# Polynomial Degree and Finite Differences

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1 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 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 , 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 Key Curriculum Press

2 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 Step 3 The three plots are shown below. (time, height ) (time, d 1) (time 3, d ) D D 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 a 0.b c a 0.b c 1.16 One way to solve this system is by writing the matri equation a b 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

4 Lesson 7. Equivalent Quadratic Forms (continued) Time (s) Distance from line (m), y Time (s) Distance from line (m), y Time (s) Distance from line (m), y 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 So, the horizontal and vertical scale factors are and 3.36, respectively. This can be represented as the single vertical scale factor 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 a 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 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

9 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

11 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 i d i.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

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13 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 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 Length Width Height Volume y 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

14 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 Cut off Cut off Cut off Cut off 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

15 Lesson 7.6 Factoring Polynomials (continued) EXAMPLE Write the equation of the quadratic function below in factored form. y 6 (, ) 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

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19 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 by 9, rewrite as and rewrite 9 as 0 9. The division problem below shows that ) 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

20 Lesson 7.8 More About Finding Solutions (continued) EXAMPLE Find the roots of Solution The graph of the function y = 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 So 3_ 7 is a root, which means 3_ 7 is a factor. Use long division to divide out this factor ) So is equivalent to 3_ To find the other roots, solve 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 Note that in the eample above you found this same quotient using long division. Known zero 3_ 7 Coefficients of Add Add Add Bring down 3_ 7 7 3_ _ The result shows that , so _ CHAPTER 7 Discovering Advanced Algebra Condensed Lessons

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