# Discrete Mathematics: Homework 7 solution. Due:

 To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video
Save this PDF as:

Size: px
Start display at page:

## Transcription

1 EE 2060 Discrete Mathematics spring 2011 Discrete Mathematics: Homework 7 solution Due: Let a n = 2 n n for n = 0, 1, 2,... (a) (2%) Find a 0, a 1, a 2, a 3 and a 4. (b) (2%) Show that a 2 = 5a 1 6a 0, a 3 = 5a 2 6a 1, and a 4 = 5a 3 6a 2. (c) (3%) Show that a n = 5a n 1 6a n 2 for all integers n with n 2. (a) We simply plug in n = 0, n = 1, n = 2, n = 3, and n = 4. Thus we have a 0 = = 6, a 1 = = 17, a 2 = = 49, a 3 = = 143, and a 4 = = 421. (b) Using our data from part (a), we see that 49 = , 143 = , and 421 = (c) This is algebra. The messiest part is factoring out a large power of 2 and a large power of 3. If we substitute n 1 for n in the definition we have a n 1 = 2 n n 1 ; similarly a n 2 = 2 n n 2. We start with the right-hand side of our desired identity: 5a n 1 6a n 2 = 5(2 n n 1 ) 6(2 n n 2 ) =2 n 2 (10 6) + 3 n 2 (75 30) =2 n n =2 n + 3 n 5 = a n 2. (a) (6%) Find a recurrence relation for the number of bit strings of length n that contain three consecutive 0s. (b) (1%) What are the initial conditions?

4 a 3 = 2a 2 + 2a = 5 a 4 = 2a 3 + 2a = 21 a 5 = 2a 4 + 2a = 79 a 6 = 2a 5 + 2a = 281 Thus there are 281 bit strings of length 6 containing two consecutive 0 s. 5. Solve these recurrence relations together with the initial conditions given (a) (3%) a n = 7a n 1 10a n 2 for n 2, a 0 = 2, a 1 = 1. (b) (3%) a n = 2a n 1 a n 2 for n 2, a 0 = 4, a 1 = 1. (c) (3%) a n = 6a n 1 9a n 2 for n 2, a 0 = 3, a 1 = 3. (a) r 2 7r + 10 = 0 r = 2, 5 a n = α 1 2 n + α 2 5 n 2 = α 1 + α 2 1 = 2α 1 + 5α 2 α 1 = 3, α 2 = 1 a n = 3 2 n 5 n (b) r 2 2r + 1 = 0 r = 1, 1 a n = α 1 1 n + α 2 n1 n 4 = α 1 1 = α 1 + α 2 α 1 = 4, α 2 = 3 a n = 4 3n (c) r 2 + 6r + 9 = 0 r = 3, 3 a n = α 1 ( 3) n + α 2 n( 3) n 3 = α 1 3 = 3α 1 3α 2 α 1 = 3, α 2 = 2 a n = (3 2n)( 3) n

5 6. (4%) Find the solution to a n = 5a n 2 4a n 4 with a 0 = 3, a 1 = 2, a 2 = 6 and a 3 = 8. The characteristic equation is r 4 5r = 0. This factor as (r 2 1)(r 2 4) = (r + 1)(r 1)(r 2)(r + 2) = 0, so the roots are 1, 1, 2, 2. Therefore the general solution is a n = α 1 + α 2 ( 1) n + α 3 2 n + α 4 ( 2) n. Plugging in initial conditions gives 3 = α 1 + α 2 + α 3 + α 4, 2 = α 1 α 2 + 2α 3 2α 4, 6 = α 1 + α 2 + 4α 3 + 4α 4, and 8 = α 1 α 2 +8α 3 8α 4. The solution to this system of equations is α 1 = α 2 = α 3 = 1 and α 4 = 0. Therefore the answer is a n = 1 + ( 1) n + 2 n. 7. (5%) Solve the recurrence relation a n = 6a n 1 12a n 2 +8a n 3 with a 0 = 5, a 1 = 4, and a 2 = 88. The characteristic equation is r 3 6r r 8 = 0. By the rational root test, the possible rational roots are ±1, ±2, ±4. We find that r = 2 is a root. Dividing r 2 into r 3 6r 2 +12r 8, we find that r 3 6r 2 +12r 8 = (r 2) 3. Hence the only root is 2, with multiplicity 3, so the general solution is (by Thm.4) a n = α 1 2 n +α 2 n2 n +α 3 n 2 2 n. Plugging the initial conditions: 5 = a 0 = α 1 4 = a 1 = 2α 1 + 2α 2 + 2α 3 88 = a 2 = 4α 1 + 8α α 3 Solving this system of equations, we have α 1 = 5, α 2 = 1/2, and α 3 = 13/2. Therefore the answer is a n = 5 2 n + (n/2) 2 n + (13n 2 /2) 2 n = 5 2 n + n 2 n n 2 2 n (2%) What is the general form of the solutions of a linear homogeneous recurrence relation if its characteristic equation has the roots 1, 1, 1, 2, 2, 5, 5, 7. We can write down the general solution using Theorem 4 of Sec.7.2. In this case there are four distinct roots, so t = 4. The multiplicities are 3, 2, 2, and 1. So the general solution is a n = (α 1,0 +α 1,1 n+α 1,2 n 2 )( 1) n +(α 2,0 +α 2,1 n)2 n +(α 3,0 +α 3,1 n)5 n +α 4,0 7 n. 9. Consider the nonhomogeneous linear recurrence relation a n = 2a n n (a) (2%) Show that a n = n2 n is a solution of this recurrence relation.

6 (b) (2%) Use Theorem 5 of Sec.7.2 to find all solutions of this recurrence relation. (c) (2%) Find the solution with a 0 = 2. (a) We compute the right-hand side of the recurrence relation: 2(n 1)2 n n = (n 1)2 n + 2 n = n2 n, which is the left-hand side. (b) The solution of the associated homogeneous equation a n = 2a n 1 is easily found to be a n = α2 n. Therefore the general solution of the inhomogeneous equation is a n = α2 n + n2 n. (c) Plugging in a 0 = 2, we obtain α = 2. Therefore the solution is a n = 2 2 n +n2 n = (n + 2)2 n 10. What is the general form of the particular solution guaranteed to exist by Theorem 6 of the linear nonhomogeneous recurrence relation a n = 6a n 1 12a n 2 +8a n 3 +F (n) if (a) (2%) F (n) = 2 n (b) (2%) F (n) = ( 2) n (c) (2%) F (n) = n 3 ( 2) n (a) Since 2 is a root with multiplicity 3 of the characteristic polynomial of the associated homogeneous recurrence relation, Theorem 6 of Sec.7.2 tells us that the particular solution will be of the form n 3 p 0 2 n. (b) Since 2 is not a root of the characteristic polynomial of the associated homogeneous recurrence relation, Theorem 6 of Sec.7.2 tells us that the particular solution will be of the form p 0 ( 2) n. (c) Since 2 is not a root of the characteristic polynomial of the associated homogeneous recurrence relation, Theorem 6 of Sec.7.2 tells us that the particular solution will be of the form (p 3 n 3 + p 2 n 2 + p 1 n + p 0 )( 2) n. 11. Find a closed form for the generating function for the sequence {a n }, where

7 (a) (2%) a n = 2 n for n = 1, 2, 3, 4,... and a 0 = 0. (b) (2%) a n = 1/(n + 1)! for n = 0, 1, 2,... ( ) 10 (c) (2%) a n = for n = 0, 1, 2,... n + 1 (a) By Table 1 of Sec.7.4, the generating function for the sequence in which a n = 2 n for all n is 1/(1 2x). Here we can either think of subtracting out the missing constant term or factoring out 2x. Therefore the answer can be written as either 1(1 2x) 1 or 2x/(1 2x), which are of course algebraically equivalent. (b) The power series for the function e x is xn /n!. That is almost what we have here; the difference is that the denominator is (n + 1)! instead of n!. So (c) we have x n (n + 1)! = 1 x x n+1 (n + 1)! = 1 x by change of variable. This last sum is (e x 1), so our answer is (e x 1)/x. C(10, n + 1)x n = C(10, n)x n 1 = 1 x n=1 n=1 x n n! C(10, n)x n = 1 x ((1 + x)10 1) 12. For each of these generating functions, provide a closed formula for the sequence it determines. (a) (2%) (3x 1) 3 (b) (2%) x 2 (1 x) 3 (c) (2%) (1 + x 3 )/(1 + x) 3 (d) (2%) e 3x2 1 (a) First we need to factor out 1 and write this as (1 3x) 3. n=1 Then by the Binomial Theorem we get a n = C(3, n)( 3) n for n = 0, 1, 2, 3, and the

8 other coefficients are all 0. Alternatively, we could just multiply out this finite polynomial and note the nonzero coefficients: a 0 = 1, a 1 = 9, a 2 = 27, a 3 = 27. (b) We know that x 2 (1 x) 3 = x 2 (1 3x + 3x 2 x 3 ) = x 2 3x 3 + 3x 4 x 5, so we have a 2 = 1, a 3 = 3, a 4 = 3, a 5 = 1 (c) We split this into two parts : 1 (1 + x) + x 3 3 (1 + x) = ( 1) n C(n + 2, 2)x n + x 3 3 = = ( 1) n C(n + 2, 2)x n + ( 1) n C(n + 2, 2)x n + ( 1) n C(n + 2, 2)x n ( 1) n C(n + 2, 2)x n+3 ( 1) n 3 C(n 1, 2)x n Note that n and n 3 have opposite parities. Therefore a n = ( 1) n C(n + 2, 2) + ( 1) n 3 C(n 1, 2) = ( 1) n (C(n + 2, 2) C(n 1, 2)) = ( 1) n 3n for n 3 and a n = ( 1) n C(n + 2, 2) = ( 1) n (n + 2)(n + 1)/2 for n < 3. (d) e x = 1 + x + x 2 /2! + x 3 /3! +... It follows that n=3 e 3x2 = 1 + 3x 2 + (3x2 ) 2 2! + (3x2 ) 3 3! +... We can therefore read off the coefficients of the generating function for e 3x2 1. First, clearly a 0 = 0. Second, a n = 0 when n is odd. Finally, when n is even, we have a 2m = 3 m /m!. 13. Find the coefficient of x 12 in the power series of each of these functions. (a) (2%) 1/(1 2x) 2 (b) (2%) 1/(1 4x) 3 (a) The coefficient of x n in this power series is 2 n C(n + 1, 1). Thus the answer is 2 12 C(12 + 1, 1) = 53, 248.

9 (b) The coefficient of x n in this power series is 4 n C(n + 2, 2). Thus the answer is 4 12 C(12 + 2, 2) = 1, 526, 726, (5%) Use generating functions to find the number of ways to choose a dozen bagels from three varieties- egg, salty, and plain- if at least two bagels of each kind but no more than three salty bagels are chosen. The factors in the generating function for choosing the egg and plain bagels are both x 2 + x 3 + x The factor for choosing the salty bagels is x 2 + x 3. Therefore the generating function for this problem is (x 2 + x 3 + x 4 +..) 2 (x 2 + x 3 ). We want to find the coefficient of x 12, since we want 12 bagels. This is equivalent to finding the coefficient of x 6 in (1 + x + x ) 2 (1 + x). This function is (1 + x)(1 x) 2, so we want the coefficient of x 6 in 1/(1 x) 2, which is 7, plus the coefficient of x 5 in 1/(1 x) 2, which is 6. Thus the answer is (a) (4%) What is the generating function for {a k }, where a k is the number of solutions of x 1 + x 2 + x 3 + x 4 = k when x 1, x 2, x 3 and x 4 are integers with x 1 3, 1 x 2 5, 0 x 3 4, and x 4 1? (b) (2%) Use your answer to part(a) to find a 7 (a) The restriction on x 1 gives us the factor x 3 + x 4 + x The restriction on x 2 gives us the factor x + x 2 + x 3 + x 4 + x 5. The restriction on x 3 gives us the factor 1 + x + x 2 + x 3 + x 4. And the restriction on x 4 gives us the factor x + x 2 + x Thus the answer is the product of these: (x 3 + x 4 + x )(x + x 2 + x 3 + x 4 + x 5 )(1 + x + x 2 + x 3 + x 4 )(x + x 2 + x ) We can use algebra to rewrite this in closed form as x 5 (1 + x + x 2 + x 3 + x 4 ) 2 /(1 x) 2 (b) We want the coefficient of x 7 in this series, which is the same as the coefficient of x 2 in the series for (1 + x + x 2 + x 3 + x 4 ) 2 = 1 + 2x + 3x2 + higher order terms (1 x) 2 (1 x) 2

10 Since the coefficient of x n in 1/(1 x) 2 is n+1, our answer is = (5%) Use generating functions to solve the recurrence relation a k = 3a k k 1 with the initial condition a 0 = 1. Let G(x) = k=0 a kx k. Then xg(x) = k=0 a kx k+1 = k=1 a k 1x k. Thus G(x) 3xG(x) = a k x k k=0 k=1 3a k 1 x k = a 0 + k=1 k=0 (a k 3a k 1 )x k = 1 + k=1 4 k 1 x k = 1 + x 4 k 1 x k 1 = 1 + x 4 k x k 1 = 1 + x 1 4x = 1 3x 1 4x Thus G(x)(1 3x) = (1 3x)/(1 4x), so G(x) = 1/(1 4x). Therefore a k = 4 k. k=1

### Math 55: Discrete Mathematics

Math 55: Discrete Mathematics UC Berkeley, Spring 2012 Homework # 9, due Wednesday, April 11 8.1.5 How many ways are there to pay a bill of 17 pesos using a currency with coins of values of 1 peso, 2 pesos,

### 8.2 Solving Linear Recurrence Relations

8.2 Solving Linear Recurrence Relations Determine if recurrence relation is homogeneous or nonhomogeneous. Determine if recurrence relation is linear or nonlinear. Determine whether or not the coefficients

### JUST 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

### Generating Functions

Generating Functions If you take f(x =/( x x 2 and expand it as a power series by long division, say, then you get f(x =/( x x 2 =+x+2x 2 +x +5x 4 +8x 5 +x 6 +. It certainly seems as though the coefficient

### 2.3. 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

### Date: 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.

### CLASS 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

### Algebra Practice Problems for Precalculus and Calculus

Algebra Practice Problems for Precalculus and Calculus Solve the following equations for the unknown x: 1. 5 = 7x 16 2. 2x 3 = 5 x 3. 4. 1 2 (x 3) + x = 17 + 3(4 x) 5 x = 2 x 3 Multiply the indicated polynomials

### Polynomials 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

### LINEAR RECURSIVE SEQUENCES. The numbers in the sequence are called its terms. The general form of a sequence is. a 1, a 2, a 3,...

LINEAR RECURSIVE SEQUENCES BJORN POONEN 1. Sequences A sequence is an infinite list of numbers, like 1) 1, 2, 4, 8, 16, 32,.... The numbers in the sequence are called its terms. The general form of a sequence

### Sequences. A sequence is a list of numbers, or a pattern, which obeys a rule.

Sequences A sequence is a list of numbers, or a pattern, which obeys a rule. Each number in a sequence is called a term. ie the fourth term of the sequence 2, 4, 6, 8, 10, 12... is 8, because it is the

### Vieta s Formulas and the Identity Theorem

Vieta s Formulas and the Identity Theorem This worksheet will work through the material from our class on 3/21/2013 with some examples that should help you with the homework The topic of our discussion

### The 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

### Zeros of Polynomial Functions

Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n-1 x n-1 + + 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

### Alum Rock Elementary Union School District Algebra I Study Guide for Benchmark III

Alum Rock Elementary Union School District Algebra I Study Guide for Benchmark III Name Date Adding and Subtracting Polynomials Algebra Standard 10.0 A polynomial is a sum of one ore more monomials. Polynomial

### 1 Lecture: Integration of rational functions by decomposition

Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.

### HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!

Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following

### Limits at Infinity Limits at Infinity for Polynomials Limits at Infinity for the Exponential Function Function Dominance More on Asymptotes

Lecture 5 Limits at Infinity and Asymptotes Limits at Infinity Horizontal Asymptotes Limits at Infinity for Polynomials Limit of a Reciprocal Power The End Behavior of a Polynomial Evaluating the Limit

### Partial Fraction Decomposition for Inverse Laplace Transform

Partial Fraction Decomposition for Inverse Laplace Transform Usually partial fractions method starts with polynomial long division in order to represent a fraction as a sum of a polynomial and an another

### Name Date Block. Algebra 1 Laws of Exponents/Polynomials Test STUDY GUIDE

Name Date Block Know how to Algebra 1 Laws of Eponents/Polynomials Test STUDY GUIDE Evaluate epressions with eponents using the laws of eponents: o a m a n = a m+n : Add eponents when multiplying powers

### Math Common Core Sampler Test

High School Algebra Core Curriculum Math Test Math Common Core Sampler Test Our High School Algebra sampler covers the twenty most common questions that we see targeted for this level. For complete tests

### Answers to Basic Algebra Review

Answers to Basic Algebra Review 1. -1.1 Follow the sign rules when adding and subtracting: If the numbers have the same sign, add them together and keep the sign. If the numbers have different signs, subtract

### 1.3 Polynomials and Factoring

1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable.

### 3.3. Solving Polynomial Equations. Introduction. Prerequisites. Learning Outcomes

Solving Polynomial Equations 3.3 Introduction Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations, called polynomial equations. These have the general

### 3. 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

### Algebraic expressions are a combination of numbers and variables. Here are examples of some basic algebraic expressions.

Page 1 of 13 Review of Linear Expressions and Equations Skills involving linear equations can be divided into the following groups: Simplifying algebraic expressions. Linear expressions. Solving linear

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### 4. Factor polynomials over complex numbers, describe geometrically, and apply to real-world situations. 5. Determine and apply relationships among syn

I The Real and Complex Number Systems 1. Identify subsets of complex numbers, and compare their structural characteristics. 2. Compare and contrast the properties of real numbers with the properties of

### 1 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,

### Zero: 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

### 5.1 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

### Rational 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

### SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION

CHAPTER 5 SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Alessandro Artale UniBZ - http://www.inf.unibz.it/ artale/ SECTION 5.2 Mathematical Induction I Copyright Cengage Learning. All rights reserved.

### a 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 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

### Monomial. 5 1 x A sum is not a monomial. 2 A monomial cannot have a. x 21. degree. 2x 3 1 x 2 2 5x Rewrite a polynomial

9.1 Add and Subtract Polynomials Before You added and subtracted integers. Now You will add and subtract polynomials. Why? So you can model trends in recreation, as in Ex. 37. Key Vocabulary monomial degree

### Higher Order Linear Differential Equations with Constant Coefficients

Higher Order Linear Differential Equations with Constant Coefficients Part I. Homogeneous Equations: Characteristic Roots Objectives: Solve n-th order homogeneous linear equations where a n,, a 1, a 0

### Linear Equations in One Variable

Linear Equations in One Variable MATH 101 College Algebra J. Robert Buchanan Department of Mathematics Summer 2012 Objectives In this section we will learn how to: Recognize and combine like terms. Solve

### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important

### Unit 3 Polynomials Study Guide

Unit Polynomials Study Guide 7-5 Polynomials Part 1: Classifying Polynomials by Terms Some polynomials have specific names based upon the number of terms they have: # of Terms Name 1 Monomial Binomial

### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic

### 3.2 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

### ModuMath Algebra Lessons

ModuMath Algebra Lessons Program Title 1 Getting Acquainted With Algebra 2 Order of Operations 3 Adding & Subtracting Algebraic Expressions 4 Multiplying Polynomials 5 Laws of Algebra 6 Solving Equations

### Section 5.0A Factoring Part 1

Section 5.0A Factoring Part 1 I. Work Together A. Multiply the following binomials into trinomials. (Write the final result in descending order, i.e., a + b + c ). ( 7)( + 5) ( + 7)( + ) ( + 7)( + 5) (

### Zeros 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

### Vocabulary Words and Definitions for Algebra

Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms

### Equations, Inequalities & Partial Fractions

Contents Equations, Inequalities & Partial Fractions.1 Solving Linear Equations 2.2 Solving Quadratic Equations 1. Solving Polynomial Equations 1.4 Solving Simultaneous Linear Equations 42.5 Solving Inequalities

### Zeros 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

### POLYNOMIAL 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

### 1.3 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,

### Polynomials Classwork

Polynomials Classwork What Is a Polynomial Function? Numerical, Analytical and Graphical Approaches Anatomy of an n th -degree polynomial function Def.: A polynomial function of degree n in the vaiable

### Integrals of Rational Functions

Integrals of Rational Functions Scott R. Fulton Overview A rational function has the form where p and q are polynomials. For example, r(x) = p(x) q(x) f(x) = x2 3 x 4 + 3, g(t) = t6 + 4t 2 3, 7t 5 + 3t

### 1. What s wrong with the following proofs by induction?

ArsDigita University Month : Discrete Mathematics - Professor Shai Simonson Problem Set 4 Induction and Recurrence Equations Thanks to Jeffrey Radcliffe and Joe Rizzo for many of the solutions. Pasted

### Factoring 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

### The Geometric Series

The Geometric Series Professor Jeff Stuart Pacific Lutheran University c 8 The Geometric Series Finite Number of Summands The geometric series is a sum in which each summand is obtained as a common multiple

### 1. Use Properties of Exponents

A. Polynomials Polynomials are one of the most fundamental types of functions used in mathematics. They are very simple to use, primarily because they are formed entirely by multiplication (exponents are

### QUADRATIC EQUATIONS EXPECTED BACKGROUND KNOWLEDGE

MODULE - 1 Quadratic Equations 6 QUADRATIC EQUATIONS In this lesson, you will study aout quadratic equations. You will learn to identify quadratic equations from a collection of given equations and write

Algebra 1: Chapter 10 Notes 1 Algebra Homework: Chapter 10 UPDATED ASSIGNMENT SHEET (Homework is listed by date assigned; homework is due the following class period) Leave all answers in reduced, radical

### What are the place values to the left of the decimal point and their associated powers of ten?

The verbal answers to all of the following questions should be memorized before completion of algebra. Answers that are not memorized will hinder your ability to succeed in geometry and algebra. (Everything

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### EAP/GWL Rev. 1/2011 Page 1 of 5. Factoring a polynomial is the process of writing it as the product of two or more polynomial factors.

EAP/GWL Rev. 1/2011 Page 1 of 5 Factoring a polynomial is the process of writing it as the product of two or more polynomial factors. Example: Set the factors of a polynomial equation (as opposed to an

### a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

### Tim Kerins. Leaving Certificate Honours Maths - Algebra. Tim Kerins. the date

Leaving Certificate Honours Maths - Algebra the date Chapter 1 Algebra This is an important portion of the course. As well as generally accounting for 2 3 questions in examination it is the basis for many

### Algebra 1-2. A. Identify and translate variables and expressions.

St. Mary's College High School Algebra 1-2 The Language of Algebra What is a variable? A. Identify and translate variables and expressions. The following apply to all the skills How is a variable used

### Factoring Polynomials and Solving Quadratic Equations

Factoring Polynomials and Solving Quadratic Equations Math Tutorial Lab Special Topic Factoring Factoring Binomials Remember that a binomial is just a polynomial with two terms. Some examples include 2x+3

### Math 115 Spring 2011 Written Homework 5 Solutions

. Evaluate each series. a) 4 7 0... 55 Math 5 Spring 0 Written Homework 5 Solutions Solution: We note that the associated sequence, 4, 7, 0,..., 55 appears to be an arithmetic sequence. If the sequence

### 1.7. 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

### Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers.

Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used

### Sample Induction Proofs

Math 3 Worksheet: Induction Proofs III, Sample Proofs A.J. Hildebrand Sample Induction Proofs Below are model solutions to some of the practice problems on the induction worksheets. The solutions given

### West Windsor-Plainsboro Regional School District Algebra I Part 2 Grades 9-12

West Windsor-Plainsboro Regional School District Algebra I Part 2 Grades 9-12 Unit 1: Polynomials and Factoring Course & Grade Level: Algebra I Part 2, 9 12 This unit involves knowledge and skills relative

Factoring A Quadratic Polynomial If we multiply two binomials together, the result is a quadratic polynomial: This multiplication is pretty straightforward, using the distributive property of multiplication

### Functions and Equations

Centre for Education in Mathematics and Computing Euclid eworkshop # Functions and Equations c 014 UNIVERSITY OF WATERLOO Euclid eworkshop # TOOLKIT Parabolas The quadratic f(x) = ax + bx + c (with a,b,c

### Discrete Mathematics: Solutions to Homework (12%) For each of the following sets, determine whether {2} is an element of that set.

Discrete Mathematics: Solutions to Homework 2 1. (12%) For each of the following sets, determine whether {2} is an element of that set. (a) {x R x is an integer greater than 1} (b) {x R x is the square

### Prep for College Algebra

Prep for College Algebra This course covers the topics shown below. Students navigate learning paths based on their level of readiness. Institutional users may customize the scope and sequence to meet

### Basic Terminology for Systems of Equations in a Nutshell. E. L. Lady. 3x 1 7x 2 +4x 3 =0 5x 1 +8x 2 12x 3 =0.

Basic Terminology for Systems of Equations in a Nutshell E L Lady A system of linear equations is something like the following: x 7x +4x =0 5x +8x x = Note that the number of equations is not required

### Cyclic Codes Introduction Binary cyclic codes form a subclass of linear block codes. Easier to encode and decode

Cyclic Codes Introduction Binary cyclic codes form a subclass of linear block codes. Easier to encode and decode Definition A n, k linear block code C is called a cyclic code if. The sum of any two codewords

### STUDY 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

### Partial Fractions. Combining fractions over a common denominator is a familiar operation from algebra:

Partial Fractions Combining fractions over a common denominator is a familiar operation from algebra: From the standpoint of integration, the left side of Equation 1 would be much easier to work with than

### Zeros of Polynomial Functions

Review: Synthetic Division Find (x 2-5x - 5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 3-5x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 3-5x 2 + x + 2. Zeros of Polynomial Functions Introduction

### is identically equal to x 2 +3x +2

Partial fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as 1 + 3

### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise

### MATH 65 NOTEBOOK CERTIFICATIONS

MATH 65 NOTEBOOK CERTIFICATIONS Review Material from Math 60 2.5 4.3 4.4a Chapter #8: Systems of Linear Equations 8.1 8.2 8.3 Chapter #5: Exponents and Polynomials 5.1 5.2a 5.2b 5.3 5.4 5.5 5.6a 5.7a 1

### Core Maths C1. Revision Notes

Core Maths C Revision Notes November 0 Core Maths C Algebra... Indices... Rules of indices... Surds... 4 Simplifying surds... 4 Rationalising the denominator... 4 Quadratic functions... 4 Completing the

### Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

### CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

### 0.8 Rational Expressions and Equations

96 Prerequisites 0.8 Rational Expressions and Equations We now turn our attention to rational expressions - that is, algebraic fractions - and equations which contain them. The reader is encouraged to

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### Recursive Algorithms. Recursion. Motivating Example Factorial Recall the factorial function. { 1 if n = 1 n! = n (n 1)! if n > 1

Recursion Slides by Christopher M Bourke Instructor: Berthe Y Choueiry Fall 007 Computer Science & Engineering 35 Introduction to Discrete Mathematics Sections 71-7 of Rosen cse35@cseunledu Recursive Algorithms

### Notes on Linear Recurrence Sequences

Notes on Linear Recurrence Sequences April 8, 005 As far as preparing for the final exam, I only hold you responsible for knowing sections,,, 6 and 7 Definitions and Basic Examples An example of a linear

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

### Actually, if you have a graphing calculator this technique can be used to find solutions to any equation, not just quadratics. All you need to do is

QUADRATIC EQUATIONS Definition ax 2 + bx + c = 0 a, b, c are constants (generally integers) Roots Synonyms: Solutions or Zeros Can have 0, 1, or 2 real roots Consider the graph of quadratic equations.

### COGNITIVE TUTOR ALGEBRA

COGNITIVE TUTOR ALGEBRA Numbers and Operations Standard: Understands and applies concepts of numbers and operations Power 1: Understands numbers, ways of representing numbers, relationships among numbers,

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### Name: Date: Algebra 2/ Trig Apps: Simplifying Square Root Radicals. Arithmetic perfect squares: 1, 4, 9,,,,,,...

RADICALS PACKET Algebra 2/ Trig Apps: Simplifying Square Root Radicals Perfect Squares Perfect squares are the result of any integer times itself. Arithmetic perfect squares: 1, 4, 9,,,,,,... Algebraic

### Math 2280 Section 002 [SPRING 2013] 1

Math 2280 Section 002 [SPRING 2013] 1 Today well learn about a method for solving systems of differential equations, the method of elimination, that is very similar to the elimination methods we learned

### 4/1/2017. PS. Sequences and Series FROM 9.2 AND 9.3 IN THE BOOK AS WELL AS FROM OTHER SOURCES. TODAY IS NATIONAL MANATEE APPRECIATION DAY

PS. Sequences and Series FROM 9.2 AND 9.3 IN THE BOOK AS WELL AS FROM OTHER SOURCES. TODAY IS NATIONAL MANATEE APPRECIATION DAY 1 Oh the things you should learn How to recognize and write arithmetic sequences

### SUNY ECC. ACCUPLACER Preparation Workshop. Algebra Skills

SUNY ECC ACCUPLACER Preparation Workshop Algebra Skills Gail A. Butler Ph.D. Evaluating Algebraic Epressions Substitute the value (#) in place of the letter (variable). Follow order of operations!!! E)

### Lesson 9: Radicals and Conjugates

Student Outcomes Students understand that the sum of two square roots (or two cube roots) is not equal to the square root (or cube root) of their sum. Students convert expressions to simplest radical form.