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Products and Factors of Polynomials
Chapter 4 Products and Factors of Polynomials
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Section 4-1 Polynomials
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Constant - A number -2, 3/5, 0
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Monomial- A constant, a variable, or a product of a constant and one or more variables u (1/3)m s2t x
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Coefficient- The constant (or numerical) factor in a monomial
3m coefficient = 3 u coefficient = 1 - s2t3 coefficient = -1
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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?
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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.
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Examples- 6xy degree = 4 -s2t degree = 5 u degree = 1 degree = 0
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Similar Monomials- Monomials that are identical or that differ only in their coefficients Also called like terms Are - s2t3 and 2s2t3 similar?
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Polynomial- A monomial or a sum of monomials.
The monomials in a polynomial are called the terms of the polynomial.
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Examples- x2 + (-4)x + 5 x2 – 4x + 5 What are the terms? x2, -4x, and 5
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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
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Are they Simplified? 2x3 – 5 + 4x + x3 3x3 + 4x – 5
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Degree of a Polynomial-
The greatest of the degrees of its terms after it has been simplified What is the degree? x4 + 3x x3 + 3x – 7 x – 5x x + 1 x4 – 2x2y3 + 6y -11
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Adding Polynomials (x2 + 4x – 3) + (x3 – 2x2 + 6x – 7)
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)
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Subtracting Polynomials
To subtract one polynomial from another, add the opposite of each term of the polynomial you’re subtracting (x3 – 5x2 + 2x – 5) – (2x2 – 3x + 5)
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Using Laws of Exponents
Section 4-2 Using Laws of Exponents
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Laws of Exponents Let a and b be real numbers and m and n be positive integers in all the following laws
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Law 1 am · an = am+n x2 · x4 = x6 y3 · y5 = ? m · m4 = ?
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Law 2 (ab)m = ambm (xy)3 = x3y3 (3st)2 = ? (xy)5 = ?
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Law 3 (am)n = amn (x3)2 = x6 (x2y3)4 = ? (2mn2)3 = ?
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Using Distributive Law
Distribute the variable using exponent laws 3t2(t3 – 2t2 + t – 4) = ? – 2x2(x3 – 3x + 4) = ?
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Multiplying Polynomials
Section 4-3 Multiplying Polynomials
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Binomial A polynomial that has two terms 2x + 3 4x – 3y
3xy – z
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Trinomial A polynomial that has three terms 2x2 – 3x + 1 14 + 32z – 3x
mn – m2 + n2
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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
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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
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Binomial A polynomial that has two terms 2x + 3 4x – 3y
3xy – z
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Trinomial A polynomial that has three terms 2x2 – 3x + 1 14 + 32z – 3x
mn – m2 + n2
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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
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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
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Example - (2a – b)(3a + 5b) F – 2a · 3a O – 2a · 5b I – (-b) ▪ 3a
L - (-b) ▪ 5b 6a2 + 10ab – 3ab – 5b2 6a2 + 7ab – 5b2
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Example – (x + 6)(x +4) F – x ▪ x O – x ▪ 4 I – 6 ▪ x L – 6 ▪ 4
x2 + 4x + 6x + 24 x2 + 10x + 24
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Special Products (a + b)2 = a2 + 2ab + b2 (a - b)2 = a2 - 2ab + b2
(a + b)(a – b) = a2 - b2
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Using Prime Factorization
Section 4-4 Using Prime Factorization
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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
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Example The number 15 can be factored in the following ways
(1)(15) (-1)(-15) (5)(3) (-3)(-5)
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Prime Number An integer greater than 1 whose only positive integral factors are itself and 1
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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
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Example 350 = 2 x 175 = 2 x 5 x 35 = 2 x 5 x 5 x 7 So the prime factorization of 350 is 2 x 52 x 7
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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.
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Example What is the GCF of 100, 120, and 90? 10
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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.
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Example What is the LCM of 100, 120, and 90? 1800
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Summary GCF – take the least power of each common prime factor.
LCM – take the greatest power of each prime factor
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Factoring Polynomials
Section 4-5 Factoring Polynomials
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Factor To factor a polynomial you express it as a product of other polynomials We will factor using polynomials with integral coefficients
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Greatest Monomial Factor
The GCF of the terms What is the GCF of 2x4 – 4x3 + 8x2? 2x2
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Now factor: 2x4 – 4x3 + 8x2 Factor out 2x2 2x2(x2 – 2x + 4)
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Perfect Square Trinomials
The polynomials in the form of a2 + 2ab + b2 and a2 – 2ab + b2 are the result of squaring a + b and a – b respectively
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Difference of Squares The polynomial a2 – b2 is the product of a + b and a - b
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Factor Each Polynomial
z2 + 6z + 9 4s2 – 4 st + t2 25x2 – 16a2
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Factored Form: (z + 3)2 (2s – t)2 (5x + 4a)(5x – 4a)
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Sum and Difference of Cubes
a3 + b3 = (a + b)(a2 - ab + b2) a3 – b3 = (a – b)(a2 + ab + b2)
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Factor Each Polynomial
8u3 + v3
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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
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Factor each Polynomial
3xy x + 2y xy + 3y + 2x + 6
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Factoring Quadratic Polynomials
Section 4-6 Factoring Quadratic Polynomials
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Quadratic Polynomials
Polynomials of the form ax2 + bx + c Also called second- degree polynomials
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Terms ax2 - quadratic term bx - linear term c - constant term
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Quadratic Trinomial A quadratic polynomial for which a, b, and c are all nonzero integers
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Factoring Quadratic Trinomials
ax2 + bx + c can be factored into the form (px + q)(rx + s) where p, q, r, and s are integers
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Factors a = pr b = ps + qr c = qs
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Factor the Polynomial x2 + 2x - 15 a = 1, so pr = 1 c = -15, so qs = -15 b = 2, so ps + qr = 2
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Factor the Polynomials
15t2 - 16t + 4 3 - 2z - z2 x2 + 4x - 3
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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 x2 + 4x - 3 is irreducible
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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
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Greatest Common Factor
The GCF of two or more polynomials is the common factor having the greatest degree and the greatest constant factor
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Least Common Multiple The LCM of two or more polynomials is the common multiple having the least degree and least positive constant factor
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Solving Polynomial Equations
Section 4-7 Solving Polynomial Equations
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Polynomial Equation An equation that is equivalent to one with a polynomial as one side and 0 as the other x2 = 5x + 24
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Root The value of a variable that satisfies the equation
Also called the solution
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Solving a polynomial Equation
You can factor the polynomial to solve the equation
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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
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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
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Example 2 Solve x2 = x + 30 1: x2 - x – 30 = 0 2: (x – 6)(x + 5) = 0
The solution set is {6, -5}
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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
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Example Find the zeros of f(x) = (x – 4)3 – 4(3x – 16) 1: simplify
2: factor 3: set each factor = 0
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Double Zero A number that occurs as a zero of a function twice
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Double Root A number that occurs twice as a root of a polynomial equation
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Solve x = 10x 12 + 4m = m2
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Problem Solving Using Polynomial Equations
Section 4-8 Problem Solving Using Polynomial Equations
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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.
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Solution Draw a diagram Let w = width and 2w = height
The dimensions are 6 in. greater than the print, so they are w + 6 and 2w + 6
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Solution 4. The area is represented by (w + 6)(2w + 6) = 680
5. Solve the equation.
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Example 2 The sum of two numbers is 9. The sum of their squares is Find the numbers.
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Solution Let x = one number Then 9 – x = the other number
x2 + (9 – x)2 = 101
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Solving Polynomial Inequalities
Section 4-9 Solving Polynomial Inequalities
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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
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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
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Example 1 Solve and graph x2 – 1 > x + 5 x2 – x – 6 > 0
Both factors must be positive or negative
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Example 2 Solve and graph 3t < 4 – t2 t2 + 3t – 4 < 0
The factors must have opposite signs
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