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Chapter 4 Products and Factors of Polynomials Section 4-1 Polynomials Constant A number -2, 3/5, 0 Monomial A constant, a variable, or a product of a constant and one or more variables -7 5u (1/3)m2 -s2t3 x Coefficient The constant (or numerical) factor in a monomial 3m2 coefficient = 3 u coefficient = 1 - s2t3 coefficient = -1 Degree of a Variable The number of times the variable occurs as a factor in the monomial For Example – 6xy3 What is the degree of x? y? Degree of a monomial The sum of the degrees of the variables in the monomial. A nonzero constant has degree 0. The constant 0 has no degree. Examples 6xy3 -s2t3 u -7 degree = 4 degree = 5 degree = 1 degree = 0 Similar Monomials Monomials that are identical or that differ only in their coefficients Also called like terms Are - s2t3 and 2s2t3 similar? Polynomial A monomial or a sum of monomials. The monomials in a polynomial are called the terms of the polynomial. Examples x2 + (-4)x + 5 x2 – 4x + 5 What are the terms? 2 x , -4x, and 5 Simplified Polynomial A polynomial in which no two terms are similar. The terms are usually arranged in order of decreasing degree of one of the variables Are they Simplified? 2x3 – 5 + 4x + x3 3x3 + 4x – 5 4x2 – x + 3x4 – 5 + x2 Degree of a Polynomial The greatest of the degrees of its terms after it has been simplified What is the degree? x4 + 3x 2x3 + 3x – 7 2 x – 5x + 1 7x + 1 x4 – 2x2y3 + 6y -11 Adding Polynomials To add two or more polynomials, write their sum and then simplify by combining like terms Add the following- (x2 + 4x – 3) + (x3 – 2x2 + 6x – 7) Subtracting Polynomials To subtract one polynomial from another, add the opposite of each term of the polynomial you’re subtracting 3 (x – 2 5x + 2x – 5) – (2x2 – 3x + 5) Section 4-2 Using Laws of Exponents Laws of Exponents Let a and b be real numbers and m and n be positive integers in all the following laws Law 1 am · an = am+n x2 · x 4 = x 6 3 y 5 y · =? m · m4 = ? Law 2 (ab)m = ambm (xy)3 = x3y3 2 (3st) =? (xy)5 = ? Law 3 (am)n = amn (x3)2 = x6 2 3 4 (x y ) =? (2mn2)3 = ? Using Distributive Law Distribute the variable using exponent laws 3t2(t3 – 2t2 + t – 4) = ? – 2x2(x3 – 3x + 4) = ? Section 4-3 Multiplying Polynomials Binomial A polynomial that has two terms 2x + 3 3xy – 14 4x – 3y 613 + 39z Trinomial A polynomial that has three terms 2x2 – 3x + 1 14 + 32z – 3x mn – m2 + n2 Multiplying binomials When multiplying two binomials both terms of each binomial must be multiplied by the other two terms Using the F.O.I.L method helps you remember the steps when multiplying F.O.I.L. Method F – multiply First terms O – multiply Outer terms I – multiply Inner terms L – multiply Last terms Add all terms to get product Binomial A polynomial that has two terms 2x + 3 3xy – 14 4x – 3y 613 + 39z Trinomial A polynomial that has three terms 2x2 – 3x + 1 14 + 32z – 3x mn – m2 + n2 Multiplying binomials When multiplying two binomials both terms of each binomial must be multiplied by the other two terms Using the F.O.I.L method helps you remember the steps when multiplying F.O.I.L. Method F – multiply First terms O – multiply Outer terms I – multiply Inner terms L – multiply Last terms Add all terms to get product Example - (2a – b)(3a + 5b) F – 2a · 3a O – 2a · 5b I – (-b) ▪ 3a L - (-b) ▪ 5b 2 2 6a + 10ab – 3ab – 5b 6a2 + 7ab – 5b2 Example – (x + 6)(x +4) F – x ▪ x O – x ▪ 4 I – 6 ▪ x L – 6 ▪ 4 x2 + 4x + 6x + 24 x2 + 10x + 24 Special Products (a + b)2 = a2 + 2ab + b2 (a - b)2 = a2 - 2ab + b2 (a + b)(a – b) = 2 a - 2 b Section 4-4 Using Prime Factorization Factor A number over a set of numbers, you write it as a product of numbers chosen from that set The set is called a factor set Example The number 15 can be factored in the following ways (1)(15) (5)(3) (-1)(-15) (-3)(-5) Prime Number An integer greater than 1 whose only positive integral factors are itself and 1 Prime Factorization If the factor set is restricted to the set of primes To find it you write the integer as a product of primes Example 350 = 2 x 175 = 2 x 5 x 35 =2x5x5x7 So the prime factorization of 350 is 2 x 52 x 7 Greatest Common Factor The greatest integer that is a factor of each number. To find the GCF, take the least power of each common prime factor. Example What is the GCF of 100, 120, and 90? 10 Least Common Multiple The least positive integer having each as a factor To find the LCM, take the greatest power of each common prime factor. Example What is the LCM of 100, 120, and 90? 1800 Summary GCF – take the least power of each common prime factor. LCM – take the greatest power of each prime factor Section 4-5 Factoring Polynomials Factor To factor a polynomial you express it as a product of other polynomials We will factor using polynomials with integral coefficients Greatest Monomial Factor The GCF of the terms What is the GCF of 4 3 2 2x – 4x + 8x ? 2 2x Now factor: 2x4 – 4x3 + 8x2 Factor out 2x2 2 2 2x (x – 2x + 4) Perfect Square Trinomials The polynomials in the form of a2 + 2ab + b2 2 2 and a – 2ab + b are the result of squaring a + b and a – b respectively Difference of Squares The polynomial a2 – b2 is the product of a + b and a-b Factor Each Polynomial z2 + 6z + 9 4s2 – 4 st + t2 2 2 25x – 16a Factored Form: (z + 3)2 (2s – t)2 (5x + 4a)(5x – 4a) Sum and Difference of Cubes a3 + b 3 = (a + b)(a2 - ab + b2) 3 a 3 b – = (a – b)(a2 + ab + b2) Factor Each Polynomial y3 - 1 8u3 + v3 Factor by Grouping Factor each polynomial by grouping terms that have a common factor Then factor out the common factor and write the polynomial as a product of two factors Factor each Polynomial 3xy - 4 - 6x + 2y xy + 3y + 2x + 6 Section 4-6 Factoring Quadratic Polynomials Quadratic Polynomials Polynomials of the 2 form ax + bx + c Also called seconddegree polynomials Terms - quadratic term bx - linear term c - constant term 2 ax Quadratic Trinomial A quadratic polynomial for which a, b, and c are all nonzero integers Factoring Quadratic Trinomials ax2 + bx + c can be factored into the form (px + q)(rx + s) where p, q, r, and s are integers Factors a = pr b = ps + qr c = qs Factor the Polynomial x2 + 2x - 15 a = 1, so pr = 1 c = -15, so qs = -15 b = 2, so ps + qr = 2 Factor the Polynomials 2 15t - 16t + 4 3 - 2z 2 x 2 z + 4x - 3 Irreducible If a polynomial has more than one term and cannot be expressed as a product of polynomials of lower degree taken from a given factor set, it is irreducible 2 x + 4x - 3 is irreducible Factored Completely A polynomial is factored completely when it is written as a product of factors and each factor is either a monomial, a prime polynomial, or a power of a prime polynomial Greatest Common Factor The GCF of two or more polynomials is the common factor having the greatest degree and the greatest constant factor Least Common Multiple The LCM of two or more polynomials is the common multiple having the least degree and least positive constant factor Section 4-7 Solving Polynomial Equations Polynomial Equation An equation that is equivalent to one with a polynomial as one side and 0 as the other 2 x = 5x + 24 Root The value of a variable that satisfies the equation Also called the solution Solving a polynomial Equation You can factor the polynomial to solve the equation Steps to Solving a polynomial Equation Write the equation with 0 as one side Factor the other side of the equation Solve the equation obtained by setting each factor equal to 0 Example 1 Solve (x – 5)(x + 2) = 0 Step 1: already = 0 Step 2: already factored Step 3: set each factor = 0 x-5=0 x+2=0 x=5 x = -2 Example 2 Solve x2 = x + 30 1: x2 - x – 30 = 0 2: (x – 6)(x + 5) = 0 3: x – 6 = 0 x+5=0 x=6 x = -5 The solution set is {6, -5} Zeros A number r is a zero of a function f if f(r) = 0 You can find zeros using the same method that is used to solve polynomial equations Example Find the zeros of f(x) = (x – 4)3 – 4(3x – 16) 1: simplify 2: factor 3: set each factor = 0 Double Zero A number that occurs as a zero of a function twice Double Root A number that occurs twice as a root of a polynomial equation Solve x2 + 25 = 10x 12 + 4m = m2 Section 4-8 Problem Solving Using Polynomial Equations Example 1 A graphic artist is designing a poster that consists of a rectangular print with a uniform border. The print is to be twice as tall as it is wide, and the border is to be 3 in. wide. If the area of the poster is to be 680 in2, find the dimensions of the print. Solution 1.Draw a diagram 2.Let w = width and 2w = height 1.The dimensions are 6 in. greater than the print, so they are w + 6 and 2w + 6 Solution 4. The area is represented by (w + 6)(2w + 6) = 680 5. Solve the equation. Example 2 The sum of two numbers is 9. The sum of their squares is 101. Find the numbers. Solution 1.Let x = one number 2.Then 9 – x = the other number 3.x2 + (9 – x)2 = 101 Section 4-9 Solving Polynomial Inequalities Polynomial Inequality An inequality that is equivalent to an inequality with a polynomial as one side and 0 as the other side. x2 > x + 6 Solve by factoring The product is positive if both factors are positive, or both factors are negative The product is negative if the factors have opposite signs Example 1 Solve and graph x2 – 1 > x + 5 x2 – x – 6 > 0 Both factors must be positive or negative Example 2 Solve and graph 3t < 4 – t2 t2 + 3t – 4 < 0 The factors must have opposite signs