Chapter 4 Products and Factors of Polynomials
Dec 25, 2015
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 degree = 4 -s2t3 degree =
5 u degree = 1 -7 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 + 5x2 – 4x + 5What 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 – 54x2 – 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 – 7x – 5x2 + 1 7x + 1x4 – 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
(x3 – 5x2 + 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 · 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) = ?
Section 4-3
Multiplying Polynomials
Binomial
A polynomial that has two terms
2x + 3 4x – 3y3xy – 14 613 + 39z
Trinomial
A polynomial that has three terms
2x2 – 3x + 1 14 + 32z – 3xmn – m2 + n2
Multiplying binomialsWhen 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. MethodF – multiply First termsO – multiply Outer terms I – multiply Inner termsL – multiply Last terms
Add all terms to get product
Binomial
A polynomial that has two terms
2x + 3 4x – 3y3xy – 14 613 + 39z
Trinomial
A polynomial that has three terms
2x2 – 3x + 1 14 + 32z – 3xmn – m2 + n2
Multiplying binomialsWhen 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. MethodF – multiply First termsO – multiply Outer terms I – multiply Inner termsL – multiply Last terms
Add all terms to get product
Example - (2a – b)(3a + 5b)
F – 2a · 3aO – 2a · 5b I – (-b) ▪ 3aL - (-b) ▪ 5b6a2 + 10ab – 3ab – 5b2 6a2 + 7ab – 5b2
Example – (x + 6)(x +4)
F – x ▪ xO – x ▪ 4 I – 6 ▪ xL – 6 ▪ 4
x2 + 4x + 6x + 24x2 + 10x + 24
Special Products
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
(a + b)(a – b) = a2 - b2
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 MultipleThe 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 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 + 94s2 – 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
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 PolynomialsPolynomials of the form ax2 + bx + c
Also called second- degree polynomials
Terms
ax2 - quadratic termbx - linear termc - constant term
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 = prb = ps + qrc = qs
Factor the Polynomial
x2 + 2x - 15a = 1, so pr = 1c = -15, so qs = -15b = 2, so ps + qr = 2
Factor the Polynomials15t2 - 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 CompletelyA 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 FactorThe 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
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) = 0Step 1: already = 0Step 2: already factoredStep 3: set each factor = 0x - 5 = 0 x + 2 = 0 x = 5 x = -2
Example 2
Solve x2 = x + 301: x2 - x – 30 = 02: (x – 6)(x + 5) = 03: x – 6 = 0 x + 5 = 0 x = 6 x = -5The 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: simplify2: factor3: 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 diagram2.Let w = width and 2w = height3.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) = 6805. 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 number2.Then 9 – x = the other
number3.x2 + (9 – x)2 = 101
Section 4-9
Solving Polynomial Inequalities
Polynomial InequalityAn 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 + 5x2 – x – 6 > 0Both factors must be positive or negative
Example 2
Solve and graph3t < 4 – t2
t2 + 3t – 4 < 0 The factors must have opposite signs