Special Products on Factoring

Transcription

1 Special Products on Factoring

2 What Is This Module About? This module is a continuation of the module on polnomials. In the module entitled Studing Polnomials, ou learned what polnomials are as well as how to add, subtract, multipl and divide them. In this module, ou will learn about using special products, factoring and finding the greatest common factor. This module is made up of two lessons: Lesson 1 Special Products Lesson 2 Factoring Special Products What Will You Learn From This Module? After studing this module, ou should be able to: multipl polnomials using special products; differentiate prime from composite numbers; find the prime factors of an integer; find the greatest common factor (GCF) of a set of monomials; and factor polnomials using the various methods of factoring. Wait! Before ou proceed reading this module, be sure to have read the module entitled Studing Polnomials first. It will help ou understand this module better. Let s See What You Alread Know Before ou start studing this module, take this simple test first to find out how much ou alread know about the topics to be discussed. A. Multipl the following expressions. 1. (x + 1)(x 1) 2. (x + 2)(2x + 1) 3. (x + 1)(x + 1) B. Put a in the box if the number is composite and an 8 if it is prime

3 C. Find the prime factors of the following numbers D. Find the GCF of the following polnomials. 1. 8x 6 7 and 5 x x 5 3 and 15x 2 3 Well, how was it? Do ou think ou fared well? Compare our answers with those in the Answer Ke on page 19 to find out. If all our answers are correct, ver good! You ma still stud the module to review what ou alread know. Who knows, ou might learn a few more new things as well. If ou got a low score, don t feel bad. This onl goes to show that this module is for ou. It will help ou understand some important concepts that ou can appl in our dail life. If ou stud this module carefull, ou will learn the answers to all the items in the test and a lot more! Are ou read? You ma go now to the next page to begin Lesson 1. 2

4 LESSON 1 Special Products In the module entitled Studing Polnomials, ou learned all about polnomials. In this lesson, ou will learn more about them. After studing this lesson, ou should be able to: multipl polnomials using special products; differentiate prime from composite numbers; find the prime factors of an integer; and find the GCF of a set of monomials. Let s Read Ita, our class was assigned to paint a mural on one of the walls of the visual arts room. Can ou help me find the dimension of a square mural with an area of x 2 + 2x + 1? Sure, Eric. x 2 + 2x + 1 is the area, which means that it is the product of the dimensions of the square mural. The width and the length of the mural are the factors. To find the mural s dimensions, ou need to factor x 2 + 2x + 1. How can I factor out x 2 + 2x + 1? You can use the special product formula a 2 + 2ab + b 2 = (a + b) 2 to find its factors. a + b is the measurement of one of the mural s sides. Using that special product formula, x 2 + 2x + 1 will be equal to (x + 1) 2. Therefore, one of the sides of the square mural measures x

5 Let s Review How did Eric factor out x 2 + 2x + 1? Compare our answer with mine below. Eric was asked to get the dimensions of the square mural. Since it is a square, the area x 2 + 2x + 1 is a product of the square of one side. To find the factors, Eric used the special product formula which states that x 2 + 2ax + a = (x + a) 2. Using this, x 2 + 2x + 1 was therefore factored out into (x + 1) 2. The measure of one side then equals x + 1. _ Let s Learn Special Products Special products are ver useful in finding the products of two polnomial factors. Working with them is just like multipling polnomials. Below is a list of the different kinds of special products and examples of each kind. 1. Difference of two squares When the sum and the difference of a binomial are multiplied to one another, the product is the difference of the square of the two terms as in: Examples: (x + a)(x a) = x 2 a 2 where: x is a variable a is a constant number (x + 3)(x 3) = x = x 2 9 (2x + )(2x ) = (2x) Perfect squares = x 2 16 When a binomial is multiplied b itself it is called a perfect square. (x + a) 2 = x 2 + 2ax + a 2 (x a) 2 = x 2 2ax + a 2

6 Examples: (x + 3) 2 = x 2 + 2(3)x + 9 = x 2 + 6x + 9 (2x 1) 2 = (2x) 2 2(1)(2x) + 1 = x 2 x Sum of two cubes When the sum of two terms is multiplied b the sum of their squares minus the product of these terms, the result is the sum of their cubes. Examples: (x + a)(x 2 ax + a 2 ) = x 3 + a 3 (x + 1)(x 2 x + 1) = x (3x + 2)(9x 2 6x + ) = [3x + 2][(3x) 2 (2)(3x) + 22] = (3x) Difference of two cubes = 27x When the difference of two terms is multiplied b the sum of their squares plus the product of these terms, the result is the difference of their cubes. (x a)(x 2 + ax + a 2 ) = x 3 a 3 Examples: (x 2)(x 2 + 2x + ) = (x 2)(x 2 + 2x ) = x = x 3 8 (2x 3)(x 2 + 6x + 9) = [2x 3][(2x) 2 + (3)(2x) ) = (2x) = 8x Trinomials which are not perfect squares (ax + b)(cx + d) = acx 2 + (ad + bc)x + bd where: Examples: a and c are numerical coefficients b and d are constants (x + 2)(x + 3) = 1(1)x 2 + [1(2) + 1(3)]x + 2(3) = x 2 + 5x + 6 (x 2)(x + 3) = [x + ( 2)](x + 3) = 1(1)x 2 + [1( 2) + 1(3)]x + ( 2)(3) = x 2 + x 6 5

7 (x 2)(x 3) = [x + ( 2)][x + ( 3)] = 1(1)x 2 + [1( 2) + 1( 3)]x + ( 2)( 3) = x 2 5x 6 (2x + 3)(x + 2) = 2()x 2 + [2(2) + 3()]x + 3(2) = 8x x + 6 Let s Tr This Fill in the blanks in the following. 1. (x + 7)(x 7) = ( ) = x 2 2. (5x ) 2 = ( x) 2 ( )( )x + 2 = 25x 2 20x (3 + )( ) = (3 + )[( ) 2 ( )(3) + 2 )] = (3) 3 3 = 3 +. (3x 5)(9x x + 25) = (3x 5)[( ) 2 + (5)(3)x + 2 )] = ( ) = ( z + 1)(5z 2) = ( )( )z 2 + [( )( 2) + (1)(5)]z + (1)( ) = z 2 + ( 6 + 5)z 2 = 20z 2 2 Compare our answers with mine below. 1. (x + 7)(x 7) = (x) = 16x (5x ) 2 = (5x) 2 (5)x + 2 = 25x 2 20x (3 + )( ) = (3 + )[(3) 2 (3) + 2 ] = (3) 3 3 = (3x 5)(9x x + 25) = (3x 5)[(3x) 2 + 5(3)x ] = (3x) = 9x ( z + 1)(5z 2) = (5)z 2 + [( )( 2) + 1(5)]z + 1( 2) = 20z 2 + ( 6 + 5)z 2 = 20z 2 z 2 6

8 Let s Learn Prime and Composite Numbers Prime numbers are numbers that are divisible b onl 1 and itself. The factors of a prime number are 1 and itself. Examples of prime numbers are 3, 5, 7, 11, 13 and 17. Composite numbers are numbers that are divisible b both prime and non-prime numbers. Examples of composite numbers are 16, 18, 20 and 2. To find its prime factors, the composite number 2 is factored out as Factors are numbers which, when multiplied to each other, equals the composite number. Let s Tr This Write P in the box if the given number is prime and C if it is composite. In the blank, write the prime factors of the number. The first has been done as an example for ou = = =. 29 = = = Compare our answers with mine below. 2. P 19 = C 56 = P 29 = C 35 = C 81 = Let s Learn Factoring Polnomials Factoring is the process of getting the polnomial factors of a given number or expression. You learned how to factor out prime and composite numbers earlier. Now, ou will learn how to factor out variables. You will also learn how to factor out polnomials b getting their greatest common factor or b using special products. 7

9 Factoring a Polnomial With a Common Factor To find out how this is done look at the given example below. EXAMPLE 1 Get the factors of the expression 12x 2 2x 3 z. SOLUTION STEP 1 Factor out each term and get the GCF of the terms in the given expression. 12x 2 = x x 2x 3 z = x x x z GCF = x x = 12x 2 STEP 2 Take out 12x 2 from the polnomial b dividing each term b 12x 2. 12x x 2x x 12x 3 z =12x 2 [(12 12)x (2 2) (2 12)x (3 2) z] =12x 2 ( 2xz) Since 2z is alread prime, the polnomial 2z is called a proper factor of 12x 2 2x 3 z. The prime factors of 12x 2 2x 3 z and therefore 12x 2 and ( 2xz). Let s Tr This Find the GCF of each of the following pairs of expressions. 1. 2x 3 and 8x 2 GCF = 2. 9x 2 and 3x 3 GCF = z 6 and 10 z 5 GCF =. 36x 5 and 2x 2 GCF = and 30 6 GCF = Compare our answers with mine below. 1. 2x 3 and 8x 2 2x 3 = 2 x x x 8x 2 = x x GCF = 2x x 2 and 3x 3 9x 2 = 3 3 x x 3x 3 =3 x x GCF = 3x 2 8

10 z 6 and 10 z z 6 = 5 z z z z z z 10 z 5 = 2 5 z z z z z GCF = 5 z 5. 36x 5 and 2x and x 5 = x x x x x 2x 2 = x x GCF = 12x = = GCF = 10 6 Let s See What You Have Learned A. Multipl the following using special products. 1. (x + )(x ) 2. (2x + 5)(2x 5) 3. (3z + 2) 2. (x 2) 2 5. (z + 2)(z 2 2z + 1) 6. (2 + 3)( ) 7. (x 2)(x 2 + 2x + ) 8. (3 2)( ) 9. (2x + 1)(x 3) 10. (6x + 2)(7x + 5) B. Find the GCF of each of the following pairs of expressions. 1. x 2 3 and x GCF = 2. 28x 6 and 16x 3 5 GCF = C. Find the prime factors of the following expressions. 1. 9z 35z 6 = 2. 20x 6 25x 6 = 3. 17x x 6 8 = Compare our answers with those in the Answer Ke on pages 19 to 21. If all our answers are correct, ver good! If not, review the items ou missed before proceeding to the next lesson. 9

11 Let s Remember Special products are ver useful in finding the product of two polnomial factors. The following are some of the kinds of special products: Difference of two squares: ( )( ) x + a x a = x Perfect squares: ( x + a) = x + 2ax + a ( x a) = x 2ax + a a Sum of two cubes: ( )( ) x + a x a + a = x + a Difference of two cubes: ( )( ) x a x + ax + a 5. Trinomials which are not perfect squares: 2 ( ax + b)( cx + d ) = acx + ( ad + bc) x + bd = x a 10

12 LESSON 2 Factoring Special Products In Lesson 1, ou learned how to multipl polnomials using special products. You also learned the difference between prime and composite numbers as well as how to find the greatest common factor of given numbers or expressions. In this lesson, ou will learn how to factor out special products. You will also learn how to factor out polnomials using various methods of factoring. Let s Learn In Lesson 1, we used special products to find the products of given expressions. In this lesson, we will use special products to find the factors of a given product. 1. Factoring out a polnomial which is a difference of two squares x 2 a 2 = (x + a)(x a) EXAMPLE Factor out 2x 3 8x 2. STEP 1 STEP 2 Get the GCF of the term in the expression. Take out the GCF as in: 2x 3 8x 2 = 2x(x 2 2 ) Factor out x 2 2 which is a difference of two squares as in: x 2 2 = (x + 2)(x 2) The proper factors of 2x 3 8x 2 are 2x, (x + 2) and (x 2). When factoring using special products, the steps given above are alwas followed. But make sure that ou alread singled out the GCF before actuall using special products. 2. Factoring a perfect square trinomial Examples: x 2 + 2ax + a 2 = (x + a) 2 x 2 2ax + a 2 = (x a) 2 x x + 16 = (2x) 2 + 2()(2x) + 2 = (2x + ) 2 25x 2 30x + 9 = (5x) 2 2(3)(5x) = (5x 3) 2 18x 2 z + 12xz + 2z = 2z(9x 2 + 6x + 1) = 2z[(3x) 2 + 2(3x) + 1] = 2z(3x + 1) 2 11

13 3. Factoring a sum of two cubes Examples: x 3 + a 3 = (x + a)(x 2 ax + a 2 ) x = (x + 2)(x 2 2x ) = (x + 2)(x 2 2x + ) 8x = 8(x 3 + 8) = 8(x + 2)(x 2 2x + ). Factoring a difference of two cubes Examples: x 3 27 = x (x a)(x 2 + ax + a 2 ) = x 3 a 3 = (x 3)(x 2 + 3x ) = (x 3)(x 2 + 3x + 9) 125x 3 9 = (5x) = (5x 3)[(5x) 2 + 5(3x) ] = (5x 3)(25x x + 9) Let s Tr This Factor out the following expressions using special products x x z Compare our answers with mine below x 2 5 = 5(x 2 9) = 5(2x + 3)(2x 3) = 2( ) = 2( + 1) = 3( ) = 3[(3) 2 2(3) + 1)] = 3(3 1) 2. x 3 27 = x = (x 3)(x 2 + 3x + 9) 5. 8z = 8(z 3 + 1) = 8(z 3 + 1) = 8(z + 1)(z 2 z + 1) 12

14 Let s Learn Factoring Out Other Trinomials Which Are Not Perfect Squares In factoring out trinomials which are not perfect squares, we will not directl use special products as in the earlier section. Factoring out trinomials which are not perfect squares needs further analsis and uses the special products onl as guides in finding an expression s factors. To learn how to do this, look at the following example. EXAMPLE Find the prime factors of 30x x STEP 1 Get the GCF of the terms in the given expression. Take out the GCF as in: 30x x = 2 2 (15x x + 8) STEP 2 Factor out 15x x Make a frame wherein ou will put each of the terms. Write the literal coefficients and the signs in-between frames as in: 15x x + 8 = ( x + )( x + ) a b c d a, b, c and d are the numerical coefficients we are looking for. 2. Write down the different dual combinations of factors of the numerical coefficient of the first term and the last term. 15 = (1 15), ( 3 5) 1 and 15 or 3 and 5 are the choices that we have for a and c. You can onl choose between the two combinations. You cannot choose 15 and 3 or 15 and 5. 8 = (1 8), (2 ) 1 and 8 or 2 and are the onl choices that we have for b and d. 3. Do ou still remember the special product for trinomials which are not perfect squares? Its middle term s numerical coefficient is equal to (ad + bc). Because of this, we choose a, b, c and d in such a wa that (ad + bc) is equal to x x + 8 = ( x + )( x + ) a b c d ad + bc = 22 This can be done b trial and error. Tr all the possible combinations until the sum of ad and bc equals 22. Write the combinations in the frames ou made. Let s use (1 15) and (1 8) and the combination a = 15, c = 1, b = 8 and d = 1. 13

15 15x 8 1x a b c d Check: ad + bc = 15(1) + 8(1) = 23 not equal to 22 Let s tr another combination. Let s interchange a and c, so a = 1, c = 15, b = 8 and d = 1. 1x 8 15x a b c d Check: ad + bc = 1(1) + 8(15) = 121 not equal to 22 Let s use another combination, (3 5) and (2 ). Let a = 3, c = 5, b = 2 and d =. 3x 2 5x + + a b c d Check: ad + bc = 3() + 2(5) = 22 Therefore (3x + 2)(5x + ) are the factors of the expression 15x x + 8. Finall, the factors of 30x x are 2 2, (3x 2 + 2) and (5x + ). Let s Tr This Find the prime factors of the following expressions. 1. 3x 2 + 9x z 2 z x 2 2x Compare our answers with mine below. 1. 3x 2 + 9x + 6 = 3(x 2 + 3x + 2) = 3(x + 1)(x + 2) 2. 2z 2 z 16 = 2(z 2 2z 8) = 2(z )(z + 2) = (2 + 1)(3 + 2). 6x 2 2x 20 = 2(3x 2 x 10) = 2(3x + 5)(x 2) = (z )(6z 3) 1

16 Let s Review A. Multipl the following polnomials using special products. 1. (x + 3)(x 3) 2. (3x + 2) 2 3. ( 2)( ). (2z + 3)(z 2 6z + 9) 5. (2 + 5)(5 + 3) B. Write C in the box if the number is composite and P if it is prime C. Find the prime factors of the following numbers D. Find the GCF of the following pairs of expressions x 2 and x x 5 and 9x x 5 6 and 50x 3 3 E. Find the prime factors of the following expressions. 1. 3x x 2 12x x x x x x x x 2 + x 8 Compare our answers with those in the Answer Ke on pages 21 and

17 Let s See What You Have Learned A. Multipl the following expressions. 1. (x + )(x ) 2. (2x 1) 2 3. (x + 1)(x 2 x + 1). (2z 5)(z z + 25) 5. (5 + 7)(6 5) B. Put a in the box if the number is composite and an 8 if it is prime C. Find the prime factors of the following numbers D. Find the GCF of the following pairs of polnomials x 2 and 2 x 2. 51x 3 2 and 17x 3. 36x 2 and 2x 2 E. Find the prime factors of the following expressions. 1. 3x x x x x 2 25x x 2 19x

18 F. Solve the following word problems. 1. A carpenter built a rectangular table with an area of x 2 9 m 2. What expressions represent the table s length and width? 2. The carpenter also built a square table with an area of x x + 25 m 2. What expression represents the length of one side of the table? Compare our answers with those in the Answer Ke on pages 22 to 2. Let s Remember Special products are also used in factoring out polnomials. Factoring is the process of getting the polnomial factors of a given product. Well, this is the end of the module! Congratulations for finishing it. Did ou like it? Did ou learn anthing useful from it? A summar of its main points is given below to help ou remember them better. Let s Sum Up This module tells us that: Special products are ver useful in finding the product of two polnomial factors. The following are some of the kinds of special products: Difference of two squares: ( )( ) x + a x a = x a Perfect squares: ( x + a) = x + 2ax + a ( x a) = x 2ax + a Sum of two cubes: ( )( ) x + a x a + a = x + a 17

19 Difference of two cubes: ( )( ) 5. Trinomials which are x a x + ax + a = x a not perfect squares: ( )( ) 2 ax + b cx + d = acx + ( ad + bc ) x + bd Special products are also used in factoring out polnomials. Factoring is the process of getting the polnomial factors of a given product. What Have You Learned? A. Write P in the box if the number is prime and C if it is composite. In the blank, write the prime factors of the number B. Find the GCF of the following pairs of expressions x 3 z and 17x x 8 3 z and 70x 5 3 z C. Find the prime factors of the following expressions. 1. 2x x x x x 2 + 5x x 2 9x 2 Compare our answers with those in the Answer Ke on page 2. 18

20 Answer Ke A. Let s See What You Alread Know (pages 1 2) A. 1. (x + 1)(x 1) = x 2 x + x 1 = x (x + 2)(2x + 1) = 2x 2 + x + x + 2 = 2x 2 + 5x (x + 1)(x + 1) = x 2 + x + x + 1 = x 2 + 2x + 1 B C = 9 20 = (3 3)(2 2 5) = = 37(2 2 5) D. 1. 8x 6 7 and 5 x 2 8x 6 7 = x x x x x x 5x2 = 2 2 x x GCF = 2 2 x x = x x 5 3 and 15x 2 3 5x x 2 3 GCF = x x x x x = 3 5 x x = 3 5 x x = 15x 2 3 B. Lesson 1 Let s See What You Have Learned (page 9) A. 1. (x + )(x ) = x 2 2 = x (2x + 5)(2x 5) = (2x) = x

21 3. (3z + 2) 2 = (3z) 2 + 2(2)(3z) = 9z z +. (x 2) 2 = x 2 2(2)x = x 2 x + 5. (z + 2)(z 2 2z + ) = (z + 2)(z 2 2z ) = z = z (2 + 3)( ) = (2 + 3)[(2) 2 3(2) ] = (2) = (x 2)(x 2 + 2x + ) = (x 2)(x 2 + 2x ) = x = x (3 2)( ) = (3 2)[(3) 2 + 2(3) ] = (3) = (2x + 1)(x 3) = 2(1)x 2 + [2( 3) + 1(1)]x + 1( 3) = 2x 2 + ( 6 + 1)x + ( 3) = 2x 2 5x (6x + 2)(7x + 5) = 6(7)x 2 + [6(5) + 2(7)]x + 2(5) = 2x 2 + (30 + 1)x + 10 = 2x 2 + x + 10 B. 1. x 2 3 and x x 2 3 = x x x = 2 2 x GCF = x 2. 28x 6 and 16x x 6 = x x x x 16x 3 5 = x x x GCF = x 3 5 C. 1. 9z 35z 6 9z = z z z z 35z 6 = z z z z z z GCF = 7z 9z 35z 6 = 7z = 7z 9z 7z 2 ( 7 + 5z ) 35z + 7z 6 20

22 2. 20x 6 25x 6 20x 6 = x x x x 25x 6 = x x x x x x GCF = 5x 20x x 5. 17x x 6 8 = 5x = 5x 6 20x 5x 2 2 ( 5x ) 6 25x + 5x 17x 6 8 = 1 17 x x x x x x 21x 6 8 = 3 7 x x x x x x GCF = x x x x = x 6 8 x = x = x = x ( ) ( ) x x 8 C. Lesson 2 Let s Review (page 15) A. 1. (x + 3)(x 3) = x = x (3x + 2) 2 = (3x) 2 + 2(3x) = 9x 2 + 6x + 3. ( 2)( ) = ( 2)[() 2 + 2() ] = () = (2z + 3)(z 2 6z + 9) = (2z + 3)[(2z) 2 3(2z) ] = (2z) = 8z (2 + 5)(5 + 3) = 2(5) 2 + [2(3) + 5(5)] + 5(3) = B C C P C C 21

23 C = = = 1 39 D x 2 = x x x 2 = 2 2 x x GCF = x x 5 = x x x x 9x 3 2 = 3 3 x x x GCF = 9x x 5 6 = x x x x x 50x 3 3 = x x x GCF = 50x 3 3 E. 1. 3x 2 12 = 3(x 2 ) = 3(x 2)(x + 2) 2. 2x 2 12x 18 = 2(x 2 6x 9) = 2[x 2 2(3)x 3 2 ] = 2(x 3) x x + 75 = 3(x x + 25) = 3[x 2 + 2(5)x ] = 3(x + 5) 2. 2x = 2(x 3 + 1) = 2(x + 1)(x 2 x + 1) 5. x 3 32 = (x 3 8) = (x 2)(x 2 + 2x + ) 6. 6x x + 15 = (2x +3)(3x + 5) 7. 2x 2 + x 8 = 2(x 2 + 2x 2) = 2(x + 6)(x ) Let s See What You Have Learned (pages 16 17) A. 1. (x + )(x ) = x (2x 1) 2 = (2x) 2 2(1)(2x) = x 2 x (x + 1)(x 2 x + 1) = x = x 3 1. (2z 5)(z z + 25) = (2z 5)[(2z) 2 + 5(2z) ] = (2z) = 8z (5 + 7)(6 5) = 5(6) 2 + [5( 5) + 7(6)] + 7( 5) =

24 B = = = = = C = = = D x 2 and 2 x 12x 2 = x x 2 x = 2 2 x GCF = x 2. 51x 3 2 and 17x 51x 3 2 = 17 3 x x x 17x = 17 1 x GCF = 17x 3. 36x 2 and 2x 2 36x 2 = x x x x 2x 2 = x x GCF = 12x 2 E. 1. 3x 3 2 = 3(x 3 8) = 3(x ) = 3(x 2)(x 2 + 2x + ) 2. 27x = (3x) = (3x + )[(3x) 2 (3x) + 2 ] = (3x + )(9x 2 12x + 16) 3. 2x x + 30 = 2(x 2 + 8x + 15) = 2(x + 3)(x + 5). 10x 2 25x 15 = 5(2x 2 5x 3) = 5(2x + 1)(x 3) 5. 6x 2 19x + 10 = (3x 2)(2x 5) 23

25 F. 1. x 2 9 = (x + 3)(x 3) The expressions that represent the length and width of the rectangular table are (x + 3) and (x 3) meters. 2. x x + 25 = x 2 + 2(5)x = (x + 5) 2 The expression that represents one side of the square table is (x + 5) meters. D. What Have You Learned? (page 18) A. 1. C 50 = C 9 = P 67 = P 53 = P 37 = 1 37 B x 3 z = 3 17 x x x z 17x 6 7 = 17 x x x x x x GCF = 17x x 8 3 z = x x x x x x x x z 70x 5 3 z = x x x x x z GCF = 10x 5 3 z C. 1. 2x 2 32 = 2(x 2 16) = 2(x )(x + ) 2. 3x x = 3(x x + 36) = 3[x 2 + 2(6)x ] = 3(x + 6) x 3 + = (27x 3 + 1) = [(3x) 3 + 1] = [(3x + 1)[(3x) 2 1(3x) ] = (3x + 1)(9x 2 3x + 1). 15x 2 + 5x 20 = 5(3x 2 + x ) = 5(3x )(x + 1) 5. 10x 2 9x 2 = 7(10x 2 7x 6) = 7(5x 6)(2x + 1) 2

26 References The Math Forum. (2001). Math Forum Internet Mathematics Librar. swathmore.edu/librar/topics/polnomials/. June 16, 2001, date accessed. The Math Forum. (2001). The Math Forum Ask Dr. Math: Questions and Answers From Our Archives. June 16, 2001, date accessed. Network Solutions, Inc. (2001). The Mental Edge. 2/ 01/chapter A. June 16, 2001, date accessed. 25