Hawkes Learning Systems: College Algebra

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Hawkes Learning Systems: College Algebra

Section 1.5: Polynomials and Factoring




Objectives:o The terminology of polynomial expressions

o The algebra of polynomials

o Common factoring methods




The Terminology of Polynomial Expressionso Coefficient: A number multiplied by a variable in any

of the terms of a polynomial.

o Degree of the term: The sum of the exponents of the variables in that term.

o Constant term: Any non-zero number that is not multiplied by a variable.

Note: Constant terms have a degree of zero.

o Degree of a polynomial: The largest degree of all the individual terms.




The Terminology of Polynomial Expressions

o Monomials: Polynomials consisting of a single term

Ex:

o Binomials: Polynomials consisting of two terms

Ex:

o Trinomials: Polynomials consisting of three terms

Ex:

217 , 5 , 3x xy

225 2, 16 1x x

2 237 4 2, 16 11 13xy x y xy



Copyright © 2010 Hawkes Learning Systems. All rights reserved.Example 1: The Terminology of Polynomial

ExpressionsExpression Terms Type Degree Explanation

3 Trinomial 10The degree of the first term is 8, the degree of the second term is 10, and the degree of the third term is 0.

2 Binomial 7The degree of the first term is 6 and the degree of the second term is 7.

5 1 Monomial 0 The degree of a constant is always 0.

4 2 3 43 5.4x y x y

3 5 102 33x y y




The Terminology of a Single Variable

Polynomials of a Single Variable

A polynomial in the variable of a degree n can be written in the form

where are real numbers, and n is a positive integer. This form is called descending order because the powers descend from left to right. The leading coefficient of this polynomial is .

11 1 0...n n

n na x a x a x a

, 1 1 0,..., ,n na a a a 0na

na




The Algebra of Polynomials

Like or similar terms: The terms among all the polynomials being added that have the same variables raised to the same powers.

Ex: What are the like terms in the polynomial below?

3 2 2 32 3x z y z y x Notice that and both include the variable x raised to the third power. These are like or similar terms. Can you find any others? 3 3 2 2 and 3 , 2 and , and x x z z y y

23 32 32 yzx z xy

3x 33x




Example 1: The Algebra of Polynomials

Add or subtract the polynomials, as indicated.

=

=

=

3 2 22 3 3 (1 3) 4x y y xz z

3 2 2 22 3 3 3 4x y y z x y z xz

3 2 22 2 4x y xz z

The first step is to identify like terms and group these together.

Note that the terms and are similar, as multiplication is commutative.

23 222 ( ) ( ) 43 33 xy zzx y y xz

Pull out the and variables and simplify.

y 2xz

2z x23xz





Add or subtract the polynomials, as indicated.

=

=

=

Again, the like terms are identified and grouped together after distributing the minus sign over all the terms in the second polynomial.

3 3 34 4ab b c b c

34 4ab

3 334 4b c bab c

3 334 4 ( )b c b cab





Multiply the polynomials, as indicated.

=

=

2 3(2 )(4 3 3 )x y z z xy

2 2 3 2 3 4 38 6 6 4 3 3x y x yz x y z z xyz

2 32 (4 3 3 ) (4 3 3 )x y z xy z z xy Use the distributive property first to multiply each term of the first polynomial by each term of the second.

None of the resulting terms are similar, so the final answer is a polynomial of 6 terms.





Multiply the polynomials, as indicated.

=

=

=

2 2(3 )( )ab a ab a

2 2 23 ( ) ( )ab ab a a ab a

2 2 3 3 43 3a b a b a b a

2 2 3 43 2a b a b a

The four terms that result from the initial multiplication contain two similar terms.

We combine these to obtain the final trinomial.




When a binomial is multiplied by a binomial, the acronym FOIL is commonly used as a reminder of the four necessary products. Consider the product:

The solution to the product above would be

First + Outer + Inner + Last


2 2(3 )( )ab a ab a

FirstOuter

InnerLast





Consider the product:

The product of the First terms is

The product of the Outer terms is

The product of the Inner Terms is

The product of the Last terms is

So, First + Outer + Inner + Last

= + + +

2 2(3 )( )ab a ab a 2 2( )(3 )ab aba a

2 23a b 33a b 3a b 4a

2 23a b2 2( )(3 )a abab a 33a b

3a b2 2(3 )( )ab a ab a 2 2(3 )( )ab aba a 4a




Common Factoring Methods: Terminology

o Factoring: Reversing the process of multiplication in order to find two or more expressions whose product is the original expression.

o Factorable: A polynomial with integer coefficients is factorable if it can be written as a product of two or more polynomials, all of which also have integer coefficients.

o Irreducible (over the integers) or prime: A polynomial that is not factorable.

o Completely Factor: To write a polynomial as a product of prime polynomials.




Common Factoring Methods

o Method 1: Greatest common factor.

o Method 2: Factoring by grouping.

o Method 3: Factoring special binomials.

o Method 4: Factoring trinomials.o Case 1: Leading coefficient is 1.

o Case 2: Leading coefficient is not 1.

o Method 5: Factoring Expressions Containing Fractional Exponents




Common Factoring Methods: Method 1

Find the greatest common factor of

GCF:

So,

4 2 3 2 312 8 4x y x y x y

34x y

Method 1: Greatest Common FactorThe Greatest Common Factor (GCF) among all the terms is simply the product of all the factors common to each. The Greatest Common Factor method is a matter of applying the distributive property to “un-distribute” the greatest common factor.

4 2 3 2 3 34 3 2 112 8 4x y x y x y x y xy y

What do the three terms in the polynomial have in common? . The product of these terms is the GCF of the polynomial.

34, , and x y




Example 5: Common Factoring Methods

Use the Greatest Common Factor method to factor thefollowing polynomial.

=

=

5 2 3 312 4 8x x x z

3 32 2 2( )(3 ) ( )(4 4 41) ( )(2 )x x x x xz

2 3 3(4 )(3 1 2 )x x xz

Applying the distributive property in reverse leads to the factored form of this degree 6 trinomial.





Use the Greatest Common Factor method to factor thefollowing polynomials.

=

=

An alternative form of the final answer is . We would have obtained this answer if we had factored out 12a initially. These two answers are equivalent.

224 60ax a

2( 12 )(2 ) ( 12 )( 5)a x a

212 (2 5)a x 212 ( 2 5)a x





Use the Greatest Common Factor method to factor thefollowing polynomials.

=

=

=

2 2( ) 3( )a b a b

2 2( )(1) ( )( 3)a b a b

2( )(1 3)a b

22( )a b

In factoring out the greatest common factor remember that it is being multiplied first by 1 and then by –3. One common error in factoring is to forget factors of 1.

2( )a b





Method 2: Factoring by Grouping.

Factoring by Grouping: A trial and error process applied when the first factoring method is not directly applicable. If the terms of the polynomial are grouped in a suitable way, the GCF method may apply to each group, and a common factor might subsequently be found among the groups.





Ex: Use Method 2, factoring by grouping, to factor the following polynomial:

=

=

=

Note: the GCF of the four terms is 1, so you cannot use the GCF method. Try rearranging the way they are grouped. Once you group and notice that the two groups each have a GCF: and 1. Don’t forget to factor out 1 from as it is a common mistake to not factor out a 1.

(3 )y

(15 5 )x xy5x

(3 )y

5 3 15xy x y

(15 5 ) (3 )x xy y

5 (3 ) 1(3 )x y y

(5 1)(3 )x y





Use the Factor by Grouping method to factor the following polynomials.

=

=

=

=

26 2 3x y x xy 2(6 2 ) ( 3 )x x y xy

2 (3 1) ( 1 3 )x x y x

2 (3 1) (3 1)x x y x

(3 1)(2 )x x y

The GCF of the four terms in the polynomial is 1, so the Greatest Common Factor Method does not directly apply. So we must use Method 2, Factoring by Grouping.





Use the Factor by Grouping method to factor the following polynomial.

=

=

=We could also group the first and third terms and the second and fourth

terms to obtain the same result.

ax ay bx by

( ) ( )a x y b x y

( )( )x y a b

=

=

=

x a b y a b

( )( )x y a b

ax bx ay by ( ) ( )ax ay bx by

OR





Method 3: Factoring Special BinomialsThree types of binomials can always be factored followingcertain patterns. In the following, A and B represent algebraicexpressions.

o Difference of Two Squares:

o Difference of Two Cubes:

o Sum of Two Cubes:

2 2 ( )( )A B A B A B

3 3 2 2( )( )A B A B A AB B

3 3 2 2( )( )A B A B A AB B





Factor the following binomials.

=

=

The first step is to realize that the binomial is a difference of two squares and to identify the two expressions that are being squared. Then follow the pattern to factor the binomial.

2 3 2(7 ) (3 )x y

3 3(7 3 )(7 3 )x y x y

2 649 9x y






=

=

=

6 12 327a b c

2 4 3 3(3 ) ( )a b c

2 4 2 4 2 2 4 2(3 ) (3 ) (3 )( ) ( )a b c a b a b c c

2 4 4 8 2 4 2(3 )(9 3 )a b c a b a b c c

First recognize the binomial as a sum of two cubes. Then follow the pattern to factor the binomial.






=

=

=

364 ( )x y

3 34 ( )x y

2 24 ( ) 4 4( ) ( )x y x y x y

2 2(4 )(16 4 4 2 )x y x y x xy y

In this difference of two cubes the second cube is itself a binomial. But the factoring pattern still applies, leading to the final factored form of the original binomial.





Method 4, Case 1: Leading Coefficient is 1.

In this case, p and r will both be 1 so we only need q and s such that .

That is, we need two integers whose sum is b, the coefficient of x, and whose product is c, the constant term.

2 2 ( )x bx c x q s x qs





Use Method 4, Case 1 to factor the following polynomial.Ex:

o Factor:o Begin by writing o We need to find two integers to replace the question

marks. The two integers we seek must have a product of 2. Because the product is positive, both integers must be either positive or negative. Therefore, the only possibilities are

o Additionally, the sum of these two integers must be 3. Therefore, they must be

o Thus,

2 ?3 2 .? x x x x

2 3 2 ( )( ). x x x x2 1

2 3 2x x

2,1 .12,

2,1 .

or





o Factor:o Begin by writing o We need to find two integers to replace the question

marks. The two integers we seek must have a product of –12. Because the product is negative, one integer must be positive and one must be negative. Therefore, the only possibilities are

o Additionally, the sum of these two integers must be 1. Therefore, they must be .

o Thus,

2 12x x 2 12 ( )( )? ? x x x x

{ 3,4}2 12 ( )( . ) x x x x

1, ,12 ,1,12 2, ,6 ,2,6 3, ,4 43,

3 4





Method 4, Case 2: Leading Coefficient is not 1.Factoring Trinomials by GroupingFor the trinomial :

Step 1: Multiply a and c. Step 2: Factor ac into two integers whose sum is b. If no such factors exist, the trinomial is irreducible over the integers.Step 3: Rewrite b in the trinomial with the two integers found in step 2. The resulting polynomial of four terms may now be factored by grouping.

2ax bx c





Factor the following trinomial by grouping:

1.

2.

3.

4.

5.

26 12x x

(6)( 12) 72 Multiply a and c.

Factor ac into two integers whose sum is b.

Rewrite b in the trinomial with the two integers found in step 2 and distribute.

3 (2 3) 4(2 3)x x x

(2 3)(3 4)x x

Group.

9 and 826 9 8 12x x x





Perfect Square Trinomials: trinomial expressions whose factored form is the square of a binomial expression.There are two forms of Perfect Square Trinomials:

2 2 2

2 2 2

2 ( )

2 ( )

A AB B A B

A AB B A B





Method 5: Factoring Expressions Containing Fractional Exponents

To factor an algebraic expression that has fractional exponents, identify the least exponent among the various terms and factor the variable raised to that least exponent from each of the terms.





Factor the following algebraic expressions:

=

=

=

2 1 43 3 33 6 3x x x

2

233 (1 2 )x x x

233 ( 1)( 1)x x x

2233 ( 1)x x

First we factor out .

Next, notice that the second factor is a second-degree trinomial and is factorable. In fact, it is a perfect square trinomial.

233x





Factor the following algebraic expression:

=

=

1 12 2( 1) ( 1)x x

1

2( 1) ( 1) 1

x x

12( 1) ( 2)x x

Factor out using the properties of exponents to obtain the terms in the second factor.

Simplify the second term.

12( 1)x

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Hawkes Learning Systems: College Algebra