Factoring Polynomials
Total Page:16
File Type:pdf, Size:1020Kb
Load more
Recommended publications
-
Discriminant Analysis
Chapter 5 Discriminant Analysis This chapter considers discriminant analysis: given p measurements w, we want to correctly classify w into one of G groups or populations. The max- imum likelihood, Bayesian, and Fisher’s discriminant rules are used to show why methods like linear and quadratic discriminant analysis can work well for a wide variety of group distributions. 5.1 Introduction Definition 5.1. In supervised classification, there are G known groups and m test cases to be classified. Each test case is assigned to exactly one group based on its measurements wi. Suppose there are G populations or groups or classes where G 2. Assume ≥ that for each population there is a probability density function (pdf) fj(z) where z is a p 1 vector and j =1,...,G. Hence if the random vector x comes × from population j, then x has pdf fj (z). Assume that there is a random sam- ple of nj cases x1,j,..., xnj ,j for each group. Let (xj , Sj ) denote the sample mean and covariance matrix for each group. Let wi be a new p 1 (observed) random vector from one of the G groups, but the group is unknown.× Usually there are many wi, and discriminant analysis (DA) or classification attempts to allocate the wi to the correct groups. The w1,..., wm are known as the test data. Let πk = the (prior) probability that a randomly selected case wi belongs to the kth group. If x1,1..., xnG,G are a random sample of cases from G the collection of G populations, thenπ ˆk = nk/n where n = i=1 ni. -
Solving Quadratic Equations by Factoring
5.2 Solving Quadratic Equations by Factoring Goals p Factor quadratic expressions and solve quadratic equations by factoring. p Find zeros of quadratic functions. Your Notes VOCABULARY Binomial An expression with two terms, such as x ϩ 1 Trinomial An expression with three terms, such as x2 ϩ x ϩ 1 Factoring A process used to write a polynomial as a product of other polynomials having equal or lesser degree Monomial An expression with one term, such as 3x Quadratic equation in one variable An equation that can be written in the form ax2 ϩ bx ϩ c ϭ 0 where a 0 Zero of a function A number k is a zero of a function f if f(k) ϭ 0. Example 1 Factoring a Trinomial of the Form x2 ؉ bx ؉ c Factor x2 Ϫ 9x ϩ 14. Solution You want x2 Ϫ 9x ϩ 14 ϭ (x ϩ m)(x ϩ n) where mn ϭ 14 and m ϩ n ϭ Ϫ9 . Factors of 14 1, Ϫ1, Ϫ 2, Ϫ2, Ϫ (m, n) 14 14 7 7 Sum of factors Ϫ Ϫ m ؉ n) 15 15 9 9) The table shows that m ϭ Ϫ2 and n ϭ Ϫ7 . So, x2 Ϫ 9x ϩ 14 ϭ ( x Ϫ 2 )( x Ϫ 7 ). Lesson 5.2 • Algebra 2 Notetaking Guide 97 Your Notes Example 2 Factoring a Trinomial of the Form ax2 ؉ bx ؉ c Factor 2x2 ϩ 13x ϩ 6. Solution You want 2x2 ϩ 13x ϩ 6 ϭ (kx ϩ m)(lx ϩ n) where k and l are factors of 2 and m and n are ( positive ) factors of 6 . -
Alicia Dickenstein
A WORLD OF BINOMIALS ——————————– ALICIA DICKENSTEIN Universidad de Buenos Aires FOCM 2008, Hong Kong - June 21, 2008 A. Dickenstein - FoCM 2008, Hong Kong – p.1/39 ORIGINAL PLAN OF THE TALK BASICS ON BINOMIALS A. Dickenstein - FoCM 2008, Hong Kong – p.2/39 ORIGINAL PLAN OF THE TALK BASICS ON BINOMIALS COUNTING SOLUTIONS TO BINOMIAL SYSTEMS A. Dickenstein - FoCM 2008, Hong Kong – p.2/39 ORIGINAL PLAN OF THE TALK BASICS ON BINOMIALS COUNTING SOLUTIONS TO BINOMIAL SYSTEMS DISCRIMINANTS (DUALS OF BINOMIAL VARIETIES) A. Dickenstein - FoCM 2008, Hong Kong – p.2/39 ORIGINAL PLAN OF THE TALK BASICS ON BINOMIALS COUNTING SOLUTIONS TO BINOMIAL SYSTEMS DISCRIMINANTS (DUALS OF BINOMIAL VARIETIES) BINOMIALS AND RECURRENCE OF HYPERGEOMETRIC COEFFICIENTS A. Dickenstein - FoCM 2008, Hong Kong – p.2/39 ORIGINAL PLAN OF THE TALK BASICS ON BINOMIALS COUNTING SOLUTIONS TO BINOMIAL SYSTEMS DISCRIMINANTS (DUALS OF BINOMIAL VARIETIES) BINOMIALS AND RECURRENCE OF HYPERGEOMETRIC COEFFICIENTS BINOMIALS AND MASS ACTION KINETICS DYNAMICS A. Dickenstein - FoCM 2008, Hong Kong – p.2/39 RESCHEDULED PLAN OF THE TALK BASICS ON BINOMIALS - How binomials “sit” in the polynomial world and the main “secrets” about binomials A. Dickenstein - FoCM 2008, Hong Kong – p.3/39 RESCHEDULED PLAN OF THE TALK BASICS ON BINOMIALS - How binomials “sit” in the polynomial world and the main “secrets” about binomials COUNTING SOLUTIONS TO BINOMIAL SYSTEMS - with a touch of complexity A. Dickenstein - FoCM 2008, Hong Kong – p.3/39 RESCHEDULED PLAN OF THE TALK BASICS ON BINOMIALS - How binomials “sit” in the polynomial world and the main “secrets” about binomials COUNTING SOLUTIONS TO BINOMIAL SYSTEMS - with a touch of complexity (Mixed) DISCRIMINANTS (DUALS OF BINOMIAL VARIETIES) and an application to real roots A. -
The Diamond Method of Factoring a Quadratic Equation
The Diamond Method of Factoring a Quadratic Equation Important: Remember that the first step in any factoring is to look at each term and factor out the greatest common factor. For example: 3x2 + 6x + 12 = 3(x2 + 2x + 4) AND 5x2 + 10x = 5x(x + 2) If the leading coefficient is negative, always factor out the negative. For example: -2x2 - x + 1 = -1(2x2 + x - 1) = -(2x2 + x - 1) Using the Diamond Method: Example 1 2 Factor 2x + 11x + 15 using the Diamond Method. +30 Step 1: Multiply the coefficient of the x2 term (+2) and the constant (+15) and place this product (+30) in the top quarter of a large “X.” Step 2: Place the coefficient of the middle term in the bottom quarter of the +11 “X.” (+11) Step 3: List all factors of the number in the top quarter of the “X.” +30 (+1)(+30) (-1)(-30) (+2)(+15) (-2)(-15) (+3)(+10) (-3)(-10) (+5)(+6) (-5)(-6) +30 Step 4: Identify the two factors whose sum gives the number in the bottom quarter of the “x.” (5 ∙ 6 = 30 and 5 + 6 = 11) and place these factors in +5 +6 the left and right quarters of the “X” (order is not important). +11 Step 5: Break the middle term of the original trinomial into the sum of two terms formed using the right and left quarters of the “X.” That is, write the first term of the original equation, 2x2 , then write 11x as + 5x + 6x (the num bers from the “X”), and finally write the last term of the original equation, +15 , to get the following 4-term polynomial: 2x2 + 11x + 15 = 2x2 + 5x + 6x + 15 Step 6: Factor by Grouping: Group the first two terms together and the last two terms together. -
Solving Cubic Polynomials
Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial monic. a 2. Then, given x2 + a x + a , substitute x = y − 1 to obtain an equation without the linear term. 1 0 2 (This is the \depressed" equation.) 3. Solve then for y as a square root. (Remember to use both signs of the square root.) a 4. Once this is done, recover x using the fact that x = y − 1 . 2 For example, let's solve 2x2 + 7x − 15 = 0: First, we divide both sides by 2 to create an equation with leading term equal to one: 7 15 x2 + x − = 0: 2 2 a 7 Then replace x by x = y − 1 = y − to obtain: 2 4 169 y2 = 16 Solve for y: 13 13 y = or − 4 4 Then, solving back for x, we have 3 x = or − 5: 2 This method is equivalent to \completing the square" and is the steps taken in developing the much- memorized quadratic formula. For example, if the original equation is our \high school quadratic" ax2 + bx + c = 0 then the first step creates the equation b c x2 + x + = 0: a a b We then write x = y − and obtain, after simplifying, 2a b2 − 4ac y2 − = 0 4a2 so that p b2 − 4ac y = ± 2a and so p b b2 − 4ac x = − ± : 2a 2a 1 The solutions to this quadratic depend heavily on the value of b2 − 4ac. -
Factoring Polynomials
2/13/2013 Chapter 13 § 13.1 Factoring The Greatest Common Polynomials Factor Chapter Sections Factors 13.1 – The Greatest Common Factor Factors (either numbers or polynomials) 13.2 – Factoring Trinomials of the Form x2 + bx + c When an integer is written as a product of 13.3 – Factoring Trinomials of the Form ax 2 + bx + c integers, each of the integers in the product is a factor of the original number. 13.4 – Factoring Trinomials of the Form x2 + bx + c When a polynomial is written as a product of by Grouping polynomials, each of the polynomials in the 13.5 – Factoring Perfect Square Trinomials and product is a factor of the original polynomial. Difference of Two Squares Factoring – writing a polynomial as a product of 13.6 – Solving Quadratic Equations by Factoring polynomials. 13.7 – Quadratic Equations and Problem Solving Martin-Gay, Developmental Mathematics 2 Martin-Gay, Developmental Mathematics 4 1 2/13/2013 Greatest Common Factor Greatest Common Factor Greatest common factor – largest quantity that is a Example factor of all the integers or polynomials involved. Find the GCF of each list of numbers. 1) 6, 8 and 46 6 = 2 · 3 Finding the GCF of a List of Integers or Terms 8 = 2 · 2 · 2 1) Prime factor the numbers. 46 = 2 · 23 2) Identify common prime factors. So the GCF is 2. 3) Take the product of all common prime factors. 2) 144, 256 and 300 144 = 2 · 2 · 2 · 3 · 3 • If there are no common prime factors, GCF is 1. 256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 300 = 2 · 2 · 3 · 5 · 5 So the GCF is 2 · 2 = 4. -
Quadratic Polynomials
Quadratic Polynomials If a>0thenthegraphofax2 is obtained by starting with the graph of x2, and then stretching or shrinking vertically by a. If a<0thenthegraphofax2 is obtained by starting with the graph of x2, then flipping it over the x-axis, and then stretching or shrinking vertically by the positive number a. When a>0wesaythatthegraphof− ax2 “opens up”. When a<0wesay that the graph of ax2 “opens down”. I Cit i-a x-ax~S ~12 *************‘s-aXiS —10.? 148 2 If a, c, d and a = 0, then the graph of a(x + c) 2 + d is obtained by If a, c, d R and a = 0, then the graph of a(x + c)2 + d is obtained by 2 R 6 2 shiftingIf a, c, the d ⇥ graphR and ofaax=⇤ 2 0,horizontally then the graph by c, and of a vertically(x + c) + byd dis. obtained (Remember by shiftingshifting the the⇥ graph graph of of axax⇤ 2 horizontallyhorizontally by by cc,, and and vertically vertically by by dd.. (Remember (Remember thatthatd>d>0meansmovingup,0meansmovingup,d<d<0meansmovingdown,0meansmovingdown,c>c>0meansmoving0meansmoving thatleft,andd>c<0meansmovingup,0meansmovingd<right0meansmovingdown,.) c>0meansmoving leftleft,and,andc<c<0meansmoving0meansmovingrightright.).) 2 If a =0,thegraphofafunctionf(x)=a(x + c) 2+ d is called a parabola. If a =0,thegraphofafunctionf(x)=a(x + c)2 + d is called a parabola. 6 2 TheIf a point=0,thegraphofafunction⇤ ( c, d) 2 is called thefvertex(x)=aof(x the+ c parabola.) + d is called a parabola. The point⇤ ( c, d) R2 is called the vertex of the parabola. -
The Quadratic Formula You May Recall the Quadratic Formula for Roots of Quadratic Polynomials Ax2 + Bx + C
For example, when we take the polynomial f (x) = x2 − 3x − 4, we obtain p 3 ± 9 + 16 2 which gives 4 and −1. Some quick terminology 2 I We say that 4 and −1 are roots of the polynomial x − 3x − 4 or solutions to the polynomial equation x2 − 3x − 4 = 0. 2 I We may factor x − 3x − 4 as (x − 4)(x + 1). 2 I If we denote x − 3x − 4 as f (x), we have f (4) = 0 and f (−1) = 0. The quadratic formula You may recall the quadratic formula for roots of quadratic polynomials ax2 + bx + c. It says that the solutions to this polynomial are p −b ± b2 − 4ac : 2a Some quick terminology 2 I We say that 4 and −1 are roots of the polynomial x − 3x − 4 or solutions to the polynomial equation x2 − 3x − 4 = 0. 2 I We may factor x − 3x − 4 as (x − 4)(x + 1). 2 I If we denote x − 3x − 4 as f (x), we have f (4) = 0 and f (−1) = 0. The quadratic formula You may recall the quadratic formula for roots of quadratic polynomials ax2 + bx + c. It says that the solutions to this polynomial are p −b ± b2 − 4ac : 2a For example, when we take the polynomial f (x) = x2 − 3x − 4, we obtain p 3 ± 9 + 16 2 which gives 4 and −1. 2 I We may factor x − 3x − 4 as (x − 4)(x + 1). 2 I If we denote x − 3x − 4 as f (x), we have f (4) = 0 and f (−1) = 0. -
A Saturation Algorithm for Homogeneous Binomial Ideals
A Saturation Algorithm for Homogeneous Binomial Ideals Deepanjan Kesh and Shashank K Mehta Indian Institute of Technology, Kanpur - 208016, India, fdeepkesh,[email protected] Abstract. Let k[x1; : : : ; xn] be a polynomial ring in n variables, and let I ⊂ k[x1; : : : ; xn] be a homogeneous binomial ideal. We describe a fast 1 algorithm to compute the saturation, I :(x1 ··· xn) . In the special case when I is a toric ideal, we present some preliminary results comparing our algorithm with Project and Lift by Hemmecke and Malkin. 1 Introduction 1.1 Problem Description Let k[x1; : : : ; xn] be a polynomial ring in n variables over the field k, and let I ⊂ k[x1; : : : ; xn] be an ideal. Ideals are said to be homogeneous, if they have a basis consisting of homogeneous polynomials. Binomials in this ring are defined as polynomials with at most two terms [5]. Thus, a binomial is a polynomial of the form cxα+dxβ, where c; d are arbitrary coefficients. Pure difference binomials are special cases of binomials of the form xα − xβ. Ideals with a binomial basis are called binomial ideals. Toric ideals, the kernel of a specific kind of polynomial ring homomorphisms, are examples of pure difference binomial ideals. Saturation of an ideal, I, by a polynomial f, denoted by I : f, is defined as the ideal I : f = hf g 2 k[x1; : : : ; xn]: f · g 2 I gi: Similarly, I : f 1 is defined as 1 a I : f = hf g 2 k[x1; : : : ; xn]: 9a 2 N; f · g 2 I gi: 1 We describe a fast algorithm to compute the saturation, I :(x1 ··· xn) , of a homogeneous binomial ideal I. -
Nature of the Discriminant
Name: ___________________________ Date: ___________ Class Period: _____ Nature of the Discriminant Quadratic − b b 2 − 4ac x = b2 − 4ac Discriminant Formula 2a The discriminant predicts the “nature of the roots of a quadratic equation given that a, b, and c are rational numbers. It tells you the number of real roots/x-intercepts associated with a quadratic function. Value of the Example showing nature of roots of Graph indicating x-intercepts Discriminant b2 – 4ac ax2 + bx + c = 0 for y = ax2 + bx + c POSITIVE Not a perfect x2 – 2x – 7 = 0 2 b – 4ac > 0 square − (−2) (−2)2 − 4(1)(−7) x = 2(1) 2 32 2 4 2 x = = = 1 2 2 2 2 Discriminant: 32 There are two real roots. These roots are irrational. There are two x-intercepts. Perfect square x2 + 6x + 5 = 0 − 6 62 − 4(1)(5) x = 2(1) − 6 16 − 6 4 x = = = −1,−5 2 2 Discriminant: 16 There are two real roots. These roots are rational. There are two x-intercepts. ZERO b2 – 4ac = 0 x2 – 2x + 1 = 0 − (−2) (−2)2 − 4(1)(1) x = 2(1) 2 0 2 x = = = 1 2 2 Discriminant: 0 There is one real root (with a multiplicity of 2). This root is rational. There is one x-intercept. NEGATIVE b2 – 4ac < 0 x2 – 3x + 10 = 0 − (−3) (−3)2 − 4(1)(10) x = 2(1) 3 − 31 3 31 x = = i 2 2 2 Discriminant: -31 There are two complex/imaginary roots. There are no x-intercepts. Quadratic Formula and Discriminant Practice 1. -
CYCLIC RESULTANTS 1. Introduction the M-Th Cyclic Resultant of A
CYCLIC RESULTANTS CHRISTOPHER J. HILLAR Abstract. We characterize polynomials having the same set of nonzero cyclic resultants. Generically, for a polynomial f of degree d, there are exactly 2d−1 distinct degree d polynomials with the same set of cyclic resultants as f. How- ever, in the generic monic case, degree d polynomials are uniquely determined by their cyclic resultants. Moreover, two reciprocal (\palindromic") polyno- mials giving rise to the same set of nonzero cyclic resultants are equal. In the process, we also prove a unique factorization result in semigroup algebras involving products of binomials. Finally, we discuss how our results yield algo- rithms for explicit reconstruction of polynomials from their cyclic resultants. 1. Introduction The m-th cyclic resultant of a univariate polynomial f 2 C[x] is m rm = Res(f; x − 1): We are primarily interested here in the fibers of the map r : C[x] ! CN given by 1 f 7! (rm)m=0. In particular, what are the conditions for two polynomials to give rise to the same set of cyclic resultants? For technical reasons, we will only consider polynomials f that do not have a root of unity as a zero. With this restriction, a polynomial will map to a set of all nonzero cyclic resultants. Our main result gives a complete answer to this question. Theorem 1.1. Let f and g be polynomials in C[x]. Then, f and g generate the same sequence of nonzero cyclic resultants if and only if there exist u; v 2 C[x] with u(0) 6= 0 and nonnegative integers l1; l2 such that deg(u) ≡ l2 − l1 (mod 2), and f(x) = (−1)l2−l1 xl1 v(x)u(x−1)xdeg(u) g(x) = xl2 v(x)u(x): Remark 1.2. -
Formula Sheet 1 Factoring Formulas 2 Exponentiation Rules
Formula Sheet 1 Factoring Formulas For any real numbers a and b, (a + b)2 = a2 + 2ab + b2 Square of a Sum (a − b)2 = a2 − 2ab + b2 Square of a Difference a2 − b2 = (a − b)(a + b) Difference of Squares a3 − b3 = (a − b)(a2 + ab + b2) Difference of Cubes a3 + b3 = (a + b)(a2 − ab + b2) Sum of Cubes 2 Exponentiation Rules p r For any real numbers a and b, and any rational numbers and , q s ap=qar=s = ap=q+r=s Product Rule ps+qr = a qs ap=q = ap=q−r=s Quotient Rule ar=s ps−qr = a qs (ap=q)r=s = apr=qs Power of a Power Rule (ab)p=q = ap=qbp=q Power of a Product Rule ap=q ap=q = Power of a Quotient Rule b bp=q a0 = 1 Zero Exponent 1 a−p=q = Negative Exponents ap=q 1 = ap=q Negative Exponents a−p=q Remember, there are different notations: p q a = a1=q p q ap = ap=q = (a1=q)p 1 3 Quadratic Formula Finally, the quadratic formula: if a, b and c are real numbers, then the quadratic polynomial equation ax2 + bx + c = 0 (3.1) has (either one or two) solutions p −b ± b2 − 4ac x = (3.2) 2a 4 Points and Lines Given two points in the plane, P = (x1; y1);Q = (x2; y2) you can obtain the following information: p 2 2 1. The distance between them, d(P; Q) = (x2 − x1) + (y2 − y1) . x + x y + y 2.