Sect 3.3 Zeros of Polynomials Functions


 Maria Banks
 1 years ago
 Views:
Transcription
1 101 Sect 3.3 Zeros of Polynomials Functions Objective 1: Using the Factor Theorem. Recall that a zero of a polynomial P(x) is a number c such that P(c) = 0. If the remainder of P(x) (x c) is zero, then c is a zero and x c is factor of P(x). Factor Theorem A number c is a zero of a polynomial P if and only if x c is a factor of P. Proof: ( ) Suppose c is a zero of P. Using the Division Algorithm, we can rewrite P as: P(x) = Q(x) (x c) + R(x) But, by the Remainder Theorem, R(x) = P(c) = 0 since c is a zero of P. This implies that P(x) = Q(x) (x c) which means that x c is a factor of P. ( ) Suppose x c is a factor of P. Then P(x) = Q(x) (x c) for some polynomial Q(x). If we evaluate P at x = c, we get: P(c) = Q(c) (c c) = 0 This implies that c is a zero of P. Show that the given values of c are zeros of P(x) and find all the other zeros of P(x). Ex. 1 P(x) = 2x 4 + 9x 3 9x 2 46x + 24; c = 3, 2 First, evaluate P at x = 3 and at x = 2: P( 3) = 2( 3) 4 + 9( 3) 3 9( 3) 2 46( 3) + 24 = = 0 P(2) = 2(2) 4 + 9(2) 3 9(2) 2 46(2) + 24 = = 0 Thus, 3 and 2 are zeros of P so by the factor theorem, x + 3 and x 2 are factors of P. We will now use synthetic division to break down P: Thus, P(x) = (x 2)(2x x x 12)
2 So, P(x) = (x 2)(2x x x 12) = (x 2)(x + 3)(2x 2 + 7x 4) Now, we can factor 2x 2 + 7x 4 using trial and error: 2x 2 + 7x 4 = (2x 1)(x + 4) Therefore, P(x) = (x 2)(x + 3)(2x 1)(x + 4) and the zeros of P are { 4, 3, ½, 2} Show that the given values of c are zeros of P(x) and find all the other zeros of P(x). Ex. 2 P(x) = 4x 4 19x 3 21x x + 20; c = 2, 2 First, evaluate P at x = 2 and at x = 2: P( 2) = 4( 2) 4 19( 2) 3 21( 2) ( 2) + 20 = = 0 P(2) = 4(2) 4 19(2) 3 21(2) (2) + 20 = = 0 Thus, 2 and 2 are zeros of P so by the factor theorem, x + 2 and x 2 are factors of P. We will now use synthetic division to break down P: Thus, P(x) = (x 2)(4x 3 11x 2 43x 10) So, P(x) = (x 2)(4x 3 11x 2 43x 10) = (x 2)(x + 2)(4x 2 19x 5) Now, we can factor 4x 2 19x 5 using trial and error: 4x 2 19x 5 = (4x + 1)(x 5) Therefore, P(x) = (x 2)(x + 2)(4x + 1)(x 5) and the zeros of P are { 2, ¼, 2, 5} Objective 2: Finding rational zeros of functions. In the last example, P(x) = 4x 4 19x 3 21x x + 20, the constant term was 20 and the leading coefficient was 4. The factors p of the constant term are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20 and the factors q of the leading coefficient are ± 1, ± 2, ± 4. If we form all possible ratios of the factors p of the constant term to the factors q of the leading coefficient, we get: 102
3 103 ± 1, ± 2, ± 4, ± 5, ± 10, ± 20, ± 1 2, ± 5 2, ± 1 4, and ± 5. This gives us all the 4 possible rational zeros of P(x). Notice that the zeros ¼, 2, 2, 5 are contained in this list. This leads to the following theorem. Rational Zeros Theorem Let P(x) = a n x n + a n 1 x n a 1 x + a 0 be a polynomial of degree n with integer coefficients. Then every rational zero of P is in the form p q where p q is in lowest terms and p is a factor of the constant term a 0 and q is a factor of the leading coefficient a n. Proof: Let p q be a zero of a P(x) and p q be in lowest terms. Then, P( p q ) = a n( p q )n + a n 1 ( p q )n a 1 ( p q ) + a 0 = 0 (simplify) a n p n p n 1 + a q n n a p q n 1 1 q + a 0 = 0 (multiply both sides by q n ) a n p n + a n 1 p n 1 q + + a 1 pq n 1 + a 0 q n = 0 (subtract a n p n from both sides) a n 1 p n 1 q + + a 1 pq n 1 + a 0 q n = a n p n (factor out q) q(a n 1 p n a 1 pq n 2 + a 0 q n 1 ) = a n p n This implies that q is a factor of a n p n. But since p and q are in lowest terms, then q has to be a factor of a n. We can use a similar argument to show that p is a factor of a 0. Find all the rational zeros of the P(x) and then factor P(x) into linear factors: Ex. 3 P(x) = x x3 2x x In order to use the Rational Zeros Theorem, all the coefficients have to be integers. Since the LCD of the coefficients is 6, we will first factor out 1/6 from all the terms: P(x) = x x3 2x x (Rewrite with a denominator of 6) = 6 6 x x x2 3 6 x (Factor out 1 6 ) = 1 6 (6x4 + 7x 3 12x 2 3x + 2) Now, find the possible rational zeros of 6x 4 + 7x 3 12x 2 3x + 2 The possible values of p are ± 1, ± 2, and the possible values of q are ± 1, ± 2, ± 3, ± 6. Thus, the possible combinations of p over q are
4 104 ± 1 1, ± 1 2, ± 1 3, ± 1 6, ± 2 1, ± 2 2, ± 2 3, ± 2. Simplifying and eliminating 6 duplication gives us the possible rational zeros of ± 1, ± 1 2, ± 1 3, ± 1 6, ± 2, ± 2 3. Now, we will use synthetic division to factor 6x 4 + 7x 3 12x 2 3x + 2 Let's try x = 1: Yes Thus, P(x) = 1 6 (x 1)(6x3 + 13x 2 + x 2) Let's try x = 1 again: No Now, let's try x = 1: No Let's try x = 2: No Now, let's try x = 2: Yes Hence, P(x) = 1 6 (x 1)(x + 2)(6x2 + x 1) Now, factor 6x 2 + x 1 using trial and error to get (3x 1)(2x + 1). Then, our answer is P(x) = 1 (x 1)(x + 2)(3x 1)(2x + 1). 6 This suggests a procedure for finding the rational zeros of a polynomial.
5 105 Steps to Finding the Rational Zeros 1) Use the Rational Zeros Theorem to list all possible rational zeros. 2) Use synthetic division evaluate each possible rational zero in the polynomial. If the reminder is zero, write down the quotient you found. 3) Repeat the process for the quotient you found in part 2 until you obtain a quotient that is degree two or easily factors. Factor or use the quadratic formula to find the remaining zeros. Find all the zeros of the P(x): Ex. 4 P(x) = 2x 4 x 3 67x 2 159x 63 The possible values of p are ± 1, ± 3, ± 7, ± 9, ± 21, ± 63 and the possible values of q are ± 1, ± 2. Thus, the possible combinations of p over q are: ± 1, ± 1 2, ± 3, ± 3 2, ± 7, ± 7 2, ± 9, ± 9 2 Now, we will use synthetic division Let's try x = 1: No Try x = No Try x = 3: No 21 63, ± 21, ±, ± 63, ± 2 2. Try x = Yes
6 106 Try x = 3 again Yes Now, factor 2x 2 13x 7 using trial and error: 2x 2 13x 7 = (2x + 1)(x 7). Solving (2x + 1)(x 7) = 0 yields x = ½ and x = 7. Thus, the zeros are { 3, ½, 7}. If we had to write P(x) in the last example as a product of linear factors, our answer would have been P(x) = (x + 3) 2 (2x + 1)(x 7). Notice that the factors (2x + 1) and (x 7) occur only once. In such a case, we say that the zeros ½ and 7 have multiplicity of one. Since the factor (x + 3) occurs twice, the zero 3 has multiplicity of two. Find all the real solutions to the following: Ex. 5 f(x) = x 4 + 2x 3 + 3x 2 2x 4 = 0 The possible values of p are ± 1, ± 2, ± 4 and the possible values of q are ± 1. Thus, the possible combinations of p over q are ± 1, ± 2, ± 4. Now, we will use synthetic division Let's try x = 1: Yes Try x = Yes Using the discriminant on x 2 + 2x + 4 = 0, we get: b 2 4ac = (2) 2 4(1)(4) = 4 16 = 12 which tells us that x 2 + 2x + 4 = 0 has no real solutions. Thus, the only real solutions are { 1, 1}. Each zero has multiplicity of one. Suppose in the last example, if we were asked to find all solutions (real and complex), we would have to use the quadratic formula to get the to complex solutions: x 2 + 2x + 4 = 0
7 107 x = b± b2 4ac 2a = (2) ± (2)2 4(1)(4) 2(1) = 2± 12 2 = 1 ± 3 i So, all the solutions would be { 1, 1, 1 3 i, i }. In factored form, f(x) would be (x + 1)(x 1)(x i)(x i). Notice that even though f(x) does not factor into linear factors in the real numbers, it does factor into linear factors in the complex numbers. Also, the complex zeros are what we call complex conjugates. Complex Conjugates If r = a + bi is a complex number, then r = a bi is the complex conjugate. Complex conjugate pairs have the following property: r r = (a + bi )(a bi ) = a 2 + b 2 Thus, (5 + 3i )(5 3i ) = 25 15i + 15i 9i 2 = 25 9( 1) = = 34. r and r are called conjugate pairs. Objective 3: Finding complex zeros of polynomials. The depressed equation x 2 + 2x + 4 = 0 from the last example had no real solutions. Such quadratic equations are called irreducible if they cannot be factored over the real numbers. Any irreducible quadratic equation in the real numbers can be factored in the complex numbers. This suggests that every complex nonconstant polynomial has at least one complex zero. This is known as the Fundamental Theorem of Algebra. Fundamental Theorem of Algebra Every complex polynomial function with degree n 1 has at least one complex zero. Using this theorem in conjunction with the Factor Theorem leads to the following theorems: Theorem Every complex polynomial function f of degree n 1 can be factored into n linear factors (not necessarily distinct) of the form: f(x) = a n (x r 1 )(x r 2 ) (x r n ) where a n, r 1, r 2,, r n are complex numbers. In other words, every nonconstant polynomial has exact n complex zeros including repeating zeros. Conjugate Pairs Theorem Let f be a polynomial function whose coefficients are real numbers. If r = a + bi is a zero of f, then the complex conjugate r = a bi is also a zero of f so long as b 0.
8 108 Find all the complex zeros of the P(x). Then write P in factored form: Ex. 6 P(x) = x 4 + 2x x x 75 The possible values of p are ± 1, ± 3, ± 5, ± 15, ± 25, ± 75 and the possible values of q are ± 1. Thus, the possible combinations of p over q are ± 1, ± 3, ± 5, ± 15, ± 25, ± 75 which gives us the possible rational zeros. Now, we will use synthetic division: Let's try x = 1: Yes Try x = No Try x = 3: No Try x = Yes Now, factor x in the complex numbers: x = (x + 5i )(x 5i ) Solving (x + 5i )(x 5i ) = 0 yields x = ± 5i. Thus, the zeros are { 5i, 5i, 3, 1}. As a product of linear factors, we get: P(x) = (x + 5i )(x 5i )(x + 3)(x 1) Solve the following: Ex. 7 f(x) = 2x 4 + 5x 3 + 9x 2 + 2x 4 = 0 The possible values of p are ± 1, ± 2, ± 4 and the possible values of q are ± 1, ± 2. Thus, the possible combinations of p over q are:
9 109 ± 1, ± 1, ± 2, ± 4. 2 Now, we will use synthetic division Let's try x = 1: No Try x = Yes Try x = 1 again No Try x = Yes Working with the depressed equation 2x 2 + 4x + 8 = 0, we can factor out a 2 to get 2(x 2 + 2x + 4) and then we will need to use the quadratic formula on x 2 + 2x + 4: x = b± b2 4ac 2a = (2) ± (2)2 4(1)(4) 2(1) = 2± 12 2 So, the solution is { 1 3 i, i, 1, 1 2 }. If we had to write f(x) in factored form, we would get: f(x) = 2(x ( 1 3 i ))(x ( i ))(x + 1)(x 1 2 ) = 1 ± 3 i Given the zeros of a polynomial, we can construct a polynomial that would have those zeros. Let's start with the answer from example #2 and work backwards to find a polynomial.
10 110 Find a 4 th degree polynomial with the given zeros: Ex. 8 ¼, 2, 2, 5 Since ¼, 2, 2, & 5 are zeros, then (x ( ¼)) = (x + ¼), (x ( 2)) = (x + 2), (x 2), & (x 5) are factors. But saying (x + ¼) is a factor is also equivalent to saying (4x + 1) is a factor since they have the "same zero." Thus, P is the product of these factors: P(x) = (4x + 1)(x + 2)(x 2)(x 5) But by FOIL, (4x + 1)(x + 2) = 4x 2 + 9x + 2 and (x 2)(x 5) = x 2 7x + 10 Thus, P(x) = (4x + 1)(x + 2)(x 2)(x 5) = (4x 2 + 9x + 2)(x 2 7x + 10) Multiplying the two trinomials, we get: 4x 2 + 9x + 2 x 2 7x x x x 3 63x 2 14x 4x 4 + 9x 3 + 2x 2 4x 4 19x 3 21x x + 20 Hence, P(x) = 4x 4 19x 3 21x x Note that if we multiply 4x 4 19x 3 21x x + 20 by any nonzero constant, we will get another 4 th degree polynomial that has the same zeros. This means that there are an infinite number of answers for the last example; all such answers are constant multiples of one and other. Find a 4 th degree polynomial f satisfying the following conditions: Ex. 9 Zeros: 4 (multiplicity of 2), (multiplicity of 1). f( 2) = 189 (multiplicity of 1), and Since 4, 1 2, & 1 3 are zeros, then (x 4), (x ( 1 2 )) = (x ), & (x 1 3 ), are factors. But saying (x + 1 ) is a factor is also equivalent 2 to saying (2x + 1) is a factor and (x 1 ) is a factor is equivalent to 3 saying (3x 1) is a factor since they have the "same zero." Since the zero of 4 has a multiplicity of 2, the factor (x 4) will occur twice. Thus, f is the product of these factors: f(x) = (x 4) 2 (2x + 1)(3x 1)
11 But by FOIL, (x 4)(x 4) = x 2 8x + 16 and (2x + 1)(3x 1) = 6x 2 + x 1 Thus, f(x) = (x 4) 2 (2x + 1)(3x 1) = (x 2 8x + 16)(6x 2 + x 1) Multiplying the two trinomials, we get: x 2 8x x 2 + x 1 x 2 + 8x 16 x 3 8x x 6x 4 48x x 2 6x 4 47x x x 16 So, f(x) = a(6x 4 47x x x 16). But, f( 2) = 189, which says: f( 2) = a(6( 2) 4 47( 2) ( 2) ( 2) 16) = 189 a( ) = a = 189 or a = ¼ Hence, f(x) = ¼(6x 4 47x x x 16) = 3 2 x x x2 6x Find a polynomial function with the given information: Ex. 10 Degree 5; Zeros: 2, 3 5i, 4i Since the polynomial has a degree of 5, then it has five zeros. The conjugate pair of 4i is 4i which is also a zero and the conjugate pair of 3 5i is 3 + 5i which is a zero as well. So, the missing zeros are 4i and 3 + 5i. This means that (x ( 2)), (x (3 5i )), (x (3 + 5i )), (x ( 4i )), & (x (4i )) are factors. Thus, f can be written as: f(x) = a(x ( 2))(x (3 5i ))(x (3 + 5i ))(x ( 4i ))(x (4i )) where a is any nonzero real number. f(x) = a(x + 2)[x (3 5i )][x (3 + 5i )](x + 4i )(x 4i ) (expand) f(x) = a(x + 2)[x 2 (3 5i i )x + (3 5i )(3 + 5i )](x 2 16i 2 ) f(x) = a(x + 2)[x 2 6x i 2 ](x 2 16i 2 ) (replace i 2 by 1) f(x) = a(x + 2)[x 2 6x ](x ) f(x) = a(x + 2)(x 2 6x + 34)(x ) f(x) = a(x 3 4x x + 68)(x ) f(x) = a(x 5 4x x 3 + 4x x )
Zeros of Polynomial Functions
Review: Synthetic Division Find (x 25x  5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 35x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 35x 2 + x + 2. Zeros of Polynomial Functions Introduction
More informationChapter 2. Polynomial and Rational Functions. 2.5 Zeros of Polynomial Functions. Copyright 2014, 2010, 2007 Pearson Education, Inc.
Chapter 2 Polynomial and Rational Functions 2.5 Zeros of Polynomial Functions Copyright 2014, 2010, 2007 Pearson Education, Inc. 1 Objectives: Use the Rational Zero Theorem to find possible rational zeros.
More informationZeros of a Polynomial Function
Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we
More informationCLASS NOTES. We bring down (copy) the leading coefficient below the line in the same column.
SYNTHETIC DIVISION CLASS NOTES When factoring or evaluating polynomials we often find that it is convenient to divide a polynomial by a linear (first degree) binomial of the form x k where k is a real
More informationPolynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if
1. Polynomials 1.1. Definitions A polynomial in x is an expression obtained by taking powers of x, multiplying them by constants, and adding them. It can be written in the form c 0 x n + c 1 x n 1 + c
More information3.2 The Factor Theorem and The Remainder Theorem
3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial
More information3.2 The Factor Theorem and The Remainder Theorem
3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial
More informationNovember 01, S4.4p Theorems about Zeros of Polynomial Functions. 4.4 Theorems about Zeros of Polynomial Functions
MAT 171 Precalculus Algebra CHAPTER 4: Polynomial and Rational Functions 4.1 Polynomial Functions and Models 4.2 Graphing Polynomial Functions 4.3 Polynomial Division; The Remainder and Factor Theorems
More informationSect 3.2 Synthetic Division
94 Objective 1: Sect 3.2 Synthetic Division Division Algorithm Recall that when dividing two numbers, we can check our answer by the whole number (quotient) times the divisor plus the remainder. This should
More informationZeros of Polynomial Functions
Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n1 x n1 + + a 1 x + a 0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of
More information3.7 Complex Zeros; Fundamental Theorem of Algebra
SECTION.7 Complex Zeros; Fundamental Theorem of Algebra 2.7 Complex Zeros; Fundamental Theorem of Algebra PREPARING FOR THIS SECTION Before getting started, review the following: Complex Numbers (Appendix,
More informationChapter 6. Polynomials and Polynomial Functions
Chapter 6 Polynomials and Polynomial Functions Lesson 61 Polynomial Functions Polynomials A polynomial is a monomial or the sum of monomials. P( x) a x a x... a x a n n1 n n1 1 0 3 P( x) x 5x x 5 The
More informationJUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials
More informationFactors of 8 are 1 and 8 or 2 and 4. Let s substitute these into our factors and see which produce the middle term, 10x.
Quadratic equations A quadratic equation in x is an equation that can be written in the standard quadratic form ax + bx + c 0, a 0. Several methods can be used to solve quadratic equations. If the quadratic
More informationSection 32 Finding Rational Zeros of Polynomials
 Finding Rational Zeros of Polynomials 95 Section  Finding Rational Zeros of Polynomials Factor Theorem Fundamental Theorem of Algebra Imaginary Zeros Rational Zeros In this section we develop some important
More informationSOLVING POLYNOMIAL EQUATIONS
C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra
More informationMarch 29, 2011. 171S4.4 Theorems about Zeros of Polynomial Functions
MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 4: Polynomial and Rational Functions 4.1 Polynomial Functions and Models 4.2 Graphing Polynomial Functions 4.3 Polynomial
More information1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x).
.7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational
More informationReview for College Algebra Midterm 1
Review for College Algebra Midterm 1 1.2 Functions and graphs Vocabulary: function, graph, domain, range, VerticalLine Test. Determine whether a given correspondence/mapping is a function. Evaluate a
More information4.1 Polynomial Functions and Models
4.1 Polynomial Functions and Models Determine the behavior of the graph of a polynomial function using the leadingterm test. Factor polynomial functions and find the zeros and their multiplicities. Use
More informationSection 5.1: Polynomial Functions
Section 5.1: Polynomial Functions Def: A polynomial function is a function of the form: f(x) = a n x n + a n 1 x n 1 +... + a 1 x + a 0 where a n, a n 1,..., a 1, a 0 are real numbers and the exponents
More informationWarm Up Lesson Presentation Lesson Quiz. Holt McDougal Algebra 2. Algebra 2
Fundamental Fundamental Theorem Theorem 36 of of Algebra Algebra Warm Up Lesson Presentation Lesson Quiz Algebra 2 Warm Up Identify all the real roots of each equation. 1. 4x 5 8x 4 32x 3 = 0 0, 2, 4
More informationSolutions to SelfTest for Chapter 4 c4sts  p1
Solutions to SelfTest for Chapter 4 c4sts  p1 1. Graph a polynomial function. Label all intercepts and describe the end behavior. a. P(x) = x 4 2x 3 15x 2. (1) Domain = R, of course (since this is a
More informationPolynomial Equations Carnegie Learning, Inc. Chapter 10 l Polynomial Equations
C H A P T ER Polynomial Equations The Fundamental Theorem of Algebra, which states that any polynomial equation of degree n must have exactly n complex roots or solutions, was first proposed by Peter Rothe
More informationQuadratic Equations and Inequalities
MA 134 Lecture Notes August 20, 2012 Introduction The purpose of this lecture is to... Introduction The purpose of this lecture is to... Learn about different types of equations Introduction The purpose
More informationIntroduction to polynomials
Worksheet 4.5 Polynomials Section 1 Introduction to polynomials A polynomial is an expression of the form p(x) = p 0 + p 1 x + p 2 x 2 + + p n x n, (n N) where p 0, p 1,..., p n are constants and x os
More information5.1 The Remainder and Factor Theorems; Synthetic Division
5.1 The Remainder and Factor Theorems; Synthetic Division In this section you will learn to: understand the definition of a zero of a polynomial function use long and synthetic division to divide polynomials
More information9.3 Solving Quadratic Equations by the Quadratic Formula
9.3 Solving Quadratic Equations by the Quadratic Formula OBJECTIVES 1 Identify the values of a, b, and c in a quadratic equation. Use the quadratic formula to solve quadratic equations. 3 Solve quadratic
More informationa 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)
ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x
More information5.1 The Remainder and Factor Theorems; Synthetic Division
5.1 The Remainder and Factor Theorems; Synthetic Division In this section you will learn to: understand the definition of a zero of a polynomial function use long and synthetic division to divide polynomials
More informationDefinition of Subtraction x  y = x + 1y2. Subtracting Real Numbers
Algebra Review Numbers FRACTIONS Addition and Subtraction i To add or subtract fractions with the same denominator, add or subtract the numerators and keep the same denominator ii To add or subtract fractions
More informationPolynomial Test 2 Review
Polynomial Test 2 Review 2.1  standard form and vertex form of a quadratic equation identify vertex and graph  convert from standard form to vertex form and from vertex form to standard form  vertical
More informationCopy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.
Algebra 2  Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers  {1,2,3,4,...}
More informationZero: If P is a polynomial and if c is a number such that P (c) = 0 then c is a zero of P.
MATH 11011 FINDING REAL ZEROS KSU OF A POLYNOMIAL Definitions: Polynomial: is a function of the form P (x) = a n x n + a n 1 x n 1 + + a x + a 1 x + a 0. The numbers a n, a n 1,..., a 1, a 0 are called
More informationZeros of Polynomial Functions
Zeros of Polynomial Functions Objectives: 1.Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions 2.Find rational zeros of polynomial functions 3.Find conjugate
More informationMy Notes MATH TIP. Activity 17 Factors of Polynomials 267
Factors of Polynomials How Many Roots? Lesson 171 Algebraic Methods Learning Targets: Determine the linear factors of polynomial functions using algebraic methods. Determine the linear or quadratic factors
More informationPolynomial and Rational Functions along with Miscellaneous Equations. Chapter 3
Polynomial and Rational Functions along with Miscellaneous Equations Chapter 3 Quadratic Functions and Inequalities 3.1 Quadratic Function A quadratic function is defined by a quadratic or seconddegree
More informationCluster: Perform arithmetic operations on polynomials. Featured Mathematical Practice:
1. Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.
More informationDate: Section P.2: Exponents and Radicals. Properties of Exponents: Example #1: Simplify. a.) 3 4. b.) 2. c.) 3 4. d.) Example #2: Simplify. b.) a.
Properties of Exponents: Section P.2: Exponents and Radicals Date: Example #1: Simplify. a.) 3 4 b.) 2 c.) 34 d.) Example #2: Simplify. a.) b.) c.) d.) 1 Square Root: Principal n th Root: Example #3: Simplify.
More informationPreCalculus Chapter 2 Practice Answers
PreCalculus Chapter 2 Practice Answers 1. To find the x and y intercepts, let x = 0 to find the y intercept and let y = 0 to find the x. y = x 2 5x + 4 y = (0) 2 5(0) + 4 y = 4, when x = 0 (0,4) 0 = x
More informationPRECALCULUS A TEST #2 POLYNOMIALS AND RATIONAL FUNCTIONS, PRACTICE
PRECALCULUS A TEST # POLYNOMIALS AND RATIONAL FUNCTIONS, PRACTICE SECTION.3 Polynomial and Synthetic Division 1) Divide using long division: ( 6x 3 + 11x 4x 9) ( 3x ) ) Divide using long division: ( x
More informationFactoring polynomials and solving higher degree equations
Factoring polynomials and solving higher degree equations Nikos Apostolakis November 15, 2008 Recall. With respect to division polynomials behave a lot like natural numbers. It is not always possible to
More informationThe Rational Zero Test. + =, and be asked to find all
The Rational Zero Test The ultimate objective for this section of the workbook is to graph polynomial functions of degree greater than. The first step in accomplishing this will be to find all real zeros
More informationMath 002 Unit 5  Student Notes
Sections 7.1 Radicals and Radical Functions Math 002 Unit 5  Student Notes Objectives: Find square roots, cube roots, nth roots. Find where a is a real number. Look at the graphs of square root and cube
More information1.3 Algebraic Expressions
1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,
More informationSolving Quadratic Equations
Chapter 7 Solving Quadratic Equations Sec 1. Zero Product Property Back in the third grade students were taught when they multiplied a number by zero, the product would be zero. In algebra, that s an extremely
More informationSolving ax 2 + bx + c = 0 Deriving the Quadratic Formula
Solving ax + bx + c = 0 SUGGESTED LEARNING STRATEGIES: Marking the Text, Group Presentation, Activating Prior Knowledge, Quickwrite Recall solving quadratic equations of the form a x + c = 0. To solve
More informationIn this lesson you will learn to find zeros of polynomial functions that are not factorable.
2.6. Rational zeros of polynomial functions. In this lesson you will learn to find zeros of polynomial functions that are not factorable. REVIEW OF PREREQUISITE CONCEPTS: A polynomial of n th degree has
More informationRational Polynomial Functions
Rational Polynomial Functions Rational Polynomial Functions and Their Domains Today we discuss rational polynomial functions. A function f(x) is a rational polynomial function if it is the quotient of
More information2.3. Finding polynomial functions. An Introduction:
2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned
More informationBasic Algebra Practice Test
1. Exponents and integers: Problem type 2 Evaluate. Basic Algebra Practice Test 2. Exponents and signed fractions Evaluate. Write your answers as fractions. 3. Exponents and order of operations Evaluate.
More informationWorksheet 4.7. Polynomials. Section 1. Introduction to Polynomials. A polynomial is an expression of the form
Worksheet 4.7 Polynomials Section 1 Introduction to Polynomials A polynomial is an expression of the form p(x) = p 0 + p 1 x + p 2 x 2 + + p n x n (n N) where p 0, p 1,..., p n are constants and x is a
More information3. Power of a Product: Separate letters, distribute to the exponents and the bases
Chapter 5 : Polynomials and Polynomial Functions 5.1 Properties of Exponents Rules: 1. Product of Powers: Add the exponents, base stays the same 2. Power of Power: Multiply exponents, bases stay the same
More informationPOLYNOMIAL FUNCTIONS
POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a
More informationKEY. i is the number you can square to get an answer of 1. i 2 = 1. Or, i is the square root of 1 i =
Real part Imaginary part Algebra Unit: 05 Lesson: 0 Solving Quadratic Equations by Formula a _ bi KEY Complex Numbers All the quadratic equations solved to this point have had two real solutions or roots.
More informationMath Analysis Chapter 2 Notes: Polynomial and Rational Functions
Math Analysis Chapter Notes: Polynomial and Rational Functions Day 14: Section 1 Complex Numbers; Sections  Quadratic Functions 1: Complex Numbers After completing section 1 you should be able to do
More informationSect 12.1 The Square Root Property and Completing the Square
Sect 1.1 The Square Root Property and Completing the Square 110 Concept #1 Solving Quadratic Equations by Using the Square Root Property. When we solved quadratic equations before, we had to get zero on
More informationChapter 1: Number Systems and Fundamental Concepts of Algebra. If n is negative, the number is small; if n is positive, the number is large
Final Exam Review Chapter 1: Number Systems and Fundamental Concepts of Algebra Scientific Notation: Numbers written as a x 10 n where 1 < a < 10 and n is an integer If n is negative, the number is small;
More information1 Algebra  Partial Fractions
1 Algebra  Partial Fractions Definition 1.1 A polynomial P (x) in x is a function that may be written in the form P (x) a n x n + a n 1 x n 1 +... + a 1 x + a 0. a n 0 and n is the degree of the polynomial,
More informationChapter 5: Rational Functions, Zeros of Polynomials, Inverse Functions, and Exponential Functions Quiz 5 Exam 4
Chapter 5: Rational Functions, Zeros of Polynomials, Inverse Functions, and QUIZ AND TEST INFORMATION: The material in this chapter is on Quiz 5 and Exam 4. You should complete at least one attempt of
More informationSTUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS
STUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS The intermediate algebra skills illustrated here will be used extensively and regularly throughout the semester Thus, mastering these skills is an
More informationASS.PROF.DR Thamer Information Theory 4th Class in Communication. Finite Field Arithmetic. (Galois field)
Finite Field Arithmetic (Galois field) Introduction: A finite field is also often known as a Galois field, after the French mathematician Pierre Galois. A Galois field in which the elements can take q
More informationSect Addition, Subtraction, Multiplication, and Division Properties of Equality
Sect.1  Addition, Subtraction, Multiplication, and Division Properties of Equality Concept #1 Definition of a Linear Equation in One Variable An equation is a statement that two quantities are equal.
More informationMore Zeroes of Polynomials. Elementary Functions. The Rational Root Test. The Rational Root Test
More Zeroes of Polynomials In this lecture we look more carefully at zeroes of polynomials. (Recall: a zero of a polynomial is sometimes called a root.) Our goal in the next few presentations is to set
More information3.4 Complex Zeros and the Fundamental Theorem of Algebra
86 Polynomial Functions.4 Complex Zeros and the Fundamental Theorem of Algebra In Section., we were focused on finding the real zeros of a polynomial function. In this section, we expand our horizons and
More informationModule MA1S11 (Calculus) Michaelmas Term 2016 Section 2: Polynomials
Module MAS (Calculus) Michaelmas Term 206 Section 2: Polynomials D. R. Wilkins Copyright c David R. Wilkins 206 Contents 2 Polynomials 30 2. Completing the Square in Quadratic Polynomials....... 30 2.2
More informationZeros of Polynomial Functions
OpenStaxCNX module: m49349 1 Zeros of Polynomial Functions OpenStax College OpenStax College Precalculus This work is produced by OpenStaxCNX and licensed under the Creative Commons Attribution License
More informationALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form
ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form Goal Graph quadratic functions. VOCABULARY Quadratic function A function that can be written in the standard form y = ax 2 + bx+ c where a 0 Parabola
More informationAlgebra. Warm Up Lesson Presentation Lesson Quiz. Holt Algebra 2 2
66 Fundamental Fundamental Theorem Theorem 66 of of Algebra Algebra Warm Up Lesson Presentation Lesson Quiz Warm Up Identify all the real roots of each equation. 1. 4x 5 8x 4 x = 0 0,, 4. x x + 9 = 9x.
More information1 I' (9) (10) (11)
1 54 CHAPTER 6. POLYNOMIALS EXERCISE 61 (f(x), g(x)). Prove that the quotient and remainder (q(x) and r(x)) are unique for each pair There is a special shorthand method called synthetic division for dividing
More informationMath 002 Intermediate Algebra
Math 002 Intermediate Algebra Student Notes & Assignments Unit 4 Rational Exponents, Radicals, Complex Numbers and Equation Solving Unit 5 Homework Topic Due Date 7.1 BOOK pg. 491: 62, 64, 66, 72, 78,
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More informationUnit III  Solving Polynomial Equations (25%) Suggested Instruction and Assessment Time for this Unit: 12 classes(assumes minute classes)
Unit III  Solving Polynomial Equations (25%) Suggested Instruction and Assessment Time for this Unit: 12 classes(assumes 5560 minute classes) Outcomes SCO: By the end of Mathematics 3103 students will
More informationStudents will understand 1. that polynomials are closed under the operations of addition, subtraction and multiplication.
High School Arithmetic with Polynomials and Rational Expressions Essential Questions: 1. How do you use patterns to understand mathematics and model situations? 2. What is algebra? 3. How are the horizontal
More informationZeroes of polynomials and long division. ElementaryaFunctions. Zeroes of polynomials and long division. Zeroes of polynomials and long division
The Fundamental Theorem of Algebra tells us that every polynomial of degree n has at most n zeroes. Indeed, if we are willing to count multiple zeroes and also count complex numbers (more on that later)
More information1.1 Solving a Linear Equation ax + b = 0
1.1 Solving a Linear Equation ax + b = 0 To solve an equation ax + b = 0 : (i) move b to the other side (subtract b from both sides) (ii) divide both sides by a Example: Solve x = 0 (i) x = 0 x = (ii)
More informationx n = 1 x n In other words, taking a negative expoenent is the same is taking the reciprocal of the positive expoenent.
Rules of Exponents: If n > 0, m > 0 are positive integers and x, y are any real numbers, then: x m x n = x m+n x m x n = xm n, if m n (x m ) n = x mn (xy) n = x n y n ( x y ) n = xn y n 1 Can we make sense
More informationAlgebra Revision Sheet Questions 2 and 3 of Paper 1
Algebra Revision Sheet Questions and of Paper Simple Equations Step Get rid of brackets or fractions Step Take the x s to one side of the equals sign and the numbers to the other (remember to change the
More information5.5 The real zeros of a polynomial function
5.5 The real zeros of a polynomial function Recall, that c is a zero of a polynomial f(), if f(c) = 0. Eample: a) Find real zeros of f()= + . We need to find for which f() = 0, that is we have to solve
More informationMath 002 Intermediate Algebra Spring 2012 Objectives & Assignments
Math 00 Intermediate Algebra Spring 01 Objectives & Assignments Unit 3 Exponents, Polynomial Operations, and Factoring I. Exponents & Scientific Notation 1. Use the properties of exponents to simplify
More informationLesson 9 Solving Quadratic Equations
Lesson 9 Solving Quadratic Equations We will continue our work with Quadratic Functions in this lesson and will learn several methods for solving quadratic equations. Graphing is the first method you will
More informationMath 002 Intermediate Algebra Summer 2012 Objectives & Assignments Unit 4 Rational Exponents, Radicals, Complex Numbers and Equation Solving
Math 002 Intermediate Algebra Summer 2012 Objectives & Assignments Unit 4 Rational Exponents, Radicals, Complex Numbers and Equation Solving I. Rational Exponents and Radicals 1. Simplify expressions with
More informationWest WindsorPlainsboro Regional School District Algebra I Part 2 Grades 912
West WindsorPlainsboro Regional School District Algebra I Part 2 Grades 912 Unit 1: Polynomials and Factoring Course & Grade Level: Algebra I Part 2, 9 12 This unit involves knowledge and skills relative
More information2.1 Algebraic Expressions and Combining like Terms
2.1 Algebraic Expressions and Combining like Terms Evaluate the following algebraic expressions for the given values of the variables. 3 3 3 Simplify the following algebraic expressions by combining like
More informationLong Division of Polynomials
SECTION 4.3 DIVI DI NG P OLYNOM IALS: LONG DIVISION AN D SYNTH ETIC DIVISION SKI LLS OBJ ECTIVES Divide polynomials with long division. Divide polynomials with synthetic division. CONCE PTUAL OBJ ECTIVES
More informationPolynomial Rings If R is a ring, then R[x], the ring of polynomials in x with coefficients in R, consists of all formal
Polynomial Rings 8172009 If R is a ring, then R[x], the ring of polynomials in x with coefficients in R, consists of all formal sums a i x i, where a i = 0 for all but finitely many values of i. If a
More informationGreatest common divisors
Greatest common divisors Robert Friedman Long division Does 5 divide 34? It is easy to see that the answer is no, for many different reasons: 1. 34 does not end in a 0 or 5. 2. Checking directly by hand,
More information2.5 Zeros of a Polynomial Functions
.5 Zeros of a Polynomial Functions Section.5 Notes Page 1 The first rule we will talk about is Descartes Rule of Signs, which can be used to determine the possible times a graph crosses the xaxis and
More informationFactoring Polynomials
Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent
More informationALGEBRA 2 CRA 2 REVIEW  Chapters 16 Answer Section
ALGEBRA 2 CRA 2 REVIEW  Chapters 16 Answer Section MULTIPLE CHOICE 1. ANS: C 2. ANS: A 3. ANS: A OBJ: 53.1 Using Vertex Form SHORT ANSWER 4. ANS: (x + 6)(x 2 6x + 36) OBJ: 64.2 Solving Equations by
More informationAlgebra Tiles Activity 1: Adding Integers
Algebra Tiles Activity 1: Adding Integers NY Standards: 7/8.PS.6,7; 7/8.CN.1; 7/8.R.1; 7.N.13 We are going to use positive (yellow) and negative (red) tiles to discover the rules for adding and subtracting
More informationTHE THEORY OF EQUATIONS
0. The Theory of Equations (09) 7 0. THE THEORY OF EQUATIONS In this section The Number of Roots to a Polynomial Equation The Conjugate Pairs Theorem Descartes Rule of Signs Bounds on the Roots The zeros
More informationCollege Algebra. Chapter 4, section 6 Created by Lauren Atkinson Mary Stangler Center for Academic Success
College Algebra Chapter 4, section 6 Created by Lauren Atkinson Mary Stangler Center for Academic Success This review is meant to highlight basic concepts from Chapter 4. It does not cover all concepts
More informationSYNTHETIC DIVISION AND THE FACTOR THEOREM
628 (11 48) Chapter 11 Functions In this section Synthetic Division The Factor Theorem Solving Polynomial Equations 11.6 SYNTHETIC DIVISION AND THE FACTOR THEOREM In this section we study functions defined
More informationSection III.6. Factorization in Polynomial Rings
III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)
More informationAdditional Resources https://www.pearsonsuccessnet.com 2. Reason abstractly and quantitatively
CCSS for Mathematical Practice: Review Standards 1. Make sense of problems and persevere in solving them (M)ajor Content, Additional Resources https://www.pearsonsuccessnet.com 2. Reason abstractly and
More informationEquations and Inequalities
Rational Equations Overview of Objectives, students should be able to: 1. Solve rational equations with variables in the denominators.. Recognize identities, conditional equations, and inconsistent equations.
More informationLeading Term: ax n. Up and Up Down and Up Down and Down Up and Down a Positive a Positive a Negative a Negative n Even n Odd n Even n Odd
5.1 Notes: Polynomial Functions monomial: a real number, variable, or product of one real number and one or more variables degree of monomial: the exponent of the variable(s) polynomial: a monomial or
More informationMTH 098. Sections 5.1 & 5.2
MTH 098 Sections 5.1 & 5.2 5.1 The Greatest Common Factor: Factoring by Grouping What is the greatest common factor of two numbers? Find the greatest common factor (GCF) for each list of numbers. 1. 50,
More information