CHAPTER · b : Factor polynomials when the terms have a common factor, factoring out the greatest common factor. c : Factor certain expressions with four terms using

CHAPTER

5 Polynomials: Factoring

Slide 2 Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

5.1 Introduction to Factoring 5.2 Factoring Trinomials of the Type x2 + bx + c 5.3 Factoring ax2 + bx + c, a ≠ 1: The FOIL Method 5.4 Factoring ax2 + bx + c, a ≠ 1: The ac-Method 5.5 Factoring Trinomial Squares and Differences of

Squares 5.6 Factoring Sums or Differences of Cubes 5.7 Factoring: A General Strategy 5.8 Solving Quadratic Equations by Factoring

CHAPTER



5.9 Applications of Quadratic Equations

OBJECTIVES

5.1 Introduction to Factoring


a Find the greatest common factor, the GCF, of monomials.

b Factor polynomials when the terms have a common factor, factoring out the greatest common factor.

c Factor certain expressions with four terms using factoring by grouping.




The greatest of the common factors of several terms is called the greatest common factor, GCF.

EXAMPLE



2 Find the GCF of 180 and 420.


EXAMPLE



4 Find the GCF of 54, 90, and 252.


EXAMPLE Solution



4





Consider the product

To factor the polynomial on the right, we reverse the process of multiplication:


Factor; Factorization


To factor a polynomial is to express it as a product. A factor of a polynomial P is a polynomial that can be used to express P as a product. A factorization of a polynomial is an expression that names that polynomial as a product.

EXAMPLE



5 Find the GCF of and


The greatest positive common factor of the coefficients is 3. Next, we find the GCF of the powers of x. That GCF is x2 because 2 is the smallest exponent of x. Thus the GCF of the set of monomials is 3x2.


To Find the GCF of Two or More Monomials


1. Find the prime factorization of the coefficients, including –1 as a factor if any coefficient is negative.

2. Determine any common prime factors of the coefficients. For each one that occurs, include it as a factor of the GCF. If none occurs, use 1 as a factor.

3. Examine each of the variables as factors. If any appear as a factor of all the monomials, include it as a factor, using the smallest exponent of the variable. If none occurs in all the monomials, use 1 as a factor.

4. The GCF is the product of the results of steps (2) and (3).




To multiply a monomial and a polynomial with more than one term, we multiply each term of the polynomial by the monomial using the distributive laws:

To factor, we do the reverse. We express a polynomial as a product using the distributive laws in reverse:




EXAMPLE



7 Factor:


EXAMPLE



8 Factor:


EXAMPLE Solution



8



Tips for Factoring


• Before doing any other kind of factoring, first try to factor out the GCF.

• Always check the result of factoring by multiplying.




Certain polynomials with four terms can be factored using a method called factoring by grouping.




Consider the four-term polynomial

There is no factor other than 1 that is common to all the terms. We can, however, factor separately:




This method of factoring is called factoring by grouping.

EXAMPLE



15 Factor by grouping.


EXAMPLE Solution



15


We think through this process as follows:

EXAMPLE



17 Factor by grouping:


EXAMPLE Solution



17


EXAMPLE Solution



17


CHAPTER





CHAPTER




OBJECTIVES

5.2 Factoring Trinomials of the Type x2 + bx + c


a Factor trinomials of the type x2 + bx + c by examining the constant term c.




Compare the following multiplications:




Note that for all four products: • The product of the two binomials is a trinomial. • The coefficient of x in the trinomial is the sum of the

constant terms in the binomials. • The constant term in the trinomial is the product of the

constant terms in the binomials.

These observations lead to a method for factoring certain trinomials.




To factor x2 + 7x + 10, we think of FOIL in reverse. We multiplied x times x to get the first term of the trinomial, so we know that the first term of each binomial factor is x. Next, we look for numbers p and q such that

To get the middle term and the last term of the trinomial, we look for two numbers and whose product is 10 and whose sum is 7. Those numbers are 2 and 5. Thus the factorization is




EXAMPLE



1 Factor:






To Factor x2 + bx + c when c Is Positive


When the constant term of a trinomial is positive, look for two numbers with the same sign. The sign is that of the middle term:

EXAMPLE



2 Factor:


EXAMPLE Solution



2


Since the constant term, 12, is positive and the coefficient of the middle term, –8, is negative, we look for a factorization of 12 in which both factors are negative. Their sum must be –8.

EXAMPLE Solution



2





The product of two binomials can have a negative constant term:

Note that when the signs of the constants in the binomials are reversed, only the sign of the middle term in the product changes.

EXAMPLE



3 Factor:


The constant term, –20, must be expressed as the product of a negative number and a positive number. Since the sum of these two numbers must be negative (specifically, –8), the negative number must have the greater absolute value.

EXAMPLE Solution



3



To Factor x2 + bx + c when c Is Negative


When the constant term of a trinomial is negative, look for two numbers whose product is negative. One must be positive and the other negative:

Consider pairs of numbers for which the number with the larger absolute value has the same sign as b, the coefficient of the middle term.

EXAMPLE



4 Factor:


EXAMPLE Solution



4


EXAMPLE



5 Factor:


Consider this trinomial as We look for numbers p and q such that

The middle-term coefficient, –1, is small compared to –110. This tells us that the desired factors are close to each other in absolute value. The numbers we want are 10 and –11. The factorization is

EXAMPLE



6 Factor:


We consider the trinomial in the equivalent form

Think of –21b2 as the “constant” term and 4b as the “coefficient” of the middle term. Then try to express –21b2 as a product of two factors whose sum is 4b. Those factors are –3b and 7b.

The factorization is

EXAMPLE



7 Factor:


There are no factors whose sum is –1. Thus the polynomial is not factorable into factors that are polynomials with rational-number coefficients.




In this text, a polynomial like x2 – x + 5 that cannot be factored further is said to be prime. In more advanced courses, polynomials like x2 – x + 5 can be factored and are not considered prime.




Often factoring requires two or more steps. In general, when told to factor, we should factor completely. This means that the final factorization should not contain any factors that can be factored further.

EXAMPLE



8 Factor:


Always look first for a common factor. This time there is one, 2x.

EXAMPLE Solution



8


Now consider

EXAMPLE Solution



8



To Factor x2 + bx + c


1. First arrange in descending order. 2. Use a trial-and-error process that looks for factors of c

whose sum is b. 3. If c is positive, the signs of the factors are the same as

the sign of b. 4. If c is negative, one factor is positive and the other is

negative. If the sum of two factors is the opposite of b, changing the sign of each factor will give the desired factors whose sum is b.

5. Check by multiplying.

EXAMPLE



9 Factor:


Write in descending order:

EXAMPLE



9 Factor:


Each of the following is also a correct answer:

CHAPTER





CHAPTER




OBJECTIVES

5.3 Factoring ax2 + bx + c, a ≠ 1: The FOIL Method


a Factor trinomials of the type ax2 + bx + c, a ≠ 1, using the FOIL method.




We want to factor trinomials of the type ax2 + bx + c. Consider the following multiplication:

We reverse the above multiplication, using what we might call an “unFOIL” process.




We look for two binomials whose product is The answer is It turns out that is also a correct answer, but we generally choose an answer in which the first coefficients are positive.


The FOIL Method



The FOIL Method


EXAMPLE



1 Factor:


1) First, we check for a common factor. Here there is none (other than 1 or –1).

2) Find two First terms whose product is

EXAMPLE



1 Factor:


3) Find two Last terms whose product is –8.

EXAMPLE



1 Factor:


4) Inspect the Outside and Inside products resulting from steps (2) and (3). Look for a combination in which the sum of the products is the middle term, –10x. 5)

EXAMPLE



2 Factor:


EXAMPLE Solution



2


1) First, we factor out the largest common factor, 4: 2) Now factor

3) There are four pairs of factors of 10 and each pair can be listed in two ways:

EXAMPLE Solution



2


4) Look for Outside and Inside products resulting from steps (2) and (3) for which the sum is the middle term, –19x.

5)




In Example 2, look at the possibility Without multiplying, we can reject such a possibility. We removed the largest common factor in the first step. If 2x – 2 were one of the factors, then 2 would have to be a common factor in addition to the original 4. Thus, (2x – 2) cannot be part of the factorization of the original trinomial.


Tips for Factoring ax2 + bx + c, a ≠ 1


• Always factor out the largest common factor first, if one exists.

• Once the common factor has been factored out of the original trinomial, no binomial factor can contain a common factor (other than 1 or –1).

• If c is positive, then the signs in both binomial factors must match the sign of b. (This assumes that a > 0)

• Reversing the signs in the binomials reverses the sign of the middle term of their product.


Tips for Factoring ax2 + bx + c, a ≠ 1


• Organize your work so that you can keep track of which possibilities have or have not been checked.

• Always check by multiplying.

EXAMPLE



4 Factor:


Other correct answers:

EXAMPLE



5 Factor:


1) Factor out a common factor, if any. There is none (other than 1 or –1). 2) Factor the first term, 6p2.

3) Factor the last term, –28q2, which has a negative coefficient.

EXAMPLE



5 Factor:


4) Try some possibilities: 5) The check is left to the students.

CHAPTER





CHAPTER




OBJECTIVES

5.4 Factoring ax2 + bx + c, a ≠ 1: The ac-Method


a Factor trinomials of the type ax2 + bx + c, a ≠ 1, using the ac-method.




Another method for factoring trinomials of the type ax2 + bx + c, a ≠ 1, involves the product, ac, of the leading coefficient a and the last term c. It is called the ac-method. Because it uses factoring by grouping, it is also referred to as the grouping method.


The ac-Method


EXAMPLE



1 Factor:


1) First, we factor out a common factor, if any. There is none (other than 1 or –1).

2) We multiply the leading coefficient, 3, and the constant, –8:

EXAMPLE



1 Factor:


3) Then we look for a factorization of in which the sum of the factors is the coefficient of the middle term, –10.

EXAMPLE



1 Factor:


4) Next, we split the middle term as a sum or a difference using the factors found in step (3):

EXAMPLE



1 Factor:


5) Finally, we factor by grouping, as follows:

EXAMPLE



1 Factor:


6)

EXAMPLE



2 Factor:


EXAMPLE Solution



2


1) First, we factor out a common factor, if any. The number 2 is common to all three terms, so we factor it out:

2) Next, we factor the trinomial . We multiply the leading coefficient and the constant, 4 and –3:

EXAMPLE Solution



2


3) We try to factor –12 so that the sum of the factors is 4.

EXAMPLE Solution



2


4) Then we split the middle term, 4x, as follows:

5) Finally, we factor by grouping:

EXAMPLE Solution



2


6)

CHAPTER





CHAPTER




OBJECTIVES

5.5 Factoring Trinomial Squares and Differences of Squares


a Recognize trinomial squares. b Factor trinomial squares. c Recognize differences of squares. d Factor differences of squares, being careful to factor

completely.


a Recognize trinomial squares.


Some trinomials are squares of binomials. For example, the trinomial x2 + 10x + 25 is the square of the binomial x + 5. A trinomial that is the square of a binomial is called a trinomial square, or a perfect square trinomial.

We can use these equations in reverse to factor trinomial squares.


Trinomial Squares





In order for an expression to be a trinomial square: a) The two expressions A2 and B2 must be squares, such as When the coefficient is a perfect square and the

power(s) of the variable(s) is (are) even, then the expression is a perfect square.

b) There must be no minus sign before A2 or B2. c) If we multiply A and B and double the result, 2·AB, we

get either the remaining term or its opposite.

EXAMPLE



1 Determine whether is a trinomial square.



Factoring Trinomial Squares


EXAMPLE


b Factor trinomial squares.

4 Factor:


EXAMPLE



6 Factor:


EXAMPLE Solution



6


EXAMPLE



7 Factor:


EXAMPLE



8 Factor:


EXAMPLE



9 Factor:



c Recognize differences of squares.


The following polynomials are differences of squares:

To factor a difference of squares such as x2 – 9, think about the formula we used in Chapter 4: Equations are reversible, so we also know the following.


Difference of Squares





In order for a binomial to be a difference of squares:

EXAMPLE



10 Is 9x2 – 64 a difference of squares?



Factoring a Difference of Squares


EXAMPLE


d Factor differences of squares, being careful to factor completely.

14 Factor:


EXAMPLE



17 Factor:


EXAMPLE



19 Factor:


When no factor can be factored further, you have factored completely. Always factor completely whenever told to factor.

EXAMPLE



21 Factor:


EXAMPLE Solution



21



Tips for Factoring


• Always look first for a common factor. If there is one, factor it out.

• Be alert for trinomial squares and differences of squares. Once recognized, they can be factored without trial and error.

• Always factor completely. • Check by multiplying.

CHAPTER





CHAPTER




OBJECTIVES

5.6 Factoring Sums or Differences of Cubes


a Factor sums or differences of cubes.




Consider the following products:


Sum or Difference of Cubes


EXAMPLE



1 Factor:


EXAMPLE



3 Factor:


EXAMPLE Solution



3


EXAMPLE



4 Factor:


We can express this polynomial as a difference of squares.

EXAMPLE Solution



4


EXAMPLE Solution



4


Had we thought of factoring first as a difference of two cubes, we would have had

In this case, we might have missed some factors; can be factored as but we probably would not have known to do such factoring.




When you can factor as a difference of squares or a difference of cubes, factor as a difference of squares first.


Factoring Summary


CHAPTER





CHAPTER




OBJECTIVES

5.7 Factoring: A General Strategy


a Factor polynomials completely using any of the methods considered in this chapter.


Factoring Strategy


To factor a polynomial: a) Always look first for a common factor. If there is one,

factor out the largest common factor. b) Then look at the number of terms. Two terms: Determine whether you have a difference

of squares, A2 – B2, or a sum or difference of cubes, A3 + B3 or A3 – B3. Do not try to factor a sum of squares: A2 + B2.


Factoring Strategy


To factor a polynomial: b) Three terms: Determine whether the trinomial is a

square. If it is, you know how to factor. If not, try trial and error, using FOIL or the ac-method.

Four terms: Try factoring by grouping. c) Always factor completely. If a factor with more than

one term can still be factored, you should factor it. When no factor can be factored further, you have finished.

d) Check by multiplying.

EXAMPLE



2 Factor:


EXAMPLE Solution



2


EXAMPLE Solution



2


c) None of these factors can be factored further, so we have factored completely.

d)

EXAMPLE



5 Factor:


EXAMPLE Solution



5


a) We look first for a common factor: b) The expression 2 – 7xy + y2 cannot be factored. c) No further factoring is possible. d) The check is left to the students.

EXAMPLE



7 Factor:


EXAMPLE Solution



7


a) We look first for a common factor. There isn’t one. b) There are four terms. We try factoring by grouping:

c) Since neither factor can be factored further, we have factored completely.

d) The check is left to the students.

EXAMPLE



12 Factor:


EXAMPLE Solution



12


a) We look first for a common factor: b) The factor 8t3 – s3 has only two terms. It is a difference

of two cubes. We factor as follows: c) No further factoring is possible. The complete

factorization is: d) The check is left to the students.

CHAPTER





CHAPTER




OBJECTIVES

5.8 Solving Quadratic Equations by Factoring


a Solve equations (already factored) using the principle of zero products.

b Solve quadratic equations by factoring and then using the principle of zero products.


Quadratic Equation


A quadratic equation is an equation equivalent to an equation of the type




The product of two numbers is 0 if one or both of the numbers is 0. Furthermore, if any product is 0, then a factor must be 0. For example:

EXAMPLE



1 Solve:


This equation will be true when either factor is 0.

Each of the numbers –3 and 2 is a solution of the original equation, as we can see in the following checks.

EXAMPLE



1 Solve:



The Principle of Zero Products


An equation ab = 0 is true if and only if a = 0 is true or b = 0 is true, or both are true. (A product is 0 if and only if one or both of the factors is 0.)

EXAMPLE



2 Solve:


EXAMPLE Solution



2


EXAMPLE Solution



2


EXAMPLE



4 Solve:


EXAMPLE



5 Solve:


EXAMPLE



6 Solve:


EXAMPLE Solution



6


EXAMPLE



8 Solve:


EXAMPLE Solution



8


EXAMPLE



9 Solve:


Remember: We must have a 0 on one side.

EXAMPLE Solution



9





The graph of y = ax2 + bx + c, a ≠ 0, is shaped like one of the following curves. Note that each x-intercept represents a solution of ax2 + bx + c = 0.

EXAMPLE



10 Find the x-intercepts of the graph of y = x2 – 4x – 5, shown here.


EXAMPLE Solution



10


EXAMPLE Solution



10


Thus the x-intercepts of the graph of y = x2 – 4x – 5 are (5, 0) and (–1, 0). We can now label them on the graph.

CHAPTER





CHAPTER




OBJECTIVES



a Solve applied problems involving quadratic equations that can be solved by factoring.

EXAMPLE



1 Kitchen Island


Lisa buys a kitchen island with a butcher block top as part of a remodeling project. The top of the island is a rectangle that is twice as long as it is wide and that has an area of 800 in2. What are the dimensions of the top of the island?

EXAMPLE Solution



1


1. Familiarize. We first make a drawing. Recall that the area of a rectangle is Length·Width. We let x = the width of the top, in inches. The length is then 2x.

EXAMPLE Solution



1


2. Translate.

EXAMPLE Solution



1


3. Solve.

EXAMPLE Solution



1


4. Check. The solutions of the equation are 20 and –20. Since the width must be positive, –20 cannot be a solution. To check 20 in., we note that if the width is 20 in., then the length is 2·20 in., or 40 in., and the area is 20 in. · 40 in., or 800 in2. Thus the solution 20 checks. 5. State. The top is 20 in. wide by 40 in. long.

EXAMPLE



4 Marathoner’s Numbers


The product of the numbers of two consecutive entrants in a marathon race is 156. Find the numbers.

EXAMPLE Solution



4


1. Familiarize. Let x = the smaller integer; then x + 1 = the larger integer.

2. Translate.

EXAMPLE Solution



4


3. Solve.

EXAMPLE Solution



4


4. Check. The solutions of the equation are 12 and –13. When x is 12, then x + 1 is 13, and 12·13 = 156. The numbers 12 and 13 are consecutive. Negative numbers are not used as entry numbers. 5. State. The entry numbers are 12 and 13.




The problems that follow involve the Pythagorean theorem, which states a relationship involving the lengths of the sides of a right triangle. A triangle is a right triangle if it has a 90°, or right, angle. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs.


The Pythagorean Theorem


In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then

EXAMPLE



5 Wood Scaffold


Jonah is building a wood scaffold to use for a home improvement project. He designs the scaffold with diagonal braces that are 5 ft long and that span a distance of 3 ft. How high does each brace reach vertically?

EXAMPLE Solution



5


1. Familiarize. Let h = the height, in feet, to which each brace rises vertically.

2. Translate. A right triangle is formed, so we can use the Pythagorean theorem:

EXAMPLE Solution



5


3. Solve.

EXAMPLE Solution



5


4. Check. Since height cannot be negative, –4 cannot be a solution. If the height is 4 ft, we have 32 + 42 = 9 + 16 = 25, which is 52. Thus, 4 checks and is the solution. 5. State. Each brace reaches a height of 4 ft.

EXAMPLE



6 Ladder Settings


A ladder of length 13 ft is placed against a building in such a way that the distance from the top of the ladder to the ground is 7 ft more than the distance from the bottom of the ladder to the building. Find both distances.

EXAMPLE Solution



6


1. Familiarize. The ladder and the missing distances form the hypotenuse and the legs of a right triangle. We let x = the length of the side (leg) across the bottom, in feet. Then x + 7 = the length of the other side (leg). The hypotenuse has length 13 ft.

EXAMPLE Solution



6


2. Translate.

EXAMPLE Solution



6


3. Solve.

EXAMPLE Solution



6


3. Solve.

EXAMPLE Solution



6


4. Check. The negative integer –12 cannot be the length of a side. When x = 5, x + 7 = 12, and 52 + 122 = 132. Thus, 5 and 12 check. 5. State. The distance from the top of the ladder to the ground is 12 ft. The distance from the bottom of the ladder to the building is 5 ft.

Documents

CHAPTER · b : Factor polynomials when the terms have a common factor, factoring out the greatest common factor. c : Factor certain expressions with four terms using