DOCSLIB.ORG

Factoring Polynomials

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Factoring Polynomials FactoringFactoringFactoring Polynomials PolynomialsPolynomials AnyAnyAny natural naturalnatural number numbernumber that thatthat is is greater greateris greater than thanthan 1 1can can1 becan be factoredbe factoredfactored into intointo a aproduct producta product ofof primeof primeprime numbers. numbers.numbers. For For exampleFor exampleexample 20 20 =20 = (2)(2)(5) (2)(2)(5)= (2)(2)(5) and and 30and 30 =30 = (2)(3)(5). (2)(3)(5).— (2)(3)(5). 20 30 /\ /\ Lf. .5 2. IS /\ /\ 22 35 InIn thisIn thisthis chapter chapterchapter we’ll we’llwe’ll learn learnlearn an an analogousan analogousanalogous way wayway to to factorto factorfactor polynomials. polynomials.polynomials. FundamentalFundamentalFundamental Theorem TheoremTheorem of of Algebraof AlgebraAlgebra AAmonicAmonicmonicpolynomialpolynomialpolynomial is is a apolynomialis polynomiala polynomial whose whosewhose leading leadingleading coe coefficoefficientfficientcient equals equalsequals 1. 1. So1. SoSo 4x4 23x3 +5x 7ismonic,andx 2ismonic,but32x2 4isnotmonic. x 2x— +52x3x+ 5x7ismonic,and— 7 is monic, andx x2ismonic,but3— 2 is monic, butx 3x24isnotmonic.— 4 is not monic. −− −− −− −− TheCarlCarl following FriedrichFriedrich result GaussGauss tells waswas us the howthe boy toboy who factorwho discovered polynomials.discovered a reallya really It essentially quickquick wayway tells toto seesee that 1 + 2 + 3 + +100=5050. us whatthat the1+2+ “prime3+··· polynomials”•••+ 100 = 5050. are: InIn 1799,1799, a grown-upa grown-up GaussGauss provedproved thethe followingfollowing theorem:theorem: Any polynomial is the product of a real number, Any polynomial is the product of a real number, and aAny collectionpolynomial of monicis the quadraticproduct polynomialsof a real number, that and a collection of monic quadratic polynomials that do notand havea collection roots, andof monic of monicquadratic linear polynomials.polynomials that dodo notnot havehave roots,roots, andand ofof monicmonic linearlinear polynomials.polynomials. ThisThis result result is is called called the theFundamentalFundamental Theorem Theorem of of Algebra Algebra.. It It is is one one of of the mostThis importantresult is called resultsthe inFundamental all of mathematics,Theorem thoughof Algebra. from theIt is formone of thethe mostmost importantimportant resultsresults inin allall ofof mathematics,mathematics, thoughthough fromfrom thethe formform it’sit’s written written in in above, above, it’s it’s probably probably di diffifficultcult to to immediately immediately understand understand its its importance.it’s written in above, it’s probably difficult to immediately understand its importance.importance. TheThe explanation explanation for for why why this this theorem theorem is is true true is is somewhat somewhat di diffifficult,cult, and and it it is beyondThe theexplanation scope offor thiswhy course.this We’lltheorem haveis totrue acceptis somewhat it on faith.difficult, and it isis beyondbeyond thethe scopescope ofof thisthis course.course. We’llWe’ll havehave toto acceptaccept it onit on faith.faith. Examples.Examples. Examples.2 4x4x2 1212x x+8canbefactoredintoaproductofanumber,4,and+8canbefactoredintoaproductofanumber,4,and • • 4x2− — 12x +8 can be factored into a product of a number, 4, and •twotwo monic− monic linear linear polynomials, polynomials,x x 1and1andx x 2.2. That That is, is, two monic linear polynomials,−x — 1 and− x — 2. That is, 4x2 2 12x +8=4(x 1)(x 2). − − 4x4~2_− 1212x+8=4(x—1)(x—2).x +8=4(x− 1)(x− 2). − − 157−130 130 x5 +2x4 7x3 +14x2 10x +20canbefactoredintoaproduct • −of a number,− 1, a monic− linear polynomial, x 2, and two monic quadratic polynomials− that don’t have roots, x2 −+2 and x2 +5. That is x5 +2x4 7x3 +14x2 10x +20= (x 2)(x2 +2)(x2 +5). (We− can check− the discriminants− of x2 +2and− −x2 +5toseethat these two quadratics don’t have roots.) 2x4 2x3 +14x2 6x +24=2(x2 +3)(x2 x + 4). Again, x2 +3 • and−x2 x +4donothaveroots.− − − Notice that in each of the above examples, the real number that appears in the product of polynomials – 4 in the first example, 1inthesecond, and 2 in the third – is the same as the leading coefficient− for the original polynomial. This always happens, so the Fundamental Theorem of Algebra can be more precisely stated as follows: n n 1 If p(x)=anx + an 1x − + + a0, − ··· then p(x)istheproductoftherealnumberan, and a collection of monic quadratic polynomials that do not have roots, and of monic linear polynomials. Completely factored Apolynomialiscompletely factored if it is written as a product of a real number (which will be the same number as the leading coefficient of the polynomial), and a collection of monic quadratic polynomials that do not have roots, and of monic linear polynomials. Looking at the examples above, 4(x 1)(x 2) and (x 2)(x2 +2)(x2 +5) and 2(x2 +3)(x2 x +4)arecompletelyfactored.− − − − − One reason it’s nice to completely factor a polynomial is because if you do, then it’s easy to read o↵ what the roots of the polynomial are. Example. Suppose p(x)= 2x5 +10x4 +2x3 38x2 +4x 48. Written in this form, its difficult to see− what the roots of−p(x)are.Butafterbeing− completely factored, p(x)= 2(x +2)(x 3)(x 4)(x2 +1). Therootsof − 158 − − this polynomial can be read from the monic linear factors. They are 2, 3, and 4. − (Notice that p(x)= 2(x +2)(x 3)(x 4)(x2 +1)iscompletelyfactored because x2 +1hasnoroots.)− − − ************* Factoring linears To completely factor a linear polynomial, just factor out its leading coeffi- cient: b ax + b = a x + a ⇣ ⌘ For example, to completely factor 2x +6,writeitastheproduct2(x +3). Factoring quadratics What a completely factored quadratic polynomial looks like will depend on how many roots it has. 0Roots.If the quadratic polynomial ax2 + bx + c has 0 roots, then it can be completely factored by factoring out the leading coefficient: b c ax2 + bx + c = a x2 + x + a a 2 2 b ⇣c ⌘ (The graphs of ax +bx+c and x + ax+ a di↵er by a vertical stretch or shrink that depends on a. A vertical stretch or shrink of a graph won’t change the 2 b c 2 number of x-intercepts, so x + ax + a won’t have any roots since ax + bx + c 2 b c doesn’t have any roots. Thus, x + ax + a is completely factored.) Example. The discriminant of 4x2 2x+2 equals ( 2)2 4(4)(2) = 4 32 = 28, a negative number. Therefore,− 4x2 2x +2− hasnoroots,anditis− − completely− factored as 4(x2 1x + 1). − − 2 2 2Roots. If the quadratic polynomial ax2 + bx + c has 2 roots, we can name them ↵1 and ↵2. Roots give linear factors, so we know that (x ↵1) 159 − 2 and (x ↵2)arefactorsofax + bx + c. That means that there is some polynomial− q(x)suchthat ax2 + bx + c = q(x)(x ↵ )(x ↵ ) − 1 − 2 The degree of ax2 + bx + c equals 2. Because the sum of the degrees of the factors equals the degree of the product, we know that the degree of q(x)plus the degree of (x ↵1)plusthedegreeof(x ↵2) equals 2. In other words, the degree of q(x−)plus1plus1equals2. − Zero is the only number that you can add to 1 + 1 to get 2, so q(x) must have degree 0, which means that q(x)isjustaconstantnumber. Because the leading term of ax2 + bx + c – namely ax2 –istheproductof the leading terms of q(x), (x ↵1), and (x ↵2)–namelythenumberq(x), x,andx –itmustbethatq(−x)=a. Therefore,− ax2 + bx + c = a(x ↵ )(x ↵ ) − 1 − 2 Example. The discriminant of 2x2 +4x 2equals42 4(2)( 2) = 16+16 = 32, a positive number, so there are two roots.− − − We can use the quadratic formula to find the two roots, but before we do, it’s best to simplify the square root of the discriminant: p32 = (4)(4)(2) = 4p2. p Now we use the quadratic formula to find that the roots are 4+4p2 4+4p2 − = − = 1+p2 2(2) 4 − and 4 4p2 4 4p2 − − = − − = 1 p2 2(2) 4 − − Therefore, 2x2 +4x 2iscompletelyfactoredas − 2 x ( 1+p2) x ( 1 p2) =2(x +1 p2)(x +1+p2) − − − − − − ⇣ ⌘⇣ ⌘ 2 1Root.If ax + bx + c has exactly 1 root (let’s call it ↵1)then(x ↵1)is afactorofax2 + bx + c. Hence, − ax2 + bx + c = g(x)(x ↵ ) − 1 for some polynomial g(x). 160 Because the degree of a product is the sum of the degrees of the factors, g(x)mustbeadegree1polynomial,anditcanbecompletelyfactoredinto something of the form λ(x β)whereλ, β R. Therefore, − 2 ax2 + bx + c = λ(x β)(x ↵ ) − − 1 2 Notice that β is a root of λ(x β)(x ↵1), so β is a root of ax + bx + c since they are the same polynomial.− But− we know that ax2 + bx + c has only one root, namely ↵1, so β must equal ↵1. That means that ax2 + bx + c = λ(x ↵ )(x ↵ ) − 1 − 1 2 2 The leading term of ax +bx+c is ax . The leading term of λ(x ↵1)(x ↵1) 2 2 − − is λx . Since ax + bx + c equals λ(x ↵1)(x ↵1), they must have the same leading term. Therefore, ax2 = λx2.− Hence, a−= λ. Replace λ with a in the equation above, and we are left with ax2 + bx + c = a(x ↵ )(x ↵ ) − 1 − 1 Example. The discriminant of 3x2 6x+3 equals ( 6)2 4(3)(3) = 36 36 = 0, so there is exactly one root. We find− the root using− the− quadratic formula:− ( 6) + p0 6 − − = =1 2(3) 6 Therefore, 3x2 6x +3iscompletelyfactoredas3(x 1)(x 1). − − − Summary. The following chart summarizes the discussion above. roots of ax2 + bx + c completely factored form of ax2 + bx + c 2 b c no roots a(x + ax + a) 2roots:↵ and ↵ a(x ↵ )(x ↵ ) 1 2 − 1 − 2 1root:↵ a(x ↵ )(x ↵ ) 1 − 1 − 1 161 ************* Factors in Z Recall that the factors of an integer n are all of the integers k such that n = mk for some third integer m.
Load more

Recommended publications
  • 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.
    [Show full text]
  • Algorithmic Factorization of Polynomials Over Number Fields
    Rose-Hulman Institute of Technology Rose-Hulman Scholar Mathematical Sciences Technical Reports (MSTR) Mathematics 5-18-2017 Algorithmic Factorization of Polynomials over Number Fields Christian Schulz Rose-Hulman Institute of Technology Follow this and additional works at: https://scholar.rose-hulman.edu/math_mstr Part of the Number Theory Commons, and the Theory and Algorithms Commons Recommended Citation Schulz, Christian, "Algorithmic Factorization of Polynomials over Number Fields" (2017). Mathematical Sciences Technical Reports (MSTR). 163. https://scholar.rose-hulman.edu/math_mstr/163 This Dissertation is brought to you for free and open access by the Mathematics at Rose-Hulman Scholar. It has been accepted for inclusion in Mathematical Sciences Technical Reports (MSTR) by an authorized administrator of Rose-Hulman Scholar. For more information, please contact [email protected]. Algorithmic Factorization of Polynomials over Number Fields Christian Schulz May 18, 2017 Abstract The problem of exact polynomial factorization, in other words expressing a poly- nomial as a product of irreducible polynomials over some field, has applications in algebraic number theory. Although some algorithms for factorization over algebraic number fields are known, few are taught such general algorithms, as their use is mainly as part of the code of various computer algebra systems. This thesis provides a summary of one such algorithm, which the author has also fully implemented at https://github.com/Whirligig231/number-field-factorization, along with an analysis of the runtime of this algorithm. Let k be the product of the degrees of the adjoined elements used to form the algebraic number field in question, let s be the sum of the squares of these degrees, and let d be the degree of the polynomial to be factored; then the runtime of this algorithm is found to be O(d4sk2 + 2dd3).
    [Show full text]
  • January 10, 2010 CHAPTER SIX IRREDUCIBILITY and FACTORIZATION §1. BASIC DIVISIBILITY THEORY the Set of Polynomials Over a Field
    January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION §1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics. A polynomials is irreducible iff it cannot be factored as a product of polynomials of strictly lower degree. Otherwise, the polynomial is reducible. Every linear polynomial is irreducible, and, when F = C, these are the only ones. When F = R, then the only other irreducibles are quadratics with negative discriminants. However, when F = Q, there are irreducible polynomials of arbitrary degree. As for the integers, we have a division algorithm, which in this case takes the form that, if f(x) and g(x) are two polynomials, then there is a quotient q(x) and a remainder r(x) whose degree is less than that of g(x) for which f(x) = q(x)g(x) + r(x) . The greatest common divisor of two polynomials f(x) and g(x) is a polynomial of maximum degree that divides both f(x) and g(x). It is determined up to multiplication by a constant, and every common divisor divides the greatest common divisor. These correspond to similar results for the integers and can be established in the same way. One can determine a greatest common divisor by the Euclidean algorithm, and by going through the equations in the algorithm backward arrive at the result that there are polynomials u(x) and v(x) for which gcd (f(x), g(x)) = u(x)f(x) + v(x)g(x) .
    [Show full text]
  • 2.4 Algebra of Polynomials ([1], P.136-142) in This Section We Will Give a Brief Introduction to the Algebraic Properties of the Polynomial Algebra C[T]
    2.4 Algebra of polynomials ([1], p.136-142) In this section we will give a brief introduction to the algebraic properties of the polynomial algebra C[t]. In particular, we will see that C[t] admits many similarities to the algebraic properties of the set of integers Z. Remark 2.4.1. Let us first recall some of the algebraic properties of the set of integers Z. - division algorithm: given two integers w, z 2 Z, with jwj ≤ jzj, there exist a, r 2 Z, with 0 ≤ r < jwj such that z = aw + r. Moreover, the `long division' process allows us to determine a, r. Here r is the `remainder'. - prime factorisation: for any z 2 Z we can write a1 a2 as z = ±p1 p2 ··· ps , where pi are prime numbers. Moreover, this expression is essentially unique - it is unique up to ordering of the primes appearing. - Euclidean algorithm: given integers w, z 2 Z there exists a, b 2 Z such that aw + bz = gcd(w, z), where gcd(w, z) is the `greatest common divisor' of w and z. In particular, if w, z share no common prime factors then we can write aw + bz = 1. The Euclidean algorithm is a process by which we can determine a, b. We will now introduce the polynomial algebra in one variable. This is simply the set of all polynomials with complex coefficients and where we make explicit the C-vector space structure and the multiplicative structure that this set naturally exhibits. Definition 2.4.2. - The C-algebra of polynomials in one variable, is the quadruple (C[t], α, σ, µ)43 where (C[t], α, σ) is the C-vector space of polynomials in t with C-coefficients defined in Example 1.2.6, and µ : C[t] × C[t] ! C[t];(f , g) 7! µ(f , g), is the `multiplication' function.
    [Show full text]
  • Elements of Chapter 9: Nonlinear Systems Examples
    Elements of Chapter 9: Nonlinear Systems To solve x0 = Ax, we use the ansatz that x(t) = eλtv. We found that λ is an eigenvalue of A, and v an associated eigenvector. We can also summarize the geometric behavior of the solutions by looking at a plot- However, there is an easier way to classify the stability of the origin (as an equilibrium), To find the eigenvalues, we compute the characteristic equation: p Tr(A) ± ∆ λ2 − Tr(A)λ + det(A) = 0 λ = 2 which depends on the discriminant ∆: • ∆ > 0: Real λ1; λ2. • ∆ < 0: Complex λ = a + ib • ∆ = 0: One eigenvalue. The type of solution depends on ∆, and in particular, where ∆ = 0: ∆ = 0 ) 0 = (Tr(A))2 − 4det(A) This is a parabola in the (Tr(A); det(A)) coordinate system, inside the parabola is where ∆ < 0 (complex roots), and outside the parabola is where ∆ > 0. We can then locate the position of our particular trace and determinant using the Poincar´eDiagram and it will tell us what the stability will be. Examples Given the system where x0 = Ax for each matrix A below, classify the origin using the Poincar´eDiagram: 1 −4 1. 4 −7 SOLUTION: Compute the trace, determinant and discriminant: Tr(A) = −6 Det(A) = −7 + 16 = 9 ∆ = 36 − 4 · 9 = 0 Therefore, we have a \degenerate sink" at the origin. 1 2 2. −5 −1 SOLUTION: Compute the trace, determinant and discriminant: Tr(A) = 0 Det(A) = −1 + 10 = 9 ∆ = 02 − 4 · 9 = −36 The origin is a center. 1 3. Given the system x0 = Ax where the matrix A depends on α, describe how the equilibrium solution changes depending on α (use the Poincar´e Diagram): 2 −5 (a) α −2 SOLUTION: The trace is 0, so that puts us on the \det(A)" axis.
    [Show full text]
  • Factoring Polynomials
    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 expression) equal to zero in order to solve for a variable: Example: To solve ,; , The flowchart below illustrates a sequence of steps for factoring polynomials. First, always factor out the Greatest Common Factor (GCF), if one exists. Is the equation a Binomial or a Trinomial? 1 Prime polynomials cannot be factored Yes No using integers alone. The Sum of Squares and the Four or more quadratic factors Special Cases? terms of the Sum and Difference of Binomial Trinomial Squares are (two terms) (three terms) Factor by Grouping: always Prime. 1. Group the terms with common factors and factor 1. Difference of Squares: out the GCF from each Perfe ct Square grouping. 1 , 3 Trinomial: 2. Sum of Squares: 1. 2. Continue factoring—by looking for Special Cases, 1 , 2 2. 3. Difference of Cubes: Grouping, etc.—until the 3 equation is in simplest form FYI: A Sum of Squares can 1 , 2 (or all factors are Prime). 4. Sum of Cubes: be factored using imaginary numbers if you rewrite it as a Difference of Squares: — 2 Use S.O.A.P to No Special √1 √1 Cases remember the signs for the factors of the 4 Completing the Square and the Quadratic Formula Sum and Difference Choose: of Cubes: are primarily methods for solving equations rather 1. Factor by Grouping than simply factoring expressions.
    [Show full text]
  • Calculus Terminology
    AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential
    [Show full text]
  • Finding Equations of Polynomial Functions with Given Zeros
    Finding Equations of Polynomial Functions with Given Zeros 푛 푛−1 2 Polynomials are functions of general form 푃(푥) = 푎푛푥 + 푎푛−1 푥 + ⋯ + 푎2푥 + 푎1푥 + 푎0 ′ (푛 ∈ 푤ℎ표푙푒 # 푠) Polynomials can also be written in factored form 푃(푥) = 푎(푥 − 푧1)(푥 − 푧2) … (푥 − 푧푖) (푎 ∈ ℝ) Given a list of “zeros”, it is possible to find a polynomial function that has these specific zeros. In fact, there are multiple polynomials that will work. In order to determine an exact polynomial, the “zeros” and a point on the polynomial must be provided. Examples: Practice finding polynomial equations in general form with the given zeros. Find an* equation of a polynomial with the Find the equation of a polynomial with the following two zeros: 푥 = −2, 푥 = 4 following zeroes: 푥 = 0, −√2, √2 that goes through the point (−2, 1). Denote the given zeros as 푧1 푎푛푑 푧2 Denote the given zeros as 푧1, 푧2푎푛푑 푧3 Step 1: Start with the factored form of a polynomial. Step 1: Start with the factored form of a polynomial. 푃(푥) = 푎(푥 − 푧1)(푥 − 푧2) 푃(푥) = 푎(푥 − 푧1)(푥 − 푧2)(푥 − 푧3) Step 2: Insert the given zeros and simplify. Step 2: Insert the given zeros and simplify. 푃(푥) = 푎(푥 − (−2))(푥 − 4) 푃(푥) = 푎(푥 − 0)(푥 − (−√2))(푥 − √2) 푃(푥) = 푎(푥 + 2)(푥 − 4) 푃(푥) = 푎푥(푥 + √2)(푥 − √2) Step 3: Multiply the factored terms together. Step 3: Multiply the factored terms together 푃(푥) = 푎(푥2 − 2푥 − 8) 푃(푥) = 푎(푥3 − 2푥) Step 4: The answer can be left with the generic “푎”, or a value for “푎”can be chosen, Step 4: Insert the given point (−2, 1) to inserted, and distributed.
    [Show full text]
  • Polynomials and Quadratics
    Higher hsn .uk.net Mathematics UNIT 2 OUTCOME 1 Polynomials and Quadratics Contents Polynomials and Quadratics 64 1 Quadratics 64 2 The Discriminant 66 3 Completing the Square 67 4 Sketching Parabolas 70 5 Determining the Equation of a Parabola 72 6 Solving Quadratic Inequalities 74 7 Intersections of Lines and Parabolas 76 8 Polynomials 77 9 Synthetic Division 78 10 Finding Unknown Coefficients 82 11 Finding Intersections of Curves 84 12 Determining the Equation of a Curve 86 13 Approximating Roots 88 HSN22100 This document was produced specially for the HSN.uk.net website, and we require that any copies or derivative works attribute the work to Higher Still Notes. For more details about the copyright on these notes, please see http://creativecommons.org/licenses/by-nc-sa/2.5/scotland/ Higher Mathematics Unit 2 – Polynomials and Quadratics OUTCOME 1 Polynomials and Quadratics 1 Quadratics A quadratic has the form ax2 + bx + c where a, b, and c are any real numbers, provided a ≠ 0 . You should already be familiar with the following. The graph of a quadratic is called a parabola . There are two possible shapes: concave up (if a > 0 ) concave down (if a < 0 ) This has a minimum This has a maximum turning point turning point To find the roots (i.e. solutions) of the quadratic equation ax2 + bx + c = 0, we can use: factorisation; completing the square (see Section 3); −b ± b2 − 4 ac the quadratic formula: x = (this is not given in the exam). 2a EXAMPLES 1. Find the roots of x2 −2 x − 3 = 0 .
    [Show full text]
  • 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.
    [Show full text]
  • 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.
    [Show full text]
  • The Determinant and the Discriminant
    CHAPTER 2 The determinant and the discriminant In this chapter we discuss two indefinite quadratic forms: the determi- nant quadratic form det(a, b, c, d)=ad bc, − and the discriminant disc(a, b, c)=b2 4ac. − We will be interested in the integral representations of a given integer n by either of these, that is the set of solutions of the equations 4 ad bc = n, (a, b, c, d) Z − 2 and 2 3 b ac = n, (a, b, c) Z . − 2 For q either of these forms, we denote by Rq(n) the set of all such represen- tations. Consider the three basic questions of the previous chapter: (1) When is Rq(n) non-empty ? (2) If non-empty, how large Rq(n)is? (3) How is the set Rq(n) distributed as n varies ? In a suitable sense, a good portion of the answers to these question will be similar to the four and three square quadratic forms; but there will be major di↵erences coming from the fact that – det and disc are indefinite quadratic forms (have signature (2, 2) and (2, 1) over the reals), – det and disc admit isotropic vectors: there exist x Q4 (resp. Q3) such that det(x)=0(resp.disc(x) = 0). 2 1. Existence and number of representations by the determinant As the name suggest, determining Rdet(n) is equivalent to determining the integral 2 2 matrices of determinant n: ⇥ (n) ab Rdet(n) M (Z)= g = M2(Z), det(g)=n . ' 2 { cd 2 } ✓ ◆ n 0 Observe that the diagonal matrix a = has determinant n, and any 01 ✓ ◆ other matrix in the orbit SL2(Z).a is integral and has the same determinant.
    [Show full text]