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