Section A3 Polynomials: Factoring APPLICATIONS. A22 Appendix A A BASIC ALGEBRA REVIEW


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1 A Appendi A A BASIC ALGEBRA REVIEW C In Problems 53 56, perform the indicated operations and simplify. 53. ( ) 3 ( ) 3( ) ( ) 3 ( ) 3( ) {[ ( )] ( )( 3)} 56. {( 3)( ) [3 ( )]} 57. Show by eample that, in general, (a b) a b. Discuss possible conditions on a and b that would make this a valid equation. 58. Show by eample that, in general, (a b) a b. Discuss possible conditions on a and b that would make this a valid equation. 59. If you are given two polynomials, one of degree m and the other of degree n, m n, what is the degree of the sum? 60. What is the degree of the product of the two polynomials in Problem 59? 6. How does the answer to Problem 59 change if the two polynomials can have the same degree? 6. How does the answer to Problem 60 change if the two polynomials can have the same degree? 65. Coin Problem. A parking meter contains nickels, dimes, and quarters. There are 5 fewer dimes than nickels, and more quarters than dimes. If represents the number of nickels, write an algebraic epression in terms of that represents the value of all the coins in the meter in cents. Simplify the epression. 66. Coin Problem. A vending machine contains dimes and quarters only. There are 4 more dimes than quarters. If represents the number of quarters, write an algebraic epression in terms of that represents the value of all the coins in the vending machine in cents. Simplify the epression. 67. Packaging. A spherical plastic container for designer wristwatches has an inner radius of centimeters (see the figure). If the plastic shell is 0.3 centimeters thick, write an algebraic epression in terms of that represents the volume of the plastic used to construct the container. Simplify the epression. [Recall: The volume V of a sphere of 4 radius r is given by V r 3.] cm cm APPLICATIONS 63. Geometry. The width of a rectangle is 5 centimeters less than its length. If represents the length, write an algebraic epression in terms of that represents the perimeter of the rectangle. Simplify the epression. 64. Geometry. The length of a rectangle is 8 meters more than its width. If represents the width of the rectangle, write an algebraic epression in terms of that represents its area. Change the epression to a form without parentheses. 68. Packaging. A cubical container for shipping computer components is formed by coating a metal mold with polystyrene. If the metal mold is a cube with sides centimeters long and the polystyrene coating is centimeters thick, write an algebraic epression in terms of that represents the volume of the polystyrene used to construct the container. Simplify the epression. [Recall: The volume V of a cube with sides of length t is given by V t 3.] Section A3 Polynomials: Factoring Factoring What Does It Mean? Common Factors and Factoring by Grouping Factoring SecondDegree Polynomials More Factoring Factoring_What Does It Mean? A factor of a number is one of two or more numbers whose product is the given number. Similarly, a factor of an algebraic epression is one of two or more algebraic epressions whose product is the given algebraic epression. For eample,
2 A3 Polynomials: Factoring A , 3, and 5 are each factors of ( )( ) ( ) and ( ) are each factors of 4. The process of writing a number or algebraic epression as the product of other numbers or algebraic epressions is called factoring. We start our discussion of factoring with the positive integers. An integer such as 30 can be represented in a factored form in many ways. The products 6 5 ( )(0)(6) all yield 30. A particularly useful way of factoring positive integers greater than is in terms of prime numbers. DEFINITION PRIME AND COMPOSITE NUMBERS An integer greater than is prime if its only positive integer factors are itself and. An integer greater than that is not prime is called a composite number. The integer is neither prime nor composite. Eamples of prime numbers:, 3, 5, 7,, 3 Eamples of composite numbers: 4, 6, 8, 9, 0, Eplore/Discuss In the array below, cross out all multiples of, ecept itself. Then cross out all multiples of 3, ecept 3 itself. Repeat this for each integer in the array that has not yet been crossed out. Describe the set of numbers that remains when this process is completed This process is referred to as the sieve of Eratosthenes. (Eratosthenes was a Greek mathematician and astronomer who was a contemporary of Archimedes, circa 00 B.C.) A composite number is said to be factored completely if it is represented as a product of prime factors. The only factoring of 30 given above that meets this condition is
3 A4 Appendi A A BASIC ALGEBRA REVIEW Factoring a Composite Number Write 60 in completely factored form. Solution or or Write 80 in completely factored form. Notice in Eample that we end up with the same prime factors for 60 irrespective of how we progress through the factoring process. This illustrates an important property of integers: THEOREM THE FUNDAMENTAL THEOREM OF ARITHMETIC Each integer greater than is either prime or can be epressed uniquely, ecept for the order of factors, as a product of prime factors. We can also write polynomials in completely factored form. A polynomial such as 6 can be written in factored form in many ways. The products ( 3)( ) ( 3) ( )( ) all yield 6. A particularly useful way of factoring polynomials is in terms of prime polynomials. 3 DEFINITION PRIME POLYNOMIALS A polynomial of degree greater than 0 is said to be prime relative to a given set of numbers if: () all of its coefficients are from that set of numbers; and () it cannot be written as a product of two polynomials, ecluding and itself, having coefficients from that set of numbers. Relative to the set of integers: is prime 9 is not prime, since 9 ( 3)( 3) [Note: The set of numbers most frequently used in factoring polynomials is the set of integers.]
4 A3 Polynomials: Factoring A5 A nonprime polynomial is said to be factored completely relative to a given set of numbers if it is written as a product of prime polynomials relative to that set of numbers. Our objective in this section is to review some of the standard factoring techniques for polynomials with integer coefficients. In Chapter 3 we treated in detail the topic of factoring polynomials of higher degree with arbitrary coefficients. Common Factors and Factoring by Grouping The net eample illustrates the use of the distributive properties in factoring. Factoring Out Common Factors Factor out, relative to the integers, all factors common to all terms. (A) 3 y 8 y 6y 3 (B) (3 ) 7(3 ) Solutions (A) 3 y 8 y 6y 3 (y) (y)4y (y)3y y( 4y 3y ) (B) (3 ) 7(3 ) (3 ) 7(3 ) ( 7)(3 ) Factor out, relative to the integers, all factors common to all terms. (A) 3 3 y 6 y 3y 3 (B) 3y(y 5) (y 5) 3 Factoring Out Common Factors Factor completely relative to the integers: 4( 7)( 3) ( 7) ( 3) Solution 4( 7)( 3) ( 7) ( 3) ( 7)( 3)[( 3) ( 7)] ( 7)( 3)( 6 7) ( 7)( 3)(4 ) 3 Factor completely relative to the integers. 4( 5)(3 ) 6( 5) (3 ) Some polynomials can be factored by first grouping terms in such a way that we obtain an algebraic epression that looks something like Eample, part B. We can then complete the factoring by the method used in that eample.
5 A6 Appendi A A BASIC ALGEBRA REVIEW 4 Factoring by Grouping Factor completely, relative to the integers, by grouping. (A) (B) wy wz y z (C) 3ac bd 3ad bc Solutions (A) (3 6) (4 8) 3( ) 4( ) (3 4)( ) (B) wy wz y z (wy wz) (y z) w( y z) ( y z) (w )( y z) Group the first two and last two terms. Remove common factors from each group. Factor out the common factor ( ). Group the first two and last two terms be careful of signs. Remove common factors from each group. Factor out the common factor (y z). (C) 3ac bd 3ad bc In parts A and B the polynomials are arranged in such a way that grouping the first two terms and the last two terms leads to common factors. In this problem neither the first two terms nor the last two terms have a common factor. Sometimes rearranging terms will lead to a factoring by grouping. In this case, we interchange the second and fourth terms to obtain a problem comparable to part B, which can be factored as follows: 3ac bc 3ad bd (3ac bc) (3ad bd) c(3a b) d(3a b) (c d)(3a b) 4 Factor completely, relative to the integers, by grouping. (A) (B) pr ps 6qr 3qs (C) 6wy z y 3wz Factoring SecondDegree Polynomials We now turn our attention to factoring seconddegree polynomials of the form 5 3 and 3y y into the product of two firstdegree polynomials with integer coefficients. The following eample will illustrate an approach to the problem.
6 A3 Polynomials: Factoring A7 5 Factoring SecondDegree Polynomials Factor each polynomial, if possible, using integer coefficients. (A) 3y y (B) 3 4 (C) 6 5y 4y Solutions (A) 3y y ( y)( y)?? Put in what we know. Signs must be opposite. (We can reverse this choice if we get 3y instead of 3y for the middle term.) Now, what are the factors of (the coefficient of y )? ( y)( y) 3y y ( y)( y) y The first choice gives us 3y for the middle term close, but not there so we reverse our choice of signs to obtain 3y y ( y)( y) (B) 3 4 ( )( ) Signs must be the same because the third term is positive and must be negative because the middle term is negative ( )( ) 4 4 ( )( 4) 5 4 ( 4)( ) 5 4 No choice produces the middle term; hence 3 4 is not factorable using integer coefficients. (C) 6 5y 4y ( y)( y)???? The signs must be opposite in the factors, because the third term is negative. We can reverse our choice of signs later if necessary. We now write all factors of 6 and of 4:
7 A8 Appendi A A BASIC ALGEBRA REVIEW and try each choice on the left with each on the right a total of combinations that give us the first and last terms in the polynomial 6 5y 4y. The question is: Does any combination also give us the middle term, 5y? After trial and error and, perhaps, some educated guessing among the choices, we find that 3 matched with 4 gives us the correct middle term. Thus, 6 5y 4y (3 4y)( y) If none of the 4 combinations (including reversing our sign choice) had produced the middle term, then we would conclude that the polynomial is not factorable using integer coefficients. 5 Factor each polynomial, if possible, using integer coefficients. (A) 8 (B) 5 (C) 7y 4y (D) 4 5y 4y More Factoring The factoring formulas listed below will enable us to factor certain polynomial forms that occur frequently. SPECIAL FACTORING FORMULAS. u uv v (u v) Perfect Square. u uv v (u v) Perfect Square 3. u v (u v)(u v) Difference of Squares 4. u 3 v 3 (u v)(u uv v ) Difference of Cubes 5. u 3 v 3 (u v)(u uv v ) Sum of Cubes The formulas in the bo can be established by multiplying the factors on the right. CAUTION Note that we did not list a special factoring formula for the sum of two squares. In general, u v (au bv)(cu dv) for any choice of real number coefficients a, b, c, and d.
8 A3 Polynomials: Factoring A9 6 Using Special Factoring Formulas Factor completely relative to the integers. (A) 6y 9y (B) 9 4y (C) 8m 3 (D) 3 y 3 z 3 Solutions (A) 6y 9y ()(3y) (3y) ( 3y) (B) 9 4y (3) (y) (3 y)(3 y) (C) 8m 3 (m) 3 3 (m )[(m) (m)() ] (m )(4m m ) (D) 3 y 3 z 3 3 (yz) 3 ( yz)( yz y z ) 6 Factor completely relative to the integers. (A) 4m mn 9n (B) 6y (C) z 3 (D) m 3 n 3 Eplore/Discuss (A) Verify the following factor formulas for u 4 v 4 : u 4 v 4 (u v)(u v)(u v ) (u v)(u 3 u v uv v 3 ) (B) Discuss the pattern in the following formulas: u v (u v)(u v) u 3 v 3 (u v)(u uv v ) u 4 v 4 (u v)(u 3 u v uv v 3 ) (C) Use the pattern you discovered in part B to write similar formulas for u 5 v 5 and u 6 v 6. Verify your formulas by multiplication. We complete this section by considering factoring that involves combinations of the preceding techniques as well as a few additional ones. Generally speaking, When asked to factor a polynomial, we first take out all factors common to all terms, if they are present, and then proceed as above until all factors are prime.
9 A30 Appendi A A BASIC ALGEBRA REVIEW 7 Combining Factoring Techniques Factor completely relative to the integers. (A) (B) 6 9 y (C) 4m 3 n m n mn 3 (D) t 4 6t (E) y 4 5y Solutions (A) (9 4) (3 )(3 ) (B) 6 9 y ( 6 9) y Group the first three terms. ( 3) y Factor 6 9. [( 3) y][( 3) y] ( 3 y)( 3 y) Difference of squares (C) 4m 3 n m n mn 3 mn(m mn n ) (D) t 4 6t t(t 3 8) t(t )(t t 4) (E) y 4 5y (y 3)( y 4) (y 3)( y )( y ) 7 Factor completely relative to the integers. (A) (B) y 4y 4 (C) 3u 4 3u 3 v 9u v (D) 3m 4 4mn 3 (E) Answers to Matched Problems (A) 3y( y y ) (B) (3y )(y 5) 3. ( 5)(3 )( 7) 4. (A) ( 5)( 3) (B) ( p 3q)(r s) (C) (3w )(y z) 5. (A) ( )( 6) (B) Not factorable using integers (C) ( y)( 4y) (D) (4 y)( 4y) 6. (A) (m 3n) (B) ( 4y)( 4y) (C) (z )(z z ) (D) (m n)(m mn n ) 7. (A) 3( 4)( 4) (B) ( y )( y ) (C) 3u (u uv 3v ) (D) 3m(m n)(m mn 4n ) (E) (3 )( )( ) EXERCISE A3 A In Problems 8, factor out, relative to the integers, all factors common to all terms m 4 9m 3 3m y 0 y 5y u 3 v 6u v 4uv ( ) 3( ) 6. 7m(m 3) 5(m 3) 7. w( y z) ( y z) 8. a(3c d ) 4b(3c d ) In Problems 9 6, factor completely relative to integers y 6y 5y 5. 6m 0m 3m y 3y 6y 4. 3a ab ab 8b
10 A3 Polynomials: Factoring A ac 3bd 6bc 4ad 6. 3pr qs qr 6ps In Problems 7 8, factor completely relative to the integers. If a polynomial is prime relative to the integers, say so y y 9. 4y y 0. u uv 5v. 4. m 6m m 6n 4. w y 5. 0y 5y 6. 9m 6mn n 7. u 8 8. y 6 B In Problems 9 44, factor completely relative to the integers. If a polynomial is prime relative to the integers, say so z 8z y 3 y 48y y 8y y 34. 4y y s 7st 3t 36. 6m mn n y 9y u 3 v uv m 3 6m 5m m 3 n 3 4. r 3 t c a 3 Problems are calculusrelated. Factor completely relative to the integers (3 5)( 3) 4(3 5) ( 3) 46. ( 3)(4 7) 8( 3) (4 7) (9 ) (9 ) ( 7) 4 3 ( 7) ( )( 5) 4( ) ( 5) 50. 4( 3) 3 ( ) 3 6( 3) 4 ( ) C In Problems 57 7, factor completely relative to the integers. In polynomials involving more than three terms, try grouping the terms in various combinations as a first step. If a polynomial is prime relative to the integers, say so a 3 a a 60. t 3 t t 6. 4(A B) 5(A B) ( y) 3( y) m 4 n y 4 3y s 4 t 4 8st 66. 7a a 5 b m mn n m n 68. y y y 69. 8a 3 8a( 8 6) 70. 5(4 y 9y ) 9a b a 4 a b b 4 a b APPLICATIONS 73. Construction. A rectangular opentopped bo is to be constructed out of 0inchsquare sheets of thin cardboard by cutting inch squares out of each corner and bending the sides up as indicated in the figure. Epress each of the following quantities as a polynomial in both factored and epanded form. (A) The area of cardboard after the corners have been removed. (B) The volume of the bo. 0 inches 0 inches In Problems 5 56, factor completely relative to the integers. In polynomials involving more than three terms, try grouping the terms in various combinations as a first step. If a polynomial is prime relative to the integers, say so. 5. (a b) 4(c d ) 5. ( ) 9y 53. am 3an bm 3bn 54. 5ac 0ad 3bc 4bd y 4y 56. 5u 4uv v
11 A3 Appendi A A BASIC ALGEBRA REVIEW 74. Construction. A rectangular opentopped bo is to be constructed out of 9 by 6inch sheets of thin cardboard by cutting inch squares out of each corner and bending the sides up. Epress each of the following quantities as a polynomial in both factored and epanded form. (A) The area of cardboard after the corners have been removed. (B) The volume of the bo. Section A4 Rational Epressions: Basic Operations Reducing to Lowest Terms Multiplication and Division Addition and Subtraction Compound Fractions We now turn our attention to fractional forms. A quotient of two algebraic epressions, division by 0 ecluded, is called a fractional epression. If both the numerator and denominator of a fractional epression are polynomials, the fractional epression is called a rational epression. Some eamples of rational epressions are the following (recall, a nonzero constant is a polynomial of degree 0): In this section we discuss basic operations on rational epressions, including multiplication, division, addition, and subtraction. Since variables represent real numbers in the rational epressions we are going to consider, the properties of real number fractions summarized in Section A play a central role in much of the work that we will do. Even though not always eplicitly stated, we always assume that variables are restricted so that division by 0 is ecluded. Reducing to Lowest Terms We start this discussion by restating the fundamental property of fractions (from Theorem 3 in Section A): FUNDAMENTAL PROPERTY OF FRACTIONS If a, b, and k are real numbers with b, k 0, then ka kb a b ( 3) ( 3) 0, 3 Using this property from left to right to eliminate all common factors from the numerator and the denominator of a given fraction is referred to as reducing
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