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8 Complex Numbers, Polar Equations, 8.1-1 8 Complex Numbers, Polar Equations, and Parametric Equations 8.1 Complex Numbers 8.2 Trigonometric (Polar) Form of Complex Numbers 8.3 The Product and Quotient Theorems 8.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers 8.5 Polar Equations and Graphs 8.1-2 8.1 Complex Numbers Basic Concepts of Complex Numbers ▪ Complex Solutions of Equations ▪ Operations on Complex Numbers 8.1-3 Basic Concepts of Complex Numbers i is called the imaginary unit. Numbers of the form a + bi are called complex numbers. a is the real part. b is the imaginary part. a + bi = c + di if and only if a = c and b = d. 8.1-4 8.1-5 Basic Concepts of Complex Numbers If a = 0 and b ≠ 0, the complex number is a pure imaginary number. Example: 3i If a ≠ 0 and b ≠ 0, the complex number is a nonreal complex number. Example: 7 + 2i A complex number written in the form a + bi or a + ib is written in standard form. 8.1-6 The Expression 8.1-7 Example 1 WRITING AS Write as the product of real number and i, using the definition of 8.1-8 Example 2 SOLVING QUADRATIC EQUATIONS FOR COMPLEX SOLUTIONS Solve each equation. Solution set: Solution set: 8.1-9 Example 3 SOLVING QUADRATIC EQUATIONS FOR COMPLEX SOLUTIONS Write the equation in standard form, then solve using the quadratic formula with a = 9, b = –6, and c = 5. 8.1-10 Caution When working with negative radicands, use the definition before using any of the other rules for radicals. In particular, the rule is valid only when c and d are not both negative. For example while so , , 8.1-11 Example 4 FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICANDS Multiply or divide as indicated. Simplify each answer. 8.1-12 Example 5 Write SIMPLIFYING A QUOTIENT INVOLVING A NEGATIVE RADICAND in standard form a + bi. Factor. 8.1-13 Addition and Subtraction of Complex Numbers For complex numbers a + bi and c + di, 8.1-14 Example 6 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference. (a) (3 – 4i) + (–2 + 6i) = [3 + (–2)] + (–4i + 6i) = 1 + 2i (b) (–9 + 7i) + (3 – 15i) = –6 – 8i (c) (–4 + 3i) – (6 – 7i) = (–4 – 6) + [3 – (–7)]i = –10 + 10i (d) (12 – 5i) – (8 – 3i) + (–4 + 2i) = (12 – 8 – 4) + (–5 +3 + 2)i = 0 + 0i = 0 8.1-15 Multiplication of Complex Numbers 8.1-16 Example 7 MULTIPLYING COMPLEX NUMBERS Find each product. (a) (2 – 3i)(3 + 4i) = 2(3) + (2)(4i) + (–3i)(3) + (–3i)(4i) = 6 + 8i – 9i – 12i2 = 6 –i – 12(–1) = 18 – i (b) (4 + 3i)2 = 42 + 2(4)(3i) + (3i)2 = 16 + 24i + 9i2 = 16 + 24i + 9(–1) = 7 + 24i 8.1-17 Example 7 (c) (2 + i)(–2 – i) MULTIPLYING COMPLEX NUMBERS = –4 – 2i – 2i – i2 = –4 – 4i – (–1) = –4 – 4i + 1 = –3 – 4i (d) (6 + 5i)(6 – 5i) = 62 – (5i)2 = 36 – 25i2 = 36 – 25(–1) = 36 + 25 = 61 or 61 + 0i 8.1-18 Example 7 MULTIPLYING COMPLEX NUMBERS (cont.) This screen shows how the TI–83/84 Plus displays the results found in parts (a), (b), and (d) in this example. 8.1-19 Example 8 SIMPLIFYING POWERS OF i Simplify each power of i. (a) (b) (c) Write the given power as a product involving or . (a) (b) (c) 8.1-20 Powers of i i1 = i i5 = i i9 = i i2 = 1 i6 = 1 i10 = 1 i 3 = i i 7 = i i11 = i i4 = 1 i8 = 1i12 = 1, and so on. 8.1-21 Property of Complex Conjugates For real numbers a and b, 8.1-22 Example 9(a) DIVIDING COMPLEX NUMBERS Write the quotient in standard form a + bi. Multiply the numerator and denominator by the complex conjugate of the denominator. Multiply. i2 = –1 Factor. Lowest terms; standard form 8.1-23 Example 9(b) DIVIDING COMPLEX NUMBERS Write the quotient in standard form a + bi. Multiply the numerator and denominator by the complex conjugate of the denominator. Multiply. i2 = –1 Standard form This screen shows how the TI–83/84 Plus displays the results in this example. 8.1-24 8.2 Trigonometric (Polar) Form of Complex Numbers The Complex Plane and Vector Representation ▪ Trigonometric (Polar) Form ▪ Converting Between Rectangular and Trigonometric (Polar) Forms ▪ An Application of Complex Numbers to Fractals 8.1-25 The Complex Plane and Vector Representation Horizontal axis: real axis Vertical axis: imaginary axis Each complex number a + bi determines a unique position vector with initial point (0, 0) and terminal point (a, b). 8.1-26 The Complex Plane and Vector Representation The sum of two complex numbers is represented by the vector that is the resultant of the vectors corresponding to the two numbers. (4 + i) + (1 + 3i) = 5 + 4i 8.1-27 Example 1 EXPRESSING THE SUM OF COMPLEX NUMBERS GRAPHICALLY Find the sum of 6 – 2i and –4 – 3i. Graph both complex numbers and their resultant. (6 – 2i) + (–4 – 3i) = 2 – 5i 8.1-28 Relationships Among x, y, r, and θ. 8.1-29 Trigonometric (Polar) Form of a Complex Number The expression r(cos θ + i sin θ) is called the trigonometric form (or polar form) of the complex number x + yi. The expression cos θ + i sin θ is sometimes abbreviated cis θ. Using this notation, r(cos θ + i sin θ) is written r cis θ. The number r is the absolute value (or modulus) of x + yi, and θ is the argument of x + yi. 8.1-30 Example 2 CONVERTING FROM TRIGONOMETRIC FORM TO RECTANGULAR FORM Express 2(cos 300° + i sin 300°) in rectangular form. The graphing calculator screen confirms the algebraic solution. The imaginary part is an approximation for 8.1-31 Converting From Rectangular Form to Trigonometric Form Step 1 Sketch a graph of the number x + yi in the complex plane. Step 2 Find r by using the equation Step 3 Find θ by using the equation choosing the quadrant indicated in Step 1. 8.1-32 Caution Be sure to choose the correct quadrant for θ by referring to the graph sketched in Step 1. 8.1-33 Example 3(a) CONVERTING FROM RECTANGULAR FORM TO TRIGONOMETRIC FORM Write measure.) in trigonometric form. (Use radian Step 1: Sketch the graph of in the complex plane. Step 2: 8.1-34 Example 3(a) CONVERTING FROM RECTANGULAR FORM TO TRIGONOMETRIC FORM (continued) Step 3: The reference angle for θ is The graph shows that θ is in quadrant II, so θ = 8.1-35 Example 3(b) CONVERTING FROM RECTANGULAR FORM TO TRIGONOMETRIC FORM Write –3i in trigonometric form. (Use degree measure.) Sketch the graph of –3i in the complex plane. We cannot find θ by using because x = 0. From the graph, a value for θ is 270°. 8.1-36 Example 4 CONVERTING BETWEEN TRIGONOMETRIC AND RECTANGULAR FORMS USING CALCULATOR APPROXIMATIONS Write each complex number in its alternative form, using calculator approximations as necessary. (a) 6(cos 115° + i sin 115°) ≈ –2.5357 + 5.4378i 8.1-37 Example 4 CONVERTING BETWEEN TRIGONOMETRIC AND RECTANGULAR FORMS USING CALCULATOR APPROXIMATIONS (continued) (b) 5 – 4i A sketch of 5 – 4i shows that θ must be in quadrant IV. The reference angle for θ is approximately –38.66°. The graph shows that θ is in quadrant IV, so θ = 360° – 38.66° = 321.34°. 8.1-38 Example 5 DECIDING WHETHER A COMPLEX NUMBER IS IN THE JULIA SET (cont.) Determine whether each number belongs to the Julia set. The calculations repeat as 0, –1, 0, –1, and so on. The absolute values are either 0 or 1, which do not exceed 2, so 0 + 0i is in the Julia set, and the point (0, 0) is part of the graph. 8.1-39 Example 5 DECIDING WHETHER A COMPLEX NUMBER IS IN THE JULIA SET (cont.) The absolute value is so 1 + 1i is not in the Julia set and (1, 1) is not part of the graph. 8.1-40 8.3 The Product and Quotient Theorems Products of Complex Numbers in Trigonometric Form ▪ Quotients of Complex Numbers in Trigonometric Form Copyright © 2009 Pearson Addison-Wesley 1.1-41 8.1-41 Let's try multiplying two complex numbers in polar form together. z1 r1 cos1 i sin 1 z2 r2 cos2 i sin 2 z1 z2 r1 cos 1 i sin 1 r2 cos 2 i sin 2 Look at where we started and where we ended up and see if r1r2a cos 1 i as sinto1what coshappens 2 i sin you can make statement to the2 r 's Must FOIL and the 's when you multiply twothese complex numbers. r1r2 cos 1 cos 2 i sin 2 cos 1 i sin 1 cos 2 i 2 sin 1 sin 2 Replace i 2 with -1 and group real terms and then imaginary terms Multiply the Moduli and Add the Arguments r1r2 cos1 cos2 sin 1 sin 2 sin 1 cos2 cos1 sin 2 i use sum formula for cos use sum formula for sin r1r2 cos1 2 i sin 1 2 Product Theorem are any two complex numbers, then In compact form, this is written Copyright © 2009 Pearson Addison-Wesley 1.1-43 8.1-43 Example 1 USING THE PRODUCT THEOREM Find the product of 3(cos 45° + i sin 45°) and 2(cos 135° + i sin 135°). Write the result in rectangular form. Copyright © 2009 Pearson Addison-Wesley 1.1-44 8.1-44 If z 4 cos 40 i sin 40 and w 6 cos 120 i sin 120 , find : (a) zw (b) z w zw 4 cos 40 i sin 40 6 cos120 i sin120 4 6 cos 40 120 i sin 40 120 multiply the moduli add the arguments (the i sine term will have same argument) 24 cos160 i sin160 24 0.93969 0.34202i 22.55 8.21i If you want the answer in rectangular coordinates simply compute the trig functions and multiply the 24 through. Let z1 r1 cos 1 i sin 1 and z 2 r2 cos 2 i sin 2 be two complex numbers. Then z1 z2 r1r2 cos1 2 i sin 1 2 (This says to multiply two complex numbers in polar form, multiply the moduli and add the arguments) If z 2 0, then z1 r1 cos1 2 i sin1 2 z2 r2 (This says to divide two complex numbers in polar form, divide the moduli and subtract the arguments) Quotient Theorem are any two complex numbers, where In compact form, this is written Copyright © 2009 Pearson Addison-Wesley 1.1-47 8.1-47 Example 2 USING THE QUOTIENT THEOREM Find the quotient rectangular form. Copyright © 2009 Pearson Addison-Wesley Write the result in 1.1-48 8.1-48 Example 3 USING THE PRODUCT AND QUOTIENT THEOREMS WITH A CALCULATOR Use a calculator to find the following. Write the results in rectangular form. Copyright © 2009 Pearson Addison-Wesley 1.1-49 8.1-49 Example 3 Copyright © 2009 Pearson Addison-Wesley USING THE PRODUCT AND QUOTIENT THEOREMS WITH A CALCULATOR (continued) 1.1-50 8.1-50 8.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers Powers of Complex Numbers (De Moivre’s Theorem) ▪ Roots of Complex Numbers Copyright © 2009 Pearson Addison-Wesley 1.1-51 8.1-51 You can repeat this process raising complex numbers to powers. Abraham DeMoivre did this and proved the following theorem: DeMoivre’s Theorem If z rcos i sin is a complex number, Abraham de Moivre (1667 - 1754) then z r cos n i sin n n n where n 1 is a positive integer. This says to raise a complex number to a power, raise the modulus to that power and multiply the argument by that power. De Moivre’s Theorem is a complex number, then In compact form, this is written Copyright © 2009 Pearson Addison-Wesley 1.1-53 8.1-53 Example 1 Find form. First write FINDING A POWER OF A COMPLEX NUMBER and express the result in rectangular in trigonometric form. Because x and y are both positive, θ is in quadrant I, so θ = 60°. Copyright © 2009 Pearson Addison-Wesley 1.1-54 8.1-54 Example 1 FINDING A POWER OF A COMPLEX NUMBER (continued) Now apply De Moivre’s theorem. 480° and 120° are coterminal. Rectangular form Copyright © 2009 Pearson Addison-Wesley 1.1-55 8.1-55 This theorem is used to raise complex numbers to powers. It would be a lot of work to find 3i 3i 3i 3i r 3 1 4 2 2 1 but in Quad II 3 tan 1 3 i you would need to FOIL and multiply all of these together and simplify powers of i --UGH! Instead let's convert to polar form and use DeMoivre's Theorem. 2 4 5 6 4 5 5 3 i 2 cos i sin 2 4 cos 4 5 i sin 4 5 6 6 6 6 4 10 16 cos 3 10 i sin 3 1 3 i 16 2 2 8 8 3i nth Root For a positive integer n, the complex number a + bi is an nth root of the complex number x + yi if Copyright © 2009 Pearson Addison-Wesley 1.1-57 8.1-57 Solve the following over the set of complex numbers: z 1 3 We know that if we cube root both sides we could get 1 but we know that there are 3 roots. So we want the complex cube roots of 1. Using DeMoivre's Theorem with the power being a rational exponent (and therefore meaning a root), we can develop a method for finding complex roots. This leads to the following formula: 2 k 2 k z k r cos i sin n n n n n where k 0, 1, 2, , n 1 nth Root Theorem If n is any positive integer, r is a positive real number, and θ is in degrees, then the nonzero complex number r(cos θ + i sin θ) has exactly n distinct nth roots, given by where Copyright © 2009 Pearson Addison-Wesley 1.1-59 8.1-59 Note In the statement of the nth root theorem, if θ is in radians, then Copyright © 2009 Pearson Addison-Wesley 1.1-60 8.1-60 Let's try this on our problem. We want the cube roots of 1. We want cube root so our n = 3. Can you convert 1 to polar form? (hint: 1 = 1 + 0i) r 1 0 2 2 1 0 tan 0 1 1 0 2 k 0 2 k z k 1 cos i sin 3 3 3 3 3 , for k 0, 1, 2 Once we build the formula, we use it first with k = 0 and get one root, then with k = 1 to get the second root and finally with k = 2 for last root. We want cube root so use 3 numbers here 2 k 2 k z k r cos i sin n n n n n 0 2k 0 2k zk 1cos i sin 3 3 3 3 3 0 20 z0 1cos 3 3 3 0 20 i sin 3 3 , for k 0, 1, 2 1cos 0 i sin 0 1 Here's the root we already knew. 0 21 0 21 z1 1 cos i sin 3 3 3 3 2 1 3 2 1cos i sin i 2 2 3 3 0 22 0 22 3 z 2 1 cos i sin 3 3 3 3 3 4 4 1cos i sin 3 3 1 3 i 2 2 If you cube any of these numbers you get 1. (Try it and see!) We found the cube roots of 1 were: Let's plot these on the complex plane 1 3 1 3 1, i, i 2 2 2 2 about 0.9 each line is 1/2 unit Notice each of the complex roots has the same magnitude (1). Also the three points are evenly spaced on a circle. This will always be true of complex roots. Example 2 FINDING COMPLEX ROOTS Find the two square roots of 4i. Write the roots in rectangular form. Write 4i in trigonometric form: The square roots have absolute value and argument Copyright © 2009 Pearson Addison-Wesley 1.1-64 8.1-64 Example 2 FINDING COMPLEX ROOTS (continued) Since there are two square roots, let k = 0 and 1. Using these values for , the square roots are Copyright © 2009 Pearson Addison-Wesley 1.1-65 8.1-65 Example 2 Copyright © 2009 Pearson Addison-Wesley FINDING COMPLEX ROOTS (continued) 1.1-66 8.1-66 Example 3 FINDING COMPLEX ROOTS Find all fourth roots of rectangular form. Write Write the roots in in trigonometric form: The fourth roots have absolute value and argument Copyright © 2009 Pearson Addison-Wesley 1.1-67 8.1-67 Example 3 FINDING COMPLEX ROOTS (continued) Since there are four roots, let k = 0, 1, 2, and 3. Using these values for α, the fourth roots are 2 cis 30°, 2 cis 120°, 2 cis 210°, and 2 cis 300°. Copyright © 2009 Pearson Addison-Wesley 1.1-68 8.1-68 Example 3 Copyright © 2009 Pearson Addison-Wesley FINDING COMPLEX ROOTS (continued) 1.1-69 8.1-69 Example 3 FINDING COMPLEX ROOTS (continued) The graphs of the roots lie on a circle with center at the origin and radius 2. The roots are equally spaced about the circle, 90° apart. Copyright © 2009 Pearson Addison-Wesley 1.1-70 8.1-70 Example 4 SOLVING AN EQUATION BY FINDING COMPLEX ROOTS Find all complex number solutions of x5 – i = 0. Graph them as vectors in the complex plane. There is one real solution, 1, while there are five complex solutions. Write 1 in trigonometric form: Copyright © 2009 Pearson Addison-Wesley 1.1-71 8.1-71 Example 4 SOLVING AN EQUATION BY FINDING COMPLEX ROOTS (continued) The fifth roots have absolute value argument and Since there are five roots, let k = 0, 1, 2, 3, and 4. Solution set: {cis 0°, cis 72°, cis 144°, cis 216°, cis 288°} Copyright © 2009 Pearson Addison-Wesley 1.1-72 8.1-72 Example 4 SOLVING AN EQUATION BY FINDING COMPLEX ROOTS (continued) The graphs of the roots lie on a unit circle. The roots are equally spaced about the circle, 72° apart. Copyright © 2009 Pearson Addison-Wesley 1.1-73 8.1-73 4 Polynomial and Rational Functions 4.3 Dividing Polynomials Long Division of Polynomials Long Division of Polynomials Dividing polynomials is much like the familiar process of dividing numbers. When we divide 38 by 7, the quotient is 5 and the remainder is 3. We write: 38 3 5 7 7 To divide polynomials, we use long division—as in the next example. E.g. 1—Long Division of Polynomials Divide 6x2 – 26x + 12 by x – 4. The dividend is 6x2 – 26x + 12 and the divisor is x – 4. We begin by arranging them as follows: x 4 6x 2 26x 12 E.g. 1—Long Division of Polynomials Next, we divide the leading term in the dividend by the leading term in the divisor to get the first term of the quotient: 6x2/x = 6x Then, we multiply the divisor by 6x and subtract the result from the dividend. 6x 2 x 4 6 x 26 x 12 6 x 24 x 2x 12 2 E.g. 1—Long Division of Polynomials We repeat the process using the last line –2x + 12 as the dividend. 6x 2 2 x 4 6 x 26 x 12 6 x 24 x 2 x 12 2 x 8 4 2 E.g. 1—Long Division of Polynomials The division process ends when the last line is of lesser degree than the divisor. The last line then contains the remainder. The top line contains the quotient. Long Division of Polynomials We summarize the long division process in the following theorem. Division Algorithm If P(x) and D(x) are polynomials, with D(x) ≠ 0, then there exist unique polynomials Q(x) and R(x), where R(x) is either 0 or of degree less than the degree of D(x), such that: P(x) = D(x) . Q(x) + R(x) Henc, Dividend= (divisor)(quotient)+ remainder. The polynomials P(x) and D(x) are called the dividend and divisor, respectively. Q(x) is the quotient and R(x) is the remainder. E.g. 2—Long Division of Polynomials Let and P(x) = 8x4 + 6x2 – 3x + 1 D(x) = 2x2 – x + 2 Find polynomials Q(x) and R(x) such that: P(x) = D(x) · Q(x) + R(x) E.g. 2—Long Division of Polynomials We use long division after first inserting the term 0x3 into the dividend to ensure that the columns line up correctly. 4 x 2 2x 2x 2 x 2 8 x 4 0 x 3 6 x 2 3 x 1 8x 4 4x 3 8x 2 4 x 3 2x 2 3 x 4 x 3 2x 2 4 x 7 x–7x 1+ 1 is The process is complete at this point as of lesser degree than the divisor 2x2 – x + 2. E.g. 2—Long Division of Polynomials From the long division, we see that: Q(x) = 4x2 + 2x and R(x) = –7x + 1 Thus, 8x4 + 6x2 – 3x + 1 = (2x2 – x + 2)(4x2 + 2x) + (–7x + 1) Synthetic Division Synthetic Division Synthetic division is a quick method of dividing polynomials. It can be used when the divisor is of the form x – c. In synthetic division, we write only the essential parts of the long division. Long Division vs. Synthetic Division Compare the following long and synthetic divisions, in which we divide 2x3 – 7x2 + 5 by x – 3. We’ll explain how to perform the synthetic division in Example 3. Long Division vs. Synthetic Division 2x 2 x 3 x 3 2x 7 x 0 x 5 3 2 32 2x 3 6 x 2 x 2 0x x 3x 2 2 7 0 5 6 3 9 1 3 4 Quotient 3 x 5 3 x 9 4 Remainder Synthetic Division Note that, in synthetic division, we: Abbreviate 2x3 – 7x2 + 5 by writing only the coefficients: 2, –7, 0, and 5. Simply write 3 instead of x – 3. (Writing 3 instead of –3 allows us to add instead of subtract. However, this changes the sign of all the numbers that appear in the gold boxes.) E.g. 3—Synthetic Division Use synthetic division to divide 2x3 – 7x2 + 5 by x – 3. We begin by writing the appropriate coefficients to represent the divisor and the dividend. 32 7 0 5 E.g. 3—Synthetic Division We repeat this process of multiplying and then adding until the table is complete. 32 2 7 0 6 3 1 3 5 E.g. 3—Synthetic Division 32 2 7 0 5 6 3 9 1 3 4 From the last line, we see that the quotient is 2x2 – x – 3 and the remainder is –4. Thus, 2x3 – 7x2 + 5 = (x – 3)(2x2 – x – 3) – 4. The Remainder and Factor Theorems Remainder Theorem The Remainder Theorem shows how synthetic division can be used to evaluate polynomials easily. If the polynomial P(x) is divided by x – c, then the remainder is the value P(c). Remainder Theorem—Proof If the divisor in the Division Algorithm is of the form x – c for some real number c, then the remainder must be a constant. This is because the degree of the remainder is less than the degree of the divisor. Remainder Theorem—Proof If we call this constant r, then P(x) = (x – c) · Q(x) + r Setting x = c in this equation, we get: P(c) = (c – c) · Q(x) + r =0+r =r That is, P(c) is the remainder r. E.g. 4—Using the Remainder Theorem Let P(x) = 3x5 + 5x4 – 4x3 + 7x + 3 (a) Find the quotient and remainder when P(x) is divided by x + 2. (b) Use the Remainder Theorem to find P(–2). E.g. 4—Using Remainder Example (a) Theorem As x + 2 = x – (–2) , the synthetic division for the problem takes the following form. 2 3 3 5 4 0 7 3 6 2 4 8 2 1 2 4 1 5 The quotient is 3x4 – x3 – 2x2 + 4x – 1. The remainder is 5. E.g. 4—Using Remainder Example (b) Theorem By the Remainder Theorem, P(–2) is the remainder when P(x) is divided by x – (–2) = x + 2. From part (a), the remainder is 5. Hence, P(–2) = 5. Factor Theorem The Factor Theorem says that zeros of polynomials correspond to factors. We used this fact in Section 4.2 to graph polynomials. c is a zero of P if and only if x – c is a factor of P(x). The Rational Zero Theorem The Rational Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will appear somewhere in the list. The Rational Zero Theorem p If f(x) anxn an-1xn-1 … a1x a0 has integer coefficients and q p (where is reduced) is a rational zero, then p is a factor of the constant q term a0 and q is a factor of the leading coefficient an. EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f(x) 15x3 14x2 3x – 2. Solution The constant term is –2 and the leading coefficient is 15. Factors of the constant term, 2 Factors of the leading coefficient, 15 1, 2 1, 3, 5, 15 Possible rational zeros 1, 2, 13 , 23 , Divide 1 and 2 by 1. Divide 1 and 2 by 3. 15 , 52 , Divide 1 and 2 by 5. 1 , 2 15 15 Divide 1 and 2 by 15. There are 16 possible rational zeros. The actual solution set to f(x) 15x3 14x2 3x – 2 = 0 is {-1, 1/3, 2/5}, which contains 3 of the 16 possible solutions. EXAMPLE: Solve: Solving a Polynomial Equation x4 6x2 8x + 24 0. Solution Because we are given an equation, we will use the word "roots," rather than "zeros," in the solution process. We begin by listing all possible rational roots. Factors of the constant term, 24 Factors of the leading coefficient, 1 1, 2 3, 4, 6, 8, 12, 24 1 1, 2 3, 4, 6, 8, 12, 24 Possible rational zeros EXAMPLE: Solve: Solving a Polynomial Equation x4 6x2 8x + 24 0. Solution The graph of f(x) x4 6x2 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. 2 1 0 6 8 24 The zero remainder 2 4 4 24 indicates that 2 is a root of x4 6x2 8x + 24 0. 1 2 2 12 0 2 2 8 12 1 4 6 0 The zero remainder indicates that 2 is a root of x3 2x2 2x 12 = 0. EXAMPLE: Solve: Solution Solving a Polynomial Equation x4 6x2 8x + 24 0. Now we can solve the original equation as follows. x4 6x2 8x + 24 0 (x – 2)(x – 2)(x2 4x 6) 0 x – 2 0 or x – 2 0 or x2 4x 6 0 x2 x2 x2 4x 6 0 This is the given equation. This was obtained from the first ,second synthetic division. Set each factor equal to zero. Solve. EXAMPLE: Solve: Solution Solving a Polynomial Equation x4 6x2 8x + 24 0. We can use the quadratic formula to solve x2 4x 6 0. b b2 4ac x 2a 4 42 4 1 6 2 1 4 8 2 4 2i 2 2 2 i 2 We use the quadratic formula because x2 4x 6 0 cannot be factored. Let a 1, b 4, and c 6. Multiply and subtract under the radical. 8 4(2)(1) 2i 2 Simplify. The solution set of the original equation is {2, 2,2 i i2, 2 i i2 }. EXAMPLE: Using the Rational Zero Theorem 2 x 4 3x 3 2 x 2 1 0 Solution The constant term is –1 and the leading coefficient is 2. possible rational zeros Factors.of .the.cons tan t.term,1 Factors.of .the.leading.coefficient ,2 1 Divide 1 Divide 1 by 1 by 2 ,1 2 possible rational zeros 1, 1 2 There are 4 possible rational zeros. The actual solution set to 2x4 3x3 2x2 – 1 = 0 is {-1,1 1/2, 1/2}, which contains 2 of the 4 possible solutions. EXAMPLE: Solving a Polynomial Equation 2 x 4 3x 3 2 x 2 1 0 Solution The graph of f(x) 2 x 3x 2 x 1 0 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. 4 -1 1 2 3 2 3 2 0 1 -2 -1 -1 1 21 1 -1 0 1 1 1 2 2 2 0 2 The zero remainder indicates that -1 is a root of 2x4 3x3 2x2 -1 0. The zero remainder indicates that1/2 is a root of 2x3 x2 x -1 0 EXAMPLE: Solve: Solution Solving a Polynomial Equation 2 x 4 3x 3 2 x 2 1 0 Now we can solve the original equation as follows. 2 x 4 3x 3 2 x 2 1 0 This is the given equation. 1 2 (x +1)(x – )(2x2 2x 2) 0 x +1 0 x -1 or x 1 – 2 x 0 or 2x2 2x 2 0 1 2 x2 x 1 0 Solve. This was obtained from the first ,second synthetic division. Set each factor equal to zero. EXAMPLE: Solution Solving a Polynomial Equation We can use the quadratic formula to solve x2 x 1 0. b b2 4ac x 2a 1 12 4(1)(1) We use the quadratic formula because x2 x 1 0 cannot be factored. Let a 1, b 1, and c 1. 2(1) 1 3 2 1 i 3 2 Multiply and subtract under the radical. 1 2 The solution set of the original equation is {-1, , 1 3i 2 } Properties of Polynomial Equations 1. If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. 2. If a bi is a root of a polynomial equation (b 0), then the non-real complex number a bi is also a root. Non-real complex roots, if they exist, occur in conjugate pairs. Descartes' Rule of Signs If f(x) anxn an1xn1 … a2x2 a1x a0 be a polynomial with real coefficients. 1. The number of positive real zeros of f is either equal to the number of sign changes of f(x) or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of f is either equal to the number of sign changes of f(x) or is less than that number by an even integer. If f(x) has only one variation in sign, then f has exactly one negative real zero. EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f(x) x3 2x2 5x + 4. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f(x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f(x). We obtain this equation by replacing x with x in the given function. f(x) x3 2x2 5x + 4 Replace x with x. f(x) (x)3 2(x)2 5x 4 x3 2x2 5x + 4 This is the given polynomial function. EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f(x) x3 2x2 5x + 4. Solution Now count the sign changes. f(x) x3 2x2 5x + 4 1 2 3 There are three variations in sign. # of negative real zeros of f is either equal to 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3 2 1 negative real zero. Properties of Polynomial Equations 1. If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. 2. If a bi is a root of a polynomial equation (b 0), then the non-real complex number a bi is also a root. Non-real complex roots, if they exist, occur in conjugate pairs. Complex Conjugates Theorem Roots/Zeros that are not Real are Complex with an Imaginary component. Complex roots with Imaginary components always exist in Conjugate Pairs. If a + bi (b ≠ 0) is a zero of a polynomial function, then its Conjugate, a - bi, is also a zero of the function. Find Roots/Zeros of a Polynomial If the known root is imaginary, we can use the Complex Conjugates Theorem. Ex: Find all the roots of f (x) x 3 5x 2 7x 51 If one root is 4 - i. Because of the Complex Conjugate Theorem, we know that another root must be 4 + i. Can the third root also be imaginary? Example (con’t) Ex: Find all the roots of f (x) x 3 5x 2 7x 51 If one root is 4 - i. If one root is 4 - i, then one factor is [x - (4 - i)], and Another root is 4 + i, & another factor is [x - (4 + i)]. Multiply these factors: x 4 i x 4 i x 2 x 4 i x 4 i 4 i 4 i x 2 4 x xi 4 x xi 16 i 2 x 2 8 x 16 (1) x 2 8 x 17 Example (con’t) Ex: Find all the roots of f (x) x 3 5x 2 7x 51 If one root is 4 - i. 2 If the product of the two non-real factors is x 8x 17 then the third factor (that gives us the neg. real root) is the quotient of P(x) divided by x2 8x 17 : x 3 x 2 8x 17 x 3 5x 2 7x 51 x 3 5x 2 7x 51 0 The third root is x = -3 So if asked to find a polynomial that has zeros, 2 and 1 – 3i, you would know another root would be 1 + 3i. Let’s find such a polynomial by putting the roots in factor form and multiplying them together. If x = the root then x - the root is the factor form. x 2x 1 3i x 1 3i x 2x 1 3i x 1 3i x 2 x 2 Multiply the last two factors together. All i terms should disappear when simplified. x 3xi x 1 3i 3xi 3i 9i x 2 x 2 2 x 10 -1 Now multiply the x – 2 through x 4 x 14 x 20 3 2 2 Here is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i Conjugate Pairs Complex Zeros Occur in Conjugate Pairs = If a + bi is a zero of the function, the conjugate a – bi is also a zero of the function (the polynomial function must have real coefficients) EXAMPLES: Find the polynomial equation with the given zeros -1, -1, 3i, -3i 2, 4 + i, 4 – i Now write a polynomial equation of least degree that has real coefficients, a leading coeff. of 1 and 1, -2+i, -2-i as zeros. 0= (x-1)(x-(-2+i))(x-(-2-i)) 0= (x-1)(x+2 - i)(x+2+ i) 0= (x-1)[(x+2) - i] [(x+2)+i] 0= (x-1)[(x+2)2 - i2] Foil 0=(x-1)(x2 + 4x + 4 – (-1)) Take care of i2 0= (x-1)(x2 + 4x + 4 + 1) 0= (x-1)(x2 + 4x + 5) Multiply 0= x3 + 4x2 + 5x – x2 – 4x – 5 0= x3 + 3x2 + x - 5 Now write a polynomial equation of least degree that has real coefficients, a leading coeff. of 1 and 4, 4, 2+i as zeros. Note: 2+i means 2 – i is also a zero 0= (x-4)(x-4)(x-(2+i))(x-(2-i)) 0= (x-4)(x-4)(x-2-i)(x-2+i) 0= (x2 – 8x +16)[(x-2) – i][(x-2)+i] 0= (x2 – 8x +16)[(x-2)2 – i2] 0= (x2 – 8x +16)(x2 – 4x + 4 – (– 1)) 0= (x2 – 8x +16)(x2 – 4x + 5) 0= x4– 4x3+5x2 – 8x3+32x2 – 40x+16x2 – 64x+80 0= x4-12x3+53x2-104x+80