Polynomial Degree and Finite Differences

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Polynomial Degree and Finite Differences

7.1 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 expression that can be written in the form an x n an1x n1 · · · a 1x 1 a 0 where x is a variable, the exponents are nonnegative integers, the coefficients are real numbers, and an 0. A function in the form f (x) an x n an1x n1 · · · a 1x 1 a 0 is a polynomial function. The degree of a polynomial or polynomial function is the power of the term with the greatest exponent. 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

2nd degree

3x 7

x 2

2x 1.8

3rd degree 9x 3

4x 2

x 11

4th degree 5x 4

A polynomial with one term, such as 5x 4, is called a monomial. A polynomial with two terms, such as 3x 7, is called a binomial. A polynomial with three terms, such as x 2 2x 1.8, is called a trinomial. Polynomials with more than three terms, such as 9x 3 4x 2 x 11, are usually just called polynomials. For linear functions, when the x-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 2nd-degree polynomials, the differences of the differences, called the second differences and abbreviated D2, are constant. For 3rd-degree polynomials, the differences of the second differences, called the third differences and abbreviated D3, are constant. This is illustrated in the tables on page 379 of your book. If you have a set of data with equally spaced x-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 example in your book. Read that example carefully. Notice that the finite differences method determines only the degree of the polynomial. To find the exact equation for the polynomial function, you need to find the coefficients by solving a system of equations or using some other method. In the example, the D2 values are equal. When you use experimental 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 2.) (continued) Discovering Advanced Algebra Condensed Lessons

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Lesson 7.1 • Polynomial Degree and Finite Differences (continued) Complete Steps 2–6 in your book. The results given are based on the sample data. The first and second differences, D1 and D2, are shown in the table at right. Step 2

For these data, we can stop with the second differences because they are nearly constant.

Step 3

The three plots are shown below.

(time, height )

Time (s) x

Height (m) y

0.00

2.000

0.05

1.988

0.10

1.951

0.15

1.890

0.20

1.804

0.25

1.694

0.30

1.559

0.35

1.400

0.40

1.216

0.45

1.008

(time 2, d 1)

D1 0.012 0.037 0.061 0.086 0.110 0.135 0.159 0.184 0.208

D2 0.025 0.024 0.025 0.024 0.025 0.024 0.025 0.024

(time 3, d 2 )

The graph of (time, height ) appears parabolic, suggesting that the correct model may be a 2nd-degree polynomial function. The graph of (time 2, d 1 ) shows that the first differences are not constant because they decrease in a linear fashion. The graph of (time 3, d 2 ) shows that the second differences are nearly constant, so the correct model should be a 2nd-degree polynomial function. Step 4

Step 5

A 2nd-degree polynomial in the form y ax 2 bx c fits the data.

To write the system, choose three data points. For each point, write an equation by substituting the time and height values for x and y in the equation y ax 2 bx c. The following system is based on the values (0, 2), (0.2, 1.804), and (0.4, 1.216). Step 6

c 2.000 0.04a 0.2b c 1.804 0.16a 0.4b c 1.216

One way to solve this system is by writing the matrix equation

0 0.04 0.16

0 0.2 0.4

1 1 1

2.000 a b 1.804 1.216 c

and solving using an inverse matrix. The solution is a 4.9, b 0, and c 2, so an equation that fits the data is y 4.9x 2 2.

Read the remainder of the lesson in your book. 94

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7.2

Equivalent Quadratic Forms

In this lesson you will ● ● ●

learn about the vertex 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 vertex, general, and factored form

A 2nd-degree polynomial function is called a quadratic function. In Lesson 7.1, you learned that the general form of a quadratic function is y ax 2 bx c. In this lesson you will explore other forms of a quadratic function. You know that every quadratic function is a transformation of y x 2. When yk xh 2 xh 2 ____ ____ a quadratic function is written in the form ____ b a or y b a k, you can tell that the vertex of the parabola is (h, k) and that the horizontal and vertical scale factors are a and b. Conversely, if you know the vertex of a parabola and you know (or can find) the scale factors, you can write its equation in one of these forms. xh The quadratic function y b ____ a k can be rewritten in the form b b __ 2 y a 2 (x h) k. The coefficient __ combines the two scale factors into a2 one vertical scale factor. In the vertex form of a quadratic equation, y a(x h)2 k, this single scale factor is simply denoted a. From this form, you can identify the vertex, (h, k), and the vertical scale factor, a. If you know the vertex of a parabola and the vertical scale factor, you can write an equation in vertex form. Work through Example A carefully. 2

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 example, if 3x (x 7) 0, then 3x 0 or x 7 0. Therefore, x 0 or x 7. The solutions to an equation in the form f (x) 0 are called the roots of the equation, so 0 and 7 are the roots of 3x (x 7) 0. The x-intercepts of a function are also called the zeros of the function (because the corresponding y-values are 0). The function y 1.4(x 5.6)(x 3.1), given in Example 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 1.4(x 5.6)(x 3.1) 0. Example 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 x r1x r2. From this form, you can identify the x-intercepts (or zeros), r1 and r2, and the vertical scale factor, a. Conversely, if you know the x-intercepts of a parabola and know (or can find) the vertical scale factor, then you can write the equation in factored form. Read Example C carefully.

Investigation: Rolling Along Read the Procedure Note and Steps 1–3 in your book. Make sure you can visualize how the experiment works. Use these sample data to complete Steps 4–8, 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

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Lesson 7.2 • Equivalent Quadratic Forms (continued) Time (s) x

Distance from line (m), y

Time (s) x

0.2

0.357

2.2

3.570

4.2

3.309

0.4

0.355

2.4

3.841

4.4

2.938

0.6

0.357

2.6

4.048

4.6

2.510

0.8

0.184

2.8

4.188

4.8

2.028

1.0

0.546

3.0

4.256

5.0

1.493

1.2

1.220

3.2

4.257

5.2

0.897

1.4

1.821

3.4

4.193

5.4

0.261

1.6

2.357

3.6

4.062

5.6

0.399

1.8

2.825

3.8

3.871

5.8

0.426

2.0

3.231

4.0

3.619

6.0

0.419


Time (s) x


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, D2, are almost constant, at around 0.06, which implies that a quadratic model is appropriate.

Step 4

The coordinates of the vertex are (3.2, 4.257). Consider (5.2, 0.897) to be the image of (1, 1). The horizontal and vertical distances of (1, 1) from the vertex of y x 2 are both 1. The horizontal distance of (5.2, 0.897) from the vertex, (3.2, 4.257), is 2, and the vertical distance is 3.36. So, the horizontal and vertical scale factors are 2 and 3.36, respectively. This can be represented as the single vertical scale factor 3.36 ____ 0.84. Therefore, the vertex form of a model for the data is 22 y 0.84(x 3.2)2 4.257. Step 5

Substituting the points (1, 0.546), (3, 4.256), and (5, 1.493) into the general form, y ax 2 bx c, gives the system

Step 6

a b c 0.546 9a 3b c 4.256 25a 5b c 1.493

The solution to this system is a 0.81, b 5.09, and c 3.74, so the general form of the equation is y 0.81x 2 5.09x 3.74. The x-intercepts are about (0.9, 0) and (5.5, 0). The scale factor, found in Step 5, is 0.84. So the factored form of the equation is y 0.84(x 0.9)(x 5.5).

Step 7

Step 8 In general, you use the vertex form when you know either

the vertex and the scale factor or the vertex 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 x-intercepts and at least one other point you can use to find the scale factor.

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7.3

Completing the Square

In this lesson you will ● ●

use the method of completing the square to find the vertex of a parabola whose equation is given in general form solve problems involving projectile motion

Many real-world problems involve finding the minimum or maximum value of a function. For quadratic functions, the maximum or minimum value occurs at the vertex. If you are given a quadratic equation in vertex form, you can easily find the coordinates of the parabola’s vertex. It is also fairly straightforward to find the vertex 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 vertex 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 1 gx 2 v x s y __ 0 0 2 where x is the time in seconds, y is the height (in m or ft), g is the acceleration due to gravity (either 9.8 m/s2 or 32 ft/s2), v0 is the initial upward velocity of the object (in either m/s or ft/s), and s0 is the initial height of the object (in m or ft). Read Example A in your book. It illustrates how to write a projectile motion equation when you know only the x-intercepts and how to use the x-intercepts to find the coordinates of the vertex.

Investigation: Complete the Square Complete the investigation in your book. When you are finished, compare your answers to those below. Step 1 a. x 2 5x 5x 25 x 2 10x 25 b. (x 8) is the binomial expression being squared, and

x 2 16x 64 is the perfect square trinomial.

c. (x 12)2

x

5

x

x2

5x

5

5x

25

x

12

x x2 12x d. a 2 2ab b 2 The first term of the trinomial, a 2, is the square of the first 12 12x 144 term of the binomial. The second term of the trinomial, 2ab, is twice the product of the binomial terms. The third term of the trinomial, b 2, is the square of the last term of the binomial. Step 2

a. You must add 9. b. x 2 6x x 2 6x 9 9 (x 3)2 9 c. Enter x 2 6x as f1(x) and (x 3)2 9 as f 2(x), and verify that the table values or the graphs are the same for both expressions. Discovering Advanced Algebra Condensed Lessons

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Lesson 7.3 • Completing the Square (continued) Step 3

a. 9 b. x 2 6x 4 x 2 6x 9 9 4 (x 3)2 13 c. Enter x 2 6x 4 as f1(x) and (x 3)2 13 as f2(x), and verify that the table values or the graphs are the same for both expressions. Step 4

a. Focus on x 2 14x. To complete a perfect square, you need to add 49. You need to subtract 49 to compensate. So x 2 14x 3 x 2 14x 49 49 3 (x 7)2 46 2 b2 b. To make x 2 bx a perfect square, you must add _2b , or __ 4 . You need to b2 __ subtract 4 to compensate. So

b 2 __ b 2 __ b 2 10 b 2 10 x __ x 2 bx 10 x 2 bx __ 4 4 4 2 Step 5

a. 2x 2 6x 1 2(x 2 3x) 1 9 2 __ 2 x 2 3x __ 94 1 4 3 2 7 2 x __ 2 2 10 x 7 b. ax 2 10x 7 a x 2 ___ a 10 x ___ 25 a ___ a x 2 ___ 7 25 a a2 a2 5 2 7 ___ a x __ 25 a a Step 6

Factor 2x 2

6x.

Complete the square. You add 2 ⭈ _94 , so you must subtract 2 ⭈ _94 . Write in the form a (x h)2 k. Factor ax 2 10x. Complete the square. You add a 25 must subtract a ⭈ __ a2.

25 ⭈ __ a , so you 2

Write in the form a (x h)2 k.

b b __ 2 The x-coordinate is __ 2a . Substitute 2a for x in y ax bx c to

b2 . find the y-coordinate, which is c ___ 4a 2

Read Example B carefully. Based on your work in the investigation and Example B, you now know two ways to find the vertex, (h, k), of a quadratic function given in general form, y ax 2 bx c. 1. You can use the process of completing the square to rewrite the equation in vertex form, y a(x h)2 k. The vertex is (h, k). b b2 __ 2. You can use the formulas h __ 2a and k c 4a to calculate the coordinates of the vertex 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. Example 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.

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7.4

The Quadratic Formula


learn how the quadratic formula is derived use the quadratic formula to solve projectile motion problems

You can use a graph to approximate the x-intercepts of a quadratic function. If you can write the equation of the function in factored form, you can find the exact values of its x-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 exact x-intercepts of any quadratic function. Read Example A in your book carefully, and then read the example below. Find the x-intercepts of y 2x 2 7x 1.

EXAMPLE 䊳

The x-intercepts are the solutions of 2x 2 7x 1 0. See if you can supply the reason for each step in the solution below.

Solution

2x 2 7x 1 0 2x 2 7x 1 7x ? 2x 2 __ 2

1

?

49 49 1 ___ 7 x ___ 2x 2 __ 2 8 16 7 2 ___ 41 2x __ 4 8 2

x __47

41 ___ 16

___

41 7 ______ x __ 4

4

___

___

7 41 41 7 _____ ________ x __ 4

___

4

4

___

7 41 7 41 3.351 and x ______ 0.149. The x-intercepts are x ______ 4 4

The series of equations after Example A in your book shows how you can derive the quadratic formula by following the same steps used above. The quadratic formula, ________

b b 4ac x _______________ 2

2a

gives the general solution to a quadratic equation in the form ax 2 bx 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 41 7 41 ______ 2x 2 7x 1 0 are x and x ______ . 4 4 (continued) Discovering Advanced Algebra Condensed Lessons

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Lesson 7.4 • The Quadratic Formula (continued) Investigation: How High Can You Go? Complete the investigation in your book, and then compare your answers to those below. The equation is y 16x 2 88x 3, where y is the height in feet and x is the time in seconds. (If you answered this question incorrectly, review the discussion of projectile motion in Lesson 7.3.)

Step 1

Step 2

The equation is 24 16x 2 88x 3.

In ax 2 bx c 0 form, the equation is 16x 2 88x 21 0. For this equation, a 16, b 88, and c 21. Substituting these values into the quadratic formula gives

Step 3

_________________

_____

88 88 2 4(16)(21) _____________ 88 80 x _________________________ 88 6400 _________ 32 32 2(16) 88 80 88 80 So x _______ 0.25 or x _______ 5.25. The ball is 24 feet above the 32 32 ground 0.25 second after it is hit (on the way up) and 5.25 seconds after it is hit (on the way down).

Step 4 The y-coordinate of the vertex is 124. The ball reaches the maximum height only once. The ball reaches other heights once on the way up and once on the way down, but the maximum point is the height where the ball changes directions, so only one x-value corresponds to this y-value.

The equation is 124 16x 2 88x 3. In ax 2 bx c 0 form, the equation is 16x 2 88x 121 0. For this equation, a 16, b 88, and c 121. Substituting these values into the quadratic formula gives Step 5

__________________

__

88 88 4(16)(121) __________ 88 2.75 x __________________________ 88 0 ____ 2

2(16)

32

32

The ball reaches a maximum height 2.75 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. The equation is 200 16x 2 88x 3. In ax 2 bx c 0 form, the equation is 16x 2 88x 197 0. For this equation, a 16, b 88, and c 197. Substituting these values into the quadratic formula gives Step 6

__________________

______

88 88 2 4(16)(197) ______________ x __________________________ 88 4864 32 2(16) 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 2 4ac, is 0, the equation ax 2 bx c 0 has only one solution, and when the value under the square root sign, b 2 4ac, is negative, the equation ax 2 bx 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 2 4ac to determine whether the graph will have zero, one, or two x-intercepts. Example B in your book shows the importance of writing an equation in general form before you attempt to apply the quadratic formula. Read the example carefully. 100

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7.5

Complex Numbers


learn that some polynomial equations have solutions that are complex numbers learn how to add, subtract, multiply, and divide complex numbers

The graph of y x 2 x 2.5 has no x-intercepts. y 4

–4

–2

2

4

x

–2 –4

If you use the quadratic formula to attempt to find the x-intercepts, you get ____________

___

1 12 4(1)(2.5) 1 9 ___________ x ____________________ 2 2(1) ___

___

1 9 1 9 The numbers ________ and ________ are not real numbers because they involve 2 2 the square root of a negative number. Numbers that include the real numbers as well as the square roots of negative numbers are called complex numbers. Defining the set of complex numbers makes it possible to solve equations such as x 2 x 2.5 0 and x 2 4 0, which have no solutions in the set of real numbers.

The square roots of negative numbers___ are expressed using an ___ imaginary__ unit___ 2 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 1 3i as ______ and ______ , or _12 _32 i and _12 _32 i. These two solutions are a 2 2 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 complex pair are complex conjugates. Roots of polynomial equations can be real numbers or nonreal complex 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 complex 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 complex numbers includes all real numbers and all imaginary numbers. Look at the diagram on page 410 of your book, which shows the relationship between these numbers, and some other sets of numbers you may be familiar with, as well as examples of numbers in each set. Then read the example in your book, which shows how to solve the equation x 2 3 0. (continued)

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Lesson 7.5 • Complex Numbers (continued) Investigation: Complex Arithmetic In this investigation you discover the rules for computing with complex numbers. Work through the investigation in your book before reading the answers below. Part 1: Addition and Subtraction Adding and subtracting complex 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 complex 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 complex numbers, add the real parts and add the imaginary parts. In symbols, (a bi ) (c di ) (a c) (b d )i. Part 2: Multiplication Multiplying the complex numbers a bi and c di is very similar to multiplying the binomials a bx and c dx. You just need to keep in mind that i 2 1. Multiply the complex numbers in Part 2a–d, and express the answers in the form a bi. The answers are below. a. (2 4i )(3 5i ) 2 ⭈ 3 2 ⭈ 5i 4i ⭈ 3 4i ⭈ 5i Expand as you would for a product of binomials. 6 10i 12i 20i 2

Multiply within each term.

6 2i 20i 2

Combine 10i and

6 2i 20(1)

i2

26 2i

Combine 6 and 20.

12i.

1

b. 16 3i c. 12 16i d. 8 16i Part 3: The Complex Conjugates Complete Part 3a–d, which involves finding either the sum or product of a complex number and its conjugate. The answers are below. a. 4 b. 14 c. 20 d. 32 Possible generalizations: The sum of a number and its conjugate is a real number: (a bi ) (a bi ) 2a. The product of a real number and its conjugate is a real number: (a bi )(a bi ) a 2 b 2. (continued)

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Lesson 7.5 • Complex Numbers (continued) Part 4: Division To divide two complex numbers, write the division problem as a fraction, conjugate of denominator (to change the denominator to a real number), multiply by ________________ conjugate of denominator and then write the result in the form a bi. Divide the numbers in Part 4a–d. Here are the answers. conjugate of denominator 7 2i ______ 1i 7 2i _____ a. ______ Multiply by _____________ . conjugate of denominator 1i 1i ⭈1i 5 9i ______ 2

Multiply. The denominator becomes a real number.

2.5 4.5i

Divide.

b. 0.5 2i c. 0.22 0.04i d. 0.6 0.8i

Complex numbers can be graphed on a complex plane, where the horizontal axis is the real axis and the vertical axis is the imaginary axis. The number a bi is represented by the point with coordinates (a, b). The numbers 3 4i and 4 i are graphed below. Imaginary axis 5 3 4i 4 i 5

–5

Real axis

–5


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7.6

Factoring Polynomials


learn about cubic functions use the x-intercepts of a polynomial function to help you write the function in factored form

The polynomial equations y x 2 6x 9 and y (x 3)(x 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 x-intercepts, or zeros, of the function. In this lesson you will learn some techniques for writing higher-degree polynomials in factored form. y

A 3rd-degree polynomial function is called a cubic function. At right is a graph of the cubic function y 4x 3 16x 2 9x 36.

50 (0, 36)

The x-intercepts of the function are 4, 1.5, and 1.5, so its factored equation must be in the form y a (x 4)(x 1.5)(x 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 (4)(1.5)(1.5). So, a 4, and the factored form of the equation is

(1.5, 0)

(–4, 0) –5

x

5 (–1.5, 0)

y 4(x 4)(x 1.5)(x 1.5)

–50

Read the text before Example A in your book and then work through Example A.

Investigation: The Box Factory You can make a box from a 16-by-20-unit sheet of paper by cutting squares of side length x from the corners and folding the sides up. Follow the Procedure Note in your book to construct boxes for several different integer values of x. Record the dimensions and volume of each box. (If you don’t want to construct the boxes, try to picture them in your mind.) Complete the investigation, and then compare your results to those below. Step 1

x

x

x

x

16

x

x x

x 20

Here are the results for integer x-values from 1 to 6.

x

Length

Width

Height

Volume y

1

18

14

1

252

2

16

12

2

384

3

14

10

3

420

4

12

8

4

384

5

10

6

5

300

6

8

4

6

192

Step 2 The dimensions of the boxes are 20 2x, 16 2x, and x. Therefore, the volume function is y (20 2x)(16 2x)(x). Discovering Advanced Algebra Condensed Lessons

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Lesson 7.6 • Factoring Polynomials (continued) Step 3

The data points lie on the graph of the function.

Step 4 If you were to expand (20 2x)(16 2x)(x), the result would be a polynomial, and the highest power of x would be 3. Also, the graph looks like a cubic function. Therefore, the function is a 3rd-degree polynomial function.

The x-intercepts of the graph are 0, 8, and 10, so the function is y x (x 8)(x 10).

Step 5

The graphs have the same x-intercepts and general shape but different vertical scale factors. A vertical scale factor of 4 makes them equivalent: y 4x (x 8)(x 10). Step 6

If x 0, there are no sides to fold up, so a box cannot be formed. For x 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 box.

Step 7

8

8

8

Cut off

Cut off

8 16

8

Cut off

Cut off

8

8

8 20

A value of x 10 is impossible because it is more than half the length of the shorter side of the sheet. Only a domain of 0 x 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 x-value of about 2.94 maximizes the volume.

Work through Example B in your book, which asks you to determine the factored form of a polynomial function by using the x-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, r1 and r2, you can write the polynomial in the form y a x r1x r2. Read the remainder of the lesson in your book, and then read the example on the next page. (continued)

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Lesson 7.6 • Factoring Polynomials (continued) EXAMPLE

Write the equation of the quadratic function below in factored form. y 2 2

–2 –2

4

6

8

x

(2, –2)

–4 –6 –8

䊳

Solution

The factored equation is in the form y a x r1x r2, where r1 and r2 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 (x 3)(x 3), or y a (x 3)2. To find the value of a, substitute (2, 2): 2 a (1)2, so a 2. The factored form of the function is y 2(x 3)2.


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7.7

Higher-Degree Polynomials


describe the extreme values and end behavior of polynomial functions solve a problem that involves maximizing a polynomial function write equations for polynomial functions with given intercepts y

Polynomials with degree 3 or higher are often referred to as higher-degree polynomials. At right is the graph of the polynomial y x (x 3)2, or y x 3 6x 2 9x. The zero-product property tells you that the zeros are x 0 and x 3. These are the values of x for which y 0. The x-intercepts of the graph confirm this.

2

The graph has other key features in addition to the x-intercepts. For example, the point (1, 4) is called a local minimum because it is lower than the other points near it. The point (3, 0) is called a local maximum 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 x takes on extreme values in the positive and negative directions. On this graph, look at values of x greater than 4. As x increases, y decreases. Now look at negative values of x. As x decreases, y increases.

2

–2

4

6

x

–2 –4

(1, –4)

–6

The introduction to Lesson 7.7 in your book gives another example 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 x-intercepts, and other features such as local maximums or minimums and end behavior. The maximums and minimums are called extreme values.

Investigation: The Largest Triangle Start with a 21.5-by-28 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 cm2, of the triangle formed in the lower left corner.

A x

A x

What distance, x, along the bottom edge of the paper produces the triangle with greatest area? To answer this question, first find areas for different values of x. Then find a formula for the area of the triangle in terms of x. 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. Here is one way to find the formula: Let h be the height of the triangle. Then the hypotenuse has length 21.5 h. (Why?)

h

21.5 h

x (continued)


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Lesson 7.7 • Higher-Degree Polynomials (continued) Use the Pythagorean Theorem to help you write h in terms of x . 462.25 x 2 x 2 h 2 (21.5 h)2, so h __________ 43 Now, you can write a formula for the area, y. 462.25 x 1 x __________ y __ 2 43 At right is a graph of the area function and some sample data points. If you trace the graph, you’ll find that the maximum point is about (12.4, 44.5). Therefore, the value of x that gives the greatest area is about 12.4 cm. The maximum area is about 44.5 cm2. 2

Example A in your book shows you how to find the equation for a polynomial with given x-intercepts and nonzero y-intercept. Read this example carefully. To test your understanding, find a polynomial function with x-intercepts 6, 2, and 1, and y-intercept 60. (One answer is y 5(x 6)(x 2)(x 1).) Graphs A–D on page 425 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 x 3. Like other parent functions you have studied, the graph can be translated, dilated, and reflected. Example B in your book shows you how to find a polynomial function with given zeros when some of the zeros are complex. The key to finding the solution is to recall that nonreal complex zeros come in conjugate pairs. Read that example carefully, and then read the example below.

EXAMPLE

䊳

Solution

Find a 4th-degree polynomial function with real coefficients and zeros x 2, x 3, and x 1 i. Nonreal complex zeros of polynomials with real coefficients occur in conjugate pairs, so x 1 i must also be a zero. So one possible function, in factored form, is y (x 2)(x 3)[x (1 i)][x (1 i)] Multiply the factors to get a polynomial in general form. y (x 2)(x 3)[x (1 i)][x (1 i)] x 2 x 6x 2 (1 i )x (1 i )x (1 i )(1 i ) x 2 x 6x 2 x ix x i x 2 x 2 x 6x 2 2x 2 x 4 2x 3 2x 2 x 3 2x 2 2x 6x 2 12x 12 x 4 3x 3 2x 2 10x 12 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 x-intercepts. 110

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7.8

More About Finding Solutions


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 approximation of real zeros and won’t work at all to find nonreal zeros. In this lesson you will learn a method for finding the exact zeros, both real and nonreal, of many higher-degree polynomials. Example 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 example, 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 (x r) is a factor of a polynomial function P (x) 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 example, to divide x 4 13x 2 36 by x 2 9, rewrite x 4 13x 2 36 as x 4 0x 3 13x 2 0x 36 and rewrite x 2 9 as x 2 0x 9. The division problem below shows that x 4 13x 2 36 x 2 9x 2 4. x 4 ________________________ 2

x 2 0x 9)x 4 0x 3 13x 2 0x 36 x 4 0x 3 9x 2 4x 2 0x 36 4x 2 0x 36 0 In Example A, you found some of the zeros by looking at the graph. If the x-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 (x) 0 has rational roots, then they are p of the form _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. Example B shows how the theorem is used. Work through that example, and then read the example on the next page. (continued)


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Lesson 7.8 • More About Finding Solutions (continued) EXAMPLE 䊳

Solution

Find the roots of 7x 3 3x 2 56x 24 0. The graph of the function y = 7x 3 3x 2 56x 24 appears at right. None of the x-intercepts are integers. The Rational Root Theorem tells you that any rational root will be a factor of 24 divided by a factor of 7. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The factors of 7 are 1 and 7. You know there are no integer roots, so you need to consider only _17 , _27 , _37 , 12 24 __ _47 , _67 , _87 , __ 7 , and 7 . The graph indicates that one of the roots is between 3 and 2. None of these possibilities are in that interval. Another root is a little less than _12 . This could be _37 . Try substituting _37 into the polynomial. 3 3 3 2 3 27 ___ 27 __ __ ___ 7 __ 7 3 7 56 7 24 49 49 24 24 0 So _37 is a root, which means x _37 is a factor. Use long division to divide out this factor. 7x 56 ___________________ 2

x _3 )7x 3 3x 2 56x 24 7

7x 3 3x 2

56x 24 56x 24 0

So 7x 3 3x 2 56x 24 0 is equivalent to x _37 7x 2 56 0. To find the __ __ 2 other roots, solve 7x __56 0. The solutions are x 8 , or 22 . So the __ 3 _ roots are x 7 , x 22 , and x 22 . 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 7x 3 3x 2 56x 24 division. Below is an example using synthetic division to find ______________ . 3 __ x7

Note that in the example above you found this same quotient using long division. Known zero

Coefficients of 7x 3 3x 2 56x 24

_3 7

7 1 Bring down

3

3 Add

3 2 _7 ⭈ 7

7

3 0

56

5 Add 3 4 _7 ⭈ 0

24 7 Add

06 _3 ⭈ ⫺56 24

56

7

0

7x 3 3x 2 56x 24 The result shows that ______________ 7x 2 56, so 7x 3 3x 2 56x 24 3 x __ 3 _ 2 7 x 7 7x 56.

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