 # Products and Factors of Polynomials

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Products and Factors of Polynomials
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 u (1/3)m s2t x

Coefficient- The constant (or numerical) factor in a monomial
3m 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- 6xy degree = 4 -s2t degree = 5 u 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? x2, -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

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

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)

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)

Using Laws of Exponents
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 · x4 = x6 y3 · y5 = ? m · m4 = ?

Law 2 (ab)m = ambm (xy)3 = x3y3 (3st)2 = ? (xy)5 = ?

Law 3 (am)n = amn (x3)2 = x6 (x2y3)4 = ? (2mn2)3 = ?

Using Distributive Law
Distribute the variable using exponent laws 3t2(t3 – 2t2 + t – 4) = ? – 2x2(x3 – 3x + 4) = ?

Multiplying Polynomials
Section 4-3 Multiplying Polynomials

Binomial A polynomial that has two terms 2x + 3 4x – 3y
3xy – z

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 4x – 3y
3xy – z

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 6a2 + 10ab – 3ab – 5b2 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) = a2 - b2

Using Prime Factorization
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) (-1)(-15) (5)(3) (-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 = 2 x 5 x 5 x 7 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

Factoring Polynomials
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 2x4 – 4x3 + 8x2? 2x2

Now factor: 2x4 – 4x3 + 8x2 Factor out 2x2 2x2(x2 – 2x + 4)

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

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 25x2 – 16a2

Factored Form: (z + 3)2 (2s – t)2 (5x + 4a)(5x – 4a)

Sum and Difference of Cubes
a3 + b3 = (a + b)(a2 - ab + b2) a3 – b3 = (a – b)(a2 + ab + b2)

Factor Each Polynomial
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 x + 2y xy + 3y + 2x + 6

Polynomials of the form ax2 + bx + c Also called second- degree polynomials

Terms ax2 - quadratic term bx - linear term c - constant term

Quadratic Trinomial A quadratic polynomial for which a, b, and c are all nonzero integers

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
15t2 - 16t + 4 3 - 2z - z2 x2 + 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 x2 + 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

Solving Polynomial Equations
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 x2 = 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
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 x = 10x 12 + 4m = m2

Problem Solving Using Polynomial Equations
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 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

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 Find the numbers.

Solution Let x = one number Then 9 – x = the other number
x2 + (9 – x)2 = 101

Solving Polynomial Inequalities
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