample below. 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 below. 7 1 0 7 1 7? 1? 7 9 16 1 9 8 7 1 8 7 1 16 7 1 7 1 7 1 The -intercepts are 7 1 3.351 and 7 1 0.19. The series of equations after Eample A in your book shows how you can derive the quadratic formula by following the same steps used above. The quadratic formula, b b ac a gives the general solution to a quadratic equation in the form a b 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 0 are 7 1 and 7 1. (continued) Discovering Advanced Algebra Condensed Lessons CHAPTER 7 99

Lesson 7. The Quadratic Formula (continued) Investigation: How High Can You Go? Complete the investigation in your book, and then compare your answers to those below. Step 1 The equation is y 16 88 3, where y is the height in feet and is the time in seconds. (If you answered this question incorrectly, review the discussion of projectile motion in Lesson 7.3.) Step The equation is 16 88 3. Step 3 In a b c 0 form, the equation is 16 88 1 0. For this equation, a 16, b 88, and c 1. Substituting these values into the quadratic formula gives 88 88 ( 16)( 1) 88 600 88 80 ( 16) 3 3 88 80 So 88 80 3 0.5 or 3 5.5. The ball is feet above 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 y-coordinate of the verte is 1. The ball reaches the maimum height only once. The ball reaches other heights once on the way up and once on the way down, but the maimum point is the height where the ball changes directions, so only one -value corresponds to this y-value. Step 5 The equation is 1 16 88 3. In a b c 0 form, the equation is 16 88 11 0. For this equation, a 16, b 88, and c 11. Substituting these values into the quadratic formula gives 88 88 ( 16)( 11) 88 0 ( 16) 3 3 88.75 The ball reaches a maimum height.75 seconds after it is hit. The fact that there is only one solution becomes 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 b c 0 form, the equation is 16 88 197 0. For this equation, a 16, b 88, and c 197. Substituting these values into the quadratic formula gives 88 88 ( 16)( 197) 88 86 ( 16) 3 The value under the square root sign is negative. Because the square root of a negative number is not a real number, the equation has no real-number solution. Your work in the investigation shows that when the value under the square root sign, b ac, is 0, the equation a b c 0 has only one solution, and when the value under the square root sign, b ac, is negative, the equation a b c 0 has no real-number solutions. This means that if you are given a quadratic equation in the general form, you can use the value of b ac to determine whether the graph will have zero, one, or two -intercepts. Eample B in your book shows the importance of writing an equation in general form before you attempt to apply the quadratic formula. Read the eample carefully. 100 CHAPTER 7 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 7.5 Comple Numbers In this lesson you will learn that some polynomial equations have solutions that are comple numbers learn how to add, subtract, multiply, and divide comple numbers 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 The numbers and 1 9 are not real numbers because they involve the square root of a negative number. Numbers that include the real numbers as well as the square roots of negative numbers are called comple numbers. Defining the set of comple numbers makes it possible to solve equations such as.5 0 and 0, which have no solutions in the set of real numbers. The square roots of negative numbers are epressed using an imaginary unit called i, defined by i 1 or i 1. You can rewrite 9 as 9 1, or 3i. Therefore, the two solutions to the quadratic equation above can be written 1 3i as 1 3i and, or _ 1 _ 3 i and _ 1 _ 3 i. These two solutions are a conjugate pair, meaning that one is in the form a + bi and the other is in the form a bi. The two numbers in a comple pair are comple conjugates. Roots of polynomial equations can be real numbers or nonreal comple numbers, or there may be some of each. However, as long as the polynomial has real-number coefficients, any nonreal roots will come in conjugate pairs, such as 3i and 3i or 6 5i and 6 5i. Your book defines a comple number as a number in the form a bi, where a and b are real numbers and i 1. The number a is called the real part, and the number bi is called the imaginary part. The set of comple numbers includes all real numbers and all imaginary numbers. Look at the diagram on page 10 of your book, which shows the relationship between these numbers, and some other sets of numbers you may be familiar with, as well as eamples of numbers in each set. Then read the eample in your book, which shows how to solve the equation 3 0. (continued) Discovering Advanced Algebra Condensed Lessons CHAPTER 7 101

Lesson 7.5 Comple Numbers (continued) Investigation: Comple Arithmetic In this investigation you discover the rules for computing with comple numbers. Work through the investigation in your book before reading the answers below. Part 1: Addition and Subtraction Adding and subtracting comple numbers is similar to combining like terms. Use your calculator to add or subtract the numbers in Part 1a d in your book. Then, make a conjecture about how to add comple numbers without a calculator. Below are the solutions and a possible conjecture. a. 5 i b. 5 3i c. 1 9i d. 3 i Possible conjecture: To add two comple numbers, add the real parts and add the imaginary parts. In symbols, (a bi ) (c di ) (a c) (b d )i. Part : Multiplication Multiplying the comple numbers a bi and c di is very similar to multiplying the binomials a b and c d. You just need to keep in mind that i 1. Multiply the comple numbers in Part a d, and epress the answers in the form a bi. The answers are below. a. ( i )(3 5i ) 3 5i i 3 i 5i Epand as you would for a product of binomials. 6 10i 1i 0i 6 i 0i 6 i 0( 1) Multiply within each term. Combine 10i and 1i. i 1 6 i Combine 6 and 0. b. 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 number and its conjugate. The answers are below. a. b. 1 c. 0 d. 3 Possible generalizations: The sum of a number and its conjugate is a real number: (a bi ) (a bi ) a. The product of a real number and its conjugate is a real number: (a bi )(a bi ) a b. (continued) 10 CHAPTER 7 Discovering Advanced Algebra Condensed Lessons

Lesson 7.5 Comple Numbers (continued) Part : Division To divide two comple numbers, write the division problem as a fraction, conjugate of denominator multiply by (to change the denominator to a real number), conjugate of denominator and then write the result in the form a bi. Divide the numbers in Part a d. Here are the answers. a. 7 i 1 i 7 i 1 i 1 i 1 i 5 9i b. 0.5 i c. 0. 0.0i d. 0.6 0.8i.5.5i Multiply by conjugate of denominator conjugate of denominator. Multiply. The denominator becomes a real number. Divide. Comple numbers can be graphed on a comple plane, where the horizontal ais is the real ais and the vertical ais is the imaginary ais. The number a bi is represented by the point with coordinates (a, b). The numbers 3 i and i are graphed below. 5 i 5 Imaginary ais 3 i Real ais 5 5 Discovering Advanced Algebra Condensed Lessons CHAPTER 7 103

CONDENSED LESSON 7.6 Factoring Polynomials In this lesson you will learn about cubic 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 cubic function. At right is a graph of the cubic function y 3 16 9 36. The -intercepts of the function are, 1.5, and 1.5, so its factored equation must be in the form y a ( )( 1.5)( 1.5). To find the value of a, you can substitute the coordinates of another point on the curve. The y-intercept is (0, 36). Substituting 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 Read the tet before Eample A in your book and then work through Eample A. Investigation: The Bo Factory You can make a bo from a 16-by-0-unit sheet of paper by cutting squares of side length from the corners and folding the sides up. Follow the Procedure Note in your book to construct boes for several different integer values of. Record the dimensions and volume of each bo. (If you don t want to construct the boes, try to picture them in your mind.) Complete the investigation, and then compare your results to those below. Step 1 Here are the results for integer -values from 1 to 6. 0 16 Length Width Height Volume 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 boes are 0, 16, and. Therefore, the volume function is y (0 )(16 )(). (continued) Discovering Advanced Algebra Condensed Lessons CHAPTER 7 105

Lesson 7.6 Factoring Polynomials (continued) Step 3 The data points lie on the graph of the function. Step If you were to epand (0 )(16 )(), the result would be a polynomial, and the highest power of would be 3. Also, the graph looks like a cubic function. 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 but 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 bo cannot be formed. For 8, 8-unit-wide strips would be cut off the sides of the sheet. Folding up the sides would mean folding the remaining strip in half, which would not form a bo. 8 8 8 Cut off Cut off 8 16 8 Cut off Cut off 8 8 8 0 A value of 10 is impossible because 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 about.9 maimizes the volume. Work through Eample B in your book, which asks you to determine the factored form of a polynomial function by 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 numbers. With quadratic functions, if you cannot find the zeros by 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 book, and then read the eample on the net page. (continued) 106 CHAPTER 7 Discovering Advanced Algebra Condensed Lessons

Lesson 7.6 Factoring Polynomials (continued) EXAMPLE Write the equation of the quadratic function below 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-number zero is 3. If the other zero were a nonreal number, then its conjugate would also be a zero. This would mean there are three zeros, which is not possible. So 3 must be a double zero. This means that the function is in the form y a ( 3)( 3), or y a ( 3). To find the value of a, substitute (, ): a ( 1), so a. The factored form of the function is y ( 3). Discovering Advanced Algebra Condensed Lessons CHAPTER 7 107

CONDENSED LESSON 7.7 Higher-Degree Polynomials In this lesson you will describe the etreme values and end behavior of polynomial functions solve a problem 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. These are 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 because it is lower than the other points near it. The point (3, 0) is called a local maimum because it is higher than the other points near it. You can also describe the end behavior of the graph that is, what happens to the graph as takes on etreme values in the positive and negative directions. On this graph, look at values of greater than. As increases, y decreases. Now look at negative values of. As decreases, y increases. The introduction to Lesson 7.7 in your book gives another eample of a 3rd-degree polynomial and its graph. The graph of a polynomial function with real coefficients has a y-intercept, possibly one or more -intercepts, and other features such as local maimums or minimums and end behavior. The maimums and minimums are called etreme values. y (1, ) 6 6 Investigation: The Largest Triangle Start with a 1.5-by-8 cm sheet of paper. Orient the paper so that the long side is horizontal. Fold the upper left corner so that it touches some point on the bottom edge. Find the area, in cm, of the triangle A A formed in the lower left corner. What distance,, along the bottom edge of the paper produces the triangle with greatest area? To answer this question, first find areas for different values of. Then find a formula for the area of the triangle in terms of. Try to do this on your own before reading on. Your answers will vary depending on the dimensions of your paper, but the logic used should still apply. h Here is one way to find the formula: Let h be the height of the triangle. Then the hypotenuse has length 1.5 h. (Why?) 1.5 h (continued) Discovering Advanced Algebra Condensed Lessons CHAPTER 7 109

Lesson 7.7 Higher-Degree Polynomials (continued) 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 and some sample data points. If you trace the graph, you ll find that the maimum point is about (1.,.5). Therefore, the value of that gives the greatest area is about 1. cm. The maimum area is about.5 cm. Eample A in your book shows you how to find the equation for a polynomial with given -intercepts and nonzero 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 5 of your book show some possible shapes for the graph of a 3rd-degree polynomial function. Graph A is the graph of the parent function y 3. Like other parent functions you have studied, the graph can be translated, dilated, and reflected. Eample B in your book 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 nonreal comple zeros come in conjugate pairs. Read that eample carefully, and then read the eample below. EXAMPLE Find a th-degree polynomial function with real coefficients and zeros, 3, and 1 i. Solution Nonreal comple zeros of polynomials with real coefficients occur in conjugate pairs, so 1 i must also be a zero. So one possible 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 by making a graph. (You will see only the real zeros.) Note that an nth-degree polynomial function always has n zeros (counting multiple roots). However, because some of the zeros may be nonreal numbers, the function may not have n -intercepts. 110 CHAPTER 7 Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON 7.8 More About Finding Solutions In this lesson you will use long division to find the roots of a higher-degree polynomial use the Rational Root Theorem to find all the possible rational roots of a polynomial use synthetic division to divide a polynomial by a linear factor You can find the zeros of a quadratic function by factoring or by using the quadratic formula. You can sometimes use a graph to find the zeros of higherdegree polynomials, but 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, both real and nonreal, of many higher-degree polynomials. Eample A in your book 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 substitute 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, be sure to write both 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 by 9, rewrite 13 36 as 0 3 13 0 36 and rewrite 9 as 0 9. The division problem below 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 by looking at the graph. If the -intercepts of a graph are not integers, identifying the zeros can be difficult. The Rational Root Theorem tells you which rational numbers might be zeros. It states that if the polynomial equation P () 0 has rational roots, then they are of the form p _ q, where p is a factor of the lowest-degree 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. (continued) Discovering Advanced Algebra Condensed Lessons CHAPTER 7 111

Lesson 7.8 More About Finding Solutions (continued) 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 be a factor of divided by 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 between 3 and. None of these possibilities are in that interval. Another root is a little less than 1_. This could be 3_ 7. Try substituting 3_ 7 into the polynomial. 7 3 7 3 3 3 7 56 3 7 7 9 7 0 9 So 3_ 7 is a root, which means 3_ 7 is a factor. Use long 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, or. So the roots are _ 3 7,, and. Synthetic division is a shortcut method for dividing a polynomial by a linear factor. Read the remainder of the lesson in your book to see how to use synthetic division. Below is an eample using synthetic division to find 7 3 3 56. 3 7 Note that in the eample above you found this same quotient using long division. Known zero 3_ 7 Coefficients of 7 3 3 56 7 3 56 Add Add Add 3 0 1 Bring 3 5 7 down 3_ 7 7 3_ 7 0 6 3_ 7 56 7 0 56 0 The result shows that 7 3 3 56 7 56, so 7 3 3 56 3 7 _ 3 7 7 56. 11 CHAPTER 7 Discovering Advanced Algebra Condensed Lessons