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Mathematics A30 Module 1 Lesson 2 Mathematics A30 Factoring Polynomials 69 Lesson 2 Factoring Polynomials Introduction Factoring polynomials is a very important concept in Algebra. The first section will be a review of factoring. This factoring will be expanded to include the factoring of special polynomials. In past courses polynomials have been linear or quadratic. In this course cubic polynomials will be introduced with the concept of factoring sums and differences of cubes. The concepts that are developed in this lesson will be the foundation for higher learning and problem solving in the real world. Mathematics A30 71 Lesson 2 Objectives After completing this lesson, you will be able to • expand polynomials. • factor trinomials. • factor the difference of squares. • factor the difference of squares of special polynomials. • factor the sum and difference of cubes. Mathematics A30 73 Lesson 2 2.1 Review of Factoring The process of determining polynomials whose product equals a given polynomial is called factoring. A review of some of the concepts that have already been taught in earlier mathematics courses is necessary at the beginning of this lesson. In order to understand factoring well, it is necessary to know how to expand (or multiply) polynomials and to recognize the fact that expanding and factoring are inverse operations, the same way that adding and subtracting are inverse operations. The following illustration shows expanding and factoring with numbers. Expanding (multiplying) 3 7 21 Factoring 3 7 21 3 and 7 are factors of 21. Expanding Polynomials Expanding polynomials may involve one or many of the following multiplication processes. Mathematics A30 75 Lesson 2 Multiplying a Polynomial by a Monomial The distributive property is used to multiply a polynomial by a monomial. The Distributive Property The product of a and (b c) is given by: a(b c) ab ac or (b c)a ba ca . The product of a and (b c) is given by: a(b c) ab ac or (b c)a ba ca . Example 1 Expand 2 x 5 xy x 2 1 . Solution: Write the expression. 2 x 5 xy x 2 1 Apply the distributive property. = – 2 x ( 5 x yx –+ 1 ) Multiply. = 2 x(5 xy) 2 x( x 2 ) 2 x(1) Simplify. = 10 x 2 y + 2 x 3 2 x 2 Multiplying Two Binomials The distributive property is used twice to multiply two binomials together. Distributive Property ( a b)( c d ) ( a b)c ( a b)d ac bc ad bd Mathematics A30 76 ( a b) is distributed. c and d are distributed. Lesson 2 The distributive property is often remembered as the FOIL method. Multiply the: F O I L First terms of the binomials. Outside terms of the binomials. Inside terms of the binomials. Last terms of the binomials. Example 2 Multiply x 3m m x . Solution: Write the expression. x 3m m x Apply the FOIL Method. O F x 3m m x I L Simplify. F O I L xm x 2 3 m 2 3 mx = 2 mx x 2 + 3 m 2 Squaring a Binomial • • This is when a binomial expression is multiplied by itself. Here is an example of the distributive property. The FOIL method could also be used. Example 3 Expand y 2 2 . Mathematics A30 77 Lesson 2 Solution: y 2 2 Write the expression. Expand. = y 2 y 2 Use the distributive property. = y y 2 2 y 2 Simplify. = y2 2 y 2 y 4 = y2 4 y 4 When a binomial is squared, a pattern shows how the answer can be expressed in a simple form. (a + b)2 = • • • a2 + 2ab + b2 Square the first term. Square the last term. The middle term is 2 times the first and last terms. Factoring Polynomials Remember that factoring polynomials is the opposite of multiplying or expanding polynomials. Greatest Common Factor Polynomials are factored using the reverse of the distributive property. Whenever you are asked to factor a polynomial, the first question should always be: "Is there a common factor in all of the terms?". The greatest common factor (GCF) of two numbers can be found by factoring each of the numbers, finding all the factors that are common to both and multiplying these common factors together. The concept can be shown first with numbers. Mathematics A30 78 Lesson 2 The greatest common factor of 18 and 42 can be determined by factoring each number separately. 18 = 2 × 3 × 3 42 = 2 × 3 × 7 The greatest common factor of 18 and 42 is 2 3 = 6 . The same is true when finding the greatest common factor of algebraic terms in a polynomial. Example 4 Factor 15 x 2 y 30 xy 35 xy 2 . Solution: Factor each algebraic term. 15 x 2 y = 3 5 x x y 30 xy = 2 3 5 x y 35 xy 2 = 5 7 x y y Determine the common factors. • • • The common factors are 5, x and y. Multiply these common factors together. The GCF is 5xy. Write the original expression. 15 x 2 y 30 xy 35 xy 2 Factor the common factor 5xy from each term. = 5 xy3 x 5 xy6 5 xy(7 y) Use the distributive property. = 5 xy3 x 6 7 y Check by expanding. 5 xy3 x 6 7 y = 15 x 2 y 30 xy 35 xy 2 Mathematics A30 79 () Lesson 2 Example 5 Factor 30 m 2 40 m 2 n 50 m 2 n 2 . Solution: Determine the GCF of the terms. • The GCF is 10 m 2 . Write the original expression. 30 m 2 40 m 2 n 50 m 2 n 2 Factor the common factor 10m 2 from each term. = 10 m 2 3 10 m 2 4 n 10 m 2 5 n 2 Use the distributive property. = 10 m 2 3 4 n 5 n 2 Check the answer by expanding. Factoring by Grouping Sometimes the GCF in all the terms of a polynomial is 1. In this case it is necessary to group terms together that do have a common factor. In most cases this type of factoring occurs when there are four (4) terms in the polynomial. Two terms are grouped together. When factoring by grouping, the steps are: • Group terms together that have a common factor. • Factor a common factor from each group. • The terms inside the two sets of brackets must be the same. This will be one factor of the polynomial. • The two common factors from each group are combined to make the other factor of the polynomial. Mathematics A30 80 Lesson 2 Example 6 Factor a 2 ab ac bc . Solution: Write the original expression a 2 + ab + ac + bc . Group terms with a common factor. = a 2 + ab + ac + bc Factor a common factor from each group. = aa + b + c a + b Write as two factors. = a + c a + b a + c a + b Check by expanding. (FOIL) = a 2 + ab + ac + bc () Another possible way of grouping the above terms is ( a 2 ac) ( ab bc) . Does this give the same factors? Try it. Example 7 Factor 4 x 2 7 yz 2 xy + 14 xz . Solution: Write the original expression. 4 x 2 7 yz 2 xy + 14 xz 4 x Group terms with a common factor. Factor a common factor from each group. = = 2 xy 7 yz + 14 xz 2 x2 x y + 7 z y + 2 x Write as two factors. = 2 x + 7 z 2 x y 2 x + 7 z 2 x Check by expanding. (FOIL) = • 2 y 4 x 2 xy 14 xz 7 zy () 2 Sometimes you may need to try different combinations of grouping in order to have the same factor inside both brackets. Mathematics A30 81 Lesson 2 Difference of Squares A difference of squares may occur when you are asked to factor a polynomial that is a binomial. There is a special way of factoring an expression where each of the two terms is a perfect square and there is a negative sign between the two terms. When this pattern occurs it is called factoring a difference of squares. Some examples of perfect squares are: 25 a2 16b2c2 5×5 a×a 4bc × 4bc Some examples of binomials which are a difference of perfect squares are: x 2 9 y 2 4 a 2 25 b 2 81 To factor a difference of squares, follow these steps: • Find the principal (positive) square root of the first term. • Find the principal (positive) square root of the last term. • Multiply the sum of these two square roots by the difference of these two square roots. Difference of Squares In general: Mathematics A30 2 2 a b = a + b a b 82 Lesson 2 Example 8 Factor y 2 81 . Solution: Find the principle square root of the first term. 2 y =y Find the principle square root of the last term. 81 = 9 Write the original expression. y Write the product of the sum and the difference of the roots. = 2 81 (y+9 )(y–9 ) su m o fth e sq u a rero o ts d iffe re n c eo fth e sq u a rero o ts Example 9 Factor 64 a 2 9 b 2 . Solution: • • The square root of the first term is 8a. The square root of the last term is 3b. Write the original expression. Write in factored form. 64 a 2 9 b 2 = 8 a + 3b8 a 3b 8 a + Check by expanding. 3b8 a 3b = 64 a 2 24 ab + 24 ab 9 b 2 = 64 a 2 9 b 2 () Mathematics A30 83 Lesson 2 Exercise 2.1 1. Expand the following polynomials. a) b) c) d) e) f) g) h) i) j) k) 2. r 7 2 k 10 2 2 a 3 2 (3 x 5 y) 2 Factor the following polynomials. a) b) c) d) e) f) g) 3. 7 a 6 a 5 3 xy8 x 4 y 2 xy 1 4 m 2 6 m 3 n 2 9 x 5 x 8 2 y 12 y 1 3m 4 n 2m n p 11 p 9 3 a 2 6 ab m2 9 n 2 2 n np 2 p 9x2 1 x 2 121 25 x 2 36 y 2 5 a 2 80 b 2 h) i) j) k) l) m) n) 34 u 9 24 u 7 18 u 5 21 w 4 14 w 2 35 2 nr 3 p 2 6 pr pn k 2 5 k mk 5 m 5 x 5 y x 2 xy 3 p 2 r 2 12 p 2 r 5 pr 2 20 pr 2 x 2 y 10 xy 6 x 2 30 x What is the first question that you ask yourself when you are factoring polynomials? Mathematics A30 84 Lesson 2 2.2 Factoring Trinomials Perfect Trinomial Square When two identical binomials are multiplied together, they produce a perfect trinomial square. To identify a perfect trinomial square, the following conditions must be true. • The first and last terms are both perfect squares such that the principle square root of either term can be found exactly. • The middle term is equal to twice the product of the square roots of the first and last term. A Perfect Trinomial Square has the form a 2 + 2ab + b 2 or a 2 2ab + b 2 . In general: a 2 + 2ab + b 2 = = a 2 2ab + b 2 = = a + ba + b a + b 2 a a ba b b 2 Example 1 Factor the perfect trinomial square, y 2 10 y 25 . Solution: • • • The square root of the first term is y. The square root of the term is 5. Two times the product last 5 and y is 10y. Write the expression. y2 + 10y + 25 Factor. Mathematics A30 = (y + 5)(y + 5) 85 Lesson 2 Example 2 Factor 9 x 2 48 x + 64 . Solution: • • The square root of the first term is 3x. The square root of the last term is 8. 23 x 8 = 48 x Multiply twice the product of the square roots. This is the middle term; therefore, this is a perfect trinomial square. Write the original expression. 9 x 2 48 x 64 Factor. 3 x 8 3 x 8 Use the negative signs in the answer because the middle term of the original expression is negative. Factored polynomials can always be checked by expanding. Mathematics A30 86 Lesson 2 Example 3 Factor 25 a 2 + 40 ab + 16 b 2 . Solution: Write the original expression. 25 a 2 + 40 ab + 16 b 2 5 a 4 b5 a 4 b Factor. 25 a 2 is the square of 5a. 25 a 2 40 ab 16 b 2 5 a 4 b5 a 4 b The middle term is 25 a 4 b . 16 b 2 is the square of 4b. The most important thing is to be able to identify a perfect trinomial square. Once you have done this, then use the pattern for factoring. Factoring Trinomials Trinomials can be factored by using a specific pattern. When the coefficient of the x2 term is equal to 1, use the following pattern. • Find two numbers such that: • their sum is the coefficient of the 2nd term. • their product is the last term. Mathematics A30 87 Lesson 2 Example 4 Factor x 2 6 x 8. Solution: Find two numbers whose sum is 6 and whose product is 8. • The product of 2 and 4 is 8 (the last term of the trinomial). • The sum of 2 and 4 is 6 (the coefficient of the middle term of the trinomial). Factor. sum x 2 6 x 8 x 4 x 2 product Example 5 Factor y 2 + 13 y + 30 . Solution: Find two numbers whose product is 30 and whose sum is 13. • The two numbers are 3 and 10. Write the original expression. 2 y + 13 y + 30 = y + 3 y + 10 Factor. Trinomials can also be in the form x 2 bx c where the middle term is negative. When looking for factors, use the same pattern of factoring the product, or last term, but remember that when the middle term is negative and the last term is positive, the two factors will both be negative. Mathematics A30 88 Lesson 2 Example 6 Factor g 2 21 g + 80 . Solution: Find two numbers whose product is 80 and whose sum is 21 . • The two numbers are 5 and 16 . Write the original expression. g 2 21 g + 80 = g 5 g 16 Factor. g 5 g 16 = g 16 g 5 g 80 = g 2 21 g 80 () Check by expanding. 2 If the last term of the trinomial is negative, then the factors of the trinomial will have different signs. Example 7 Factor a 2 4 ab 32 b 2 . Solution: Find two numbers whose product is 32 and whose sum is 4 . • The two numbers are 4 and 8 . Write the original expression. a 2 4 ab 32 b 2 Remember to factor the b2 in the last term. = a 8 ba 4 b Factor. Mathematics A30 89 Lesson 2 There are many different ways to factor a trinomial where the coefficient of the x2 term is not equal to 1. One method of factoring these trinomials is to use trial and error. • This can be done by finding the factors of the first term, finding the factors of the last term and writing down all the possible combinations. Expand each of the expressions to see which set of factors gives you the correct middle term. Trinomials of this form can also be factored using decomposition. • A pattern is used to factor trinomials where the x2 term is not equal to 1. Factoring by grouping is one of the steps. The steps are: • Multiply the coefficients of the first and last terms. • Find two numbers whose product is this number and whose sum is the coefficient of the middle term. • These two numbers now become the middle two terms. • Group the first two terms together and the last two terms together. • Remove a common factor from each. • Factor by grouping. Example 8 Factor 15 y 2 7 y 4 by the trial and error method. Solution: • Pairs of factors of 15 are: (15, 1) (1, 15) (5, 3) (3, 5) Mathematics A30 • 90 Pairs of factors of 4 are: (4 , 1) (1, 4 ) (4, 1 ) ( 1 , 4) (2, 2 ) ( 2 , 2) Lesson 2 Create a table of possible solutions. Middle term 11y 59 y 11 y 59y 28y 28 y 7 y 17 y 7y 17y 4y 4y (15 y 4 )(1 y 1) (15 y 1)(1 y 4 ) (15 y 4 )(1 y 1) (15 y 1)(1 y 4 ) (15 y 2)(1 y 2) (15 y 2)(1 y 2) (5 y 4 )( 3 y 1) (5 y 1)( 3 y 4 ) 5 y 4 3 y 1 (5 y 1)( 3 y 4 ) (5 y 2)( 3 y 2) (5 y 2)( 3 y 2) correct middle term Therefore, 15 y 2 7 y 4 (5 y 4 )( 3 y 1) . Example 9 Factor 15 y 2 7 y 4 by decomposition. Solution: Multiply the first and last terms. 15 ( 4 ) 60 Find two numbers whose product is 60 and whose sum is 7 (coefficient of middle term). • These two numbers are 12 and 5 . Rewrite the expression with these two numbers as coefficients for the middle terms. = 15y2+7y–4 15y2+12y–5y–4 Group terms together and factor. = 3 y(5 y 4 ) 1(5 y 4 ) = (3 y 1)( 5 y 4 ) Check by expanding. (3 y 1)( 5 y 4 ) = 15 y 2 12 y 5 y 4 = 15 y 2 7 y 4 () Mathematics A30 91 Lesson 2 Example 10 Factor 6 m 2+ mn 15 n 2 by decomposition. Solution: 6 15 = 90 Multiply the first and last terms. Find two numbers whose product is 90 and whose sum is 1. • These two numbers are 9 and 10 . Rewrite the expression with these two numbers as coefficients for the middle terms. 6 m 2 + mn 15 n 2 = Group terms together and factor. 6 m 2 9 mn + 10 mn 15 n 2 = Write in factored form. = 3m + 3 m (2 m 3 n ) 5 n (2 m 3 n ) 5n 2m 3n The greatest common factor can also be determined from mathematical expressions. The greatest common factor of two expressions can be found by factoring each of the expressions, finding all the factors that are common to both and multiplying these common factors together. Example 11 Find the greatest common factor from the expressions 6 c 3 6 c 2 36 c and 3 c2 27 c 66 . Solution Factor each expression. See if you can get the listed factors for each expression by using the decomposition method! 6 c 3 6 c 2 36 c 6 c ( c 2 c 6 ) 3 2 c ( c 2) ( c 3) 3 c 2 27 c 66 3 ( c 11 ) ( c 2) The greatest common factor of 6 c 3 6 c 2 36 c and 3 c2 27 c 66 is 3 ( c 2) 3( c 2) . Mathematics A30 92 Lesson 2 Exercise 2.2 1. Determine which of the following are perfect trinomial squares. State the reason why if they are not. a) b) c) d) e) f) g) h) 2. a 2 8 a 16 x 2 28 x 196 1 2a a 2 100 10 w w 2 4 w 2 36 wx 81 x 2 m 2 16 m 64 9k 2 6k 1 a 2 24 a 169 In the following trinomials, determine the two numbers needed to factor the expression by first finding the sum of the middle term and the product of the last term. a) b) c) d) e) x 13 x 42 p 2 70 p 600 32 12 m m 2 c 2 15 c 250 15 2w w 2 2 Sum ________________ ________________ ________________ ________________ ________________ Product ________________ ________________ ________________ ________________ ________________ Numbers ______________ ______________ ______________ ______________ ______________ 3. Factor 2 y 2 8 y 5 . What is your conclusion? 4. Factor the following polynomials. (Remember to first look for a common factor.) a) b) c) d) e) f) x2 6x 9 5 n 2 15 n 20 x2 3x 2 3 y2 7 y 6 g) h) i) j) k) 4 a 2 12 ab 9 b 2 2 y y 20 m 2 12 m 36 6 n 2 n 15 6 a 2 x 2 19 ax 2 10 x 2 6 a 4 19 a 3 10 a 2 y4 2 y2 1 l) 4 y 4 37 y 2 9 Mathematics A30 93 Lesson 2 2.3 Factoring Special Polynomials Difference of Squares of Special Polynomials The two processes of grouping and factoring a difference of squares will be incorporated into this section. In Section 2.1, when grouping within a polynomial, pairs of terms were grouped together. It is also possible to group together three terms that form a special polynomial. This special polynomial will usually be a trinomial square. It is therefore necessary to be able to recognize a perfect trinomial square at a glance. This is the first step. The second step involves factoring a difference of squares. Example 1 Factor x 2 4 x 4 25 . Solution: Write the original expression. x 2 4 x 4 25 Group together the terms that form a special polynomial. = x 2 4 x 4 25 Factor the perfect trinomial square. = x + 2 2 25 Factor as a difference of squares. = ( x + 2) + 5 ( x + 2) 5 = x + 2 + 5 x + 2 5 Simplify. = ( x 7)( x 3) Mathematics A30 Perfect Trinomial Square 94 Lesson 2 Example 2 Factor x 2 6 x 9 x 4 2 . Solution: x 2 6 x 9 x 4 2 Write the original expression. Group together terms that form a special polynomial. = Factor the perfect trinomial square. = x 2 6 x 9 x 4 2 Perfect Trinomial Square Factor as a difference of squares. x 3 2 x 4 2 = ( x 3) ( x 4)( x 3) ( x 4) Simplify. = = = x 3 x 4 x 3 x 4 2 x 7 (1) 2x 7 Mental Math Calculations • Multiply 33 27 mentally. • Think of the question as: 30 3 30 3 • Hint: Square the first terms, square the last terms and subtract. • Now try: 42 58 55 65 26 34 71 69 Exercise 2.3 1. Factor the following special polynomials. a) x 2 y2 2 y 1 b) c) d) e) f) x 2 2 xy y 2 z 2 a 2 10 a 25 b 2 9 m 2 6 mn n 2 36 25 x 2 4 xy 4 y 2 t 2 4 t 4 81 s 2 Mathematics A30 95 Lesson 2 2.4 Sum and Difference of Cubes The "cube" of a number is a number b is b b b b 3 . • • 4 4 4 4 64 3 xxx x 3 The cube root of a number is a factor of the number that when multiplied by itself three times, the result is the original number. The cube root of a term is a factor of the term that can be multiplied by itself three times and the result is the original term. • • • The cube root of 8 is 2 because 8 2 2 2 . The cube root of a 3 is a because a 3 a a a . The cube root of 64 t 3 is 4 t because 64 t 3 4 t 4 t 4 t . The volume of a cube visually shows the concept of cubes and cube roots. Volume V x x x x 3 Length of a side 3 V x 3 is the symbol for cube root. A sum or a difference of two cubed terms forms a special polynomial. Example • Sum of Cubes: • Difference of Cubes: 3 3 x + y 3 3 x y a 3 27 8 m 3 125 These two special polynomials can be factored using a special pattern. Sum of Cubes Difference of Cubes x 3 y3 x + y x 2 xy y 2 Mathematics A30 x 3 y3 x y x 2 xy y 2 96 Lesson 2 The volume of a cube with sides equal to x cm can illustrate the process that determines the rule or pattern for factoring a difference of cubes. • • • Start with the larger cube whose volume is x 3 cm 3 . Remove a smaller cube whose volume is y3 cm 3 . The remaining volume will be x 3 y 3 . y (x – y ) y B x C y y A y x x The portion of the cube that is remaining can be divided into three sections. 1) The volume of section A would be: x x x y cm 3 = x2 x y 2) The volume of section B would be: x y x y cm 3 = xyx y 3) The volume of section C would be: y y x y cm 3 = y2 x y The total volume would be: 3 2 2 3 x y x x y xy x y y x y x y x 2 xy y 2 Mathematics A30 97 Lesson 2 Example 1 Factor a 3 8b 3 . Solution: • • • • The cube root of a 3 is a. x=a The cube root of 8b 3 is 2b. y = 2b x 3 y3 x y x 2 xy y 2 Write the formula for a difference of cubes. a 2 b a 2 ab 4 b a 3 8 b 3 a 2 b a 2 2 ab 2 b 2 Substitute the values, x = a, y = 2b. 2 2 Remember to square the entire term. Example 2 Factor 125 p3 1 . Solution: • • • • The cube root of 125 p3 is 5p. x 5p The cube root of 1 is 1. y 1 3 2 3 2 x y x y x xy y Write the formula for a sum of cubes. 5 p 1 25 p 5 p 1 125 p3 1 5 p 1 5 p 2 5 p 12 Substitute the values, x = 5p, y = 1. Mathematics A30 98 2 Lesson 2 Example 3 Factor 64 m 3 729 n 3 . Solution: • • • • The cube root of 64 m 3 is 4m. x 4m The cube root of 729 n 3 is 9n. y 9n 3 2 3 2 x y x y x xy y Write the formula for a difference of cubes. Substitute the values. 64 m 3 729 n 3 4 m 9 n 4 m 2 36 mn 9 n 2 4 m 9 n 16 m 2 36 mn 81 n 2 Exercise 2.4 1. Determine the cube root of the following terms. If the term is not a perfect cube, state that. a) b) c) d) e) f) 2. 8 x3 125 m 3 42 y 3 27 a 6 9 r 1 000 b 8 Use the distributive property to show that each of the statements is true. a) b) a ba 2 ab b 2 a3 b3 x y x 2 xy y2 x 3 y3 Mathematics A30 99 Lesson 2 3. Factor completely. a) b) c) d) e) x 3 64 a 3 8b 3 4 t 3 108 729 s 3 125 t 3 1 000 m 3 n 3 p 3 f) x 6 y6 g) 1 27 b 3 64 8 Mathematics A30 100 Lesson 2 Answers to Exercises Exercise 2.1 Mathematics A30 1. a. b. c. d. e. f. g. h. i. j. k. 42 a 2 35 a 24 x 2 y 12 xy 2 6 x 2 y 2 3 xy 24 m 3 12 m 2 n 2 36 m 2 x 2 3 x 40 4 y2 1 6 m 2 11 mn 4 n 2 p 2 20 p 99 r 2 14 r 49 k 2 20 k 100 4 a 2 12 a 9 9 x 2 30 xy 25 y 2 2. a. b. c. d. e. f. g. 3 a( a 2 b) (m 3)( m 3) n ( n 2) p( n 2) ( n 2)( n p) (3 x 1)( 3 x 1) ( x 11)( x 11 ) (5 x 6 y)( 5 x 6 y) 5( a 4 b)( a 4 b) h. 34 u 9 24 u 7 18 u 5 2u 5 (17 u 4 12 u 2 9) i. 7(3w 4 2w 2 5) j. 2 nr 6 pr 3 p 2 pn = 2 r ( n 3 p) p(3 p n ) = (2 r p)( 3 p n ) k. k 2 5 k mk 5 m = k ( k 5) m ( k 5) = ( k m )( k 5) l. 5 x 5 y x 2 xy = 5( x y) x( x y) = (5 x )( x y) 101 Lesson 2 Exercise 2.2 m. pr (3 pr 12 p 5 r 20 ) = pr 3 p(r 4) 5(r 4) = pr (3 p 5)( r 4 ) n. 2 x( xy 5 y 3 x 15 ) = 2 xxy 3 x 5 y 15 = 2 xx( y 3) 5( y 3) = 2 x( x 5)( y 3) 3. “Is there a common factor in all of the terms?” 1. a. b. c. d. e. f. g. h. yes yes yes no yes no yes no a. b. c. d. e. Sum 13 70 12 15 2 2. 3. 2(10 )( w ) 20 w this should be the middle term. The last term cannot be a negative. 2(13)(a) = 26a this should be the middle term. Product 42 600 32 250 15 Numbers 6, 7 60 , 10 8, 4 25 , 10 5, 3 Factoring by decomposition, the first step is: 2 5 10 • • Find two numbers whose product is 10 and whose sum is 8 . • This is impossible. This expression cannot be factored further. It is therefore prime. 4. Mathematics A30 a. b. c. d. e. ( x 3) 2 5(n 2 3n 4) 5(n 4)( n 1) ( x 2)( x 1) (3 y 2)( y 3) (2 a 3 b)( 2 a 3 b) 102 Lesson 2 Exercise 2.3 1. f. ( y 5)( y 4 ) g. h. i. j. k. l. (m 6) 2 (2 n 3)( 3 n 5) x 2 (2 a 5)( 3 a 2) a 2 (2 a 5)( 3 a 2) ( y 2 1)( y 2 1) ( y 1) 2 ( y 1) 2 (4 y 2 1)( y 2 9) (2 y 1)( 2 y 1)( y 3)( y 3) a. x 2 ( y 2 2 y 1) x 2 ( y 1) 2 ( x y 1)( x y 1) b. ( x 2 2 xy y 2 ) z 2 ( x y) 2 z 2 ( x y z )( x y z ) d. e. f. ( a 2 10 a 25 ) b 2 ( a 5) 2 b 2 ( a 5 b)( a 5 b) (3 m n 6)( 3 m n 6) (5 x 2 y)( 5 x 2 y) (t 2 9 s)( t 2 9 s) a. b. c. d. e. 2x 5m 3 42 , 42 is not a perfect cube. 3a 2 r3 f. 10 3 b 8 , b 8 is not a perfect cube. a. b. ( a b)( a 2 ab b 2 ) a 3 a 2 b ab 2 ba 2 ab 2 b 3 a 3 b3 ( x y)( x 2 xy y 2 ) x 3 x 2 y xy 2 yx 2 xy 2 y 3 x3 y3 a. b. c. ( x 4)( x 2 4 x 16 ) ( a 2b)( a 2 2 ab 4 b 2 ) 4(t 3 27 ) 4(t 3)( t 2 3t 9) c. Exercise 2.4 1. 2. 3. Mathematics A30 103 Lesson 2 d. e. f. (9 s 5t )(81 s 2 45 st 25 t 2 ) (10 mn p)(100 m 2 n 2 10 mnp p 2 ) ( x 2 y 2 )( x 4 x 2 y 2 y 4 ) ( x y)( x y)( x 4 x 2 y 2 y 4 ) g. Mathematics A30 2 1 3 b 1 3 b 9 b 4 4 2 16 8 104 Lesson 2 Mathematics A30 Module 1 Assignment 2 Mathematics A30 105 Assignment 2 Staple here to the upper left corner of your assignment Before submitting your assignment, please complete the following procedures: Print your name and address, with postal code. This address sheet will be used when mailing back your corrected assignment. 1. Write your name and address and the course name and assignment number in the upper right corner of the first page of each assignment. Student Number 2. Number all the pages and place them in order. 3. Complete the required information details on this address sheet. 4. Staple this address sheet to the appropriately numbered assignment. Use one address sheet for each assignment. 5. Staple the appropriately numbered Assignment Submission Sheet to the upper left corner, on top of this address sheet. Name 8404 Course Number 02 Assignment Number Street Address or P.O. Box Mathematics A30 Course Title City/Town, Province Postal Code: Country Distance-Learning Teacher’s Name Mark Assigned: Assignment 2 Values (40) A. Multiple Choice: Select the correct answer for each of the following and place a () beside it. 1. ____ ____ ____ ____ 2. a. b. c. d. 5 xy 2 x 2 3 y 2 6 xy 2 x 2 6 y 6 xy 2 x 2 3 y 2 xy 2 a. b. c. d. 4a2 4a2 4a2 4a2 49 49 14 a 49 28 a 49 The greatest common factor of each of the terms in the expression, 12 a 2 b 3 16 ab 3 8 a 3 b is ***. ____ ____ ____ ____ Mathematics A30 6 a 2 15 ax 6 a 2 x 2 5 a 8 ax 6 ax 2 6 a 2 15 ax 6 a 2 x 2 6 a 2 5 x 2 ax 2 The expanded form of 2 a 7 is ***. ____ ____ ____ ____ 4. a. b. c. d. The expanded form of 2 x y 3 y x is ***. ____ ____ ____ ____ 3. The expanded form of 3 a 2 a 5 x 2 ax 2 is ***. a. b. c. d. 16 a 3 b 3 4 ab 4 a 3 b3 48 a 3 b 3 109 Assignment 2 5. To factor 7 x 7 xc 2 b 2 bc , a correct grouping of the terms is ***. ____ ____ ____ ____ 6. 8. ____ ____ a. b. ____ c. ____ d. 2 x 12 x a 4 xx a a4 x a 2 x a 2 2 12 x a 2 a. b. c. d. 0 21 ab 42 ab 42 ab One of the factors of 2 x 2 + 11 x 6 is ***. a. b. c. d. x 6 x 6 2x 1 2x 6 Another way of writing 8 x 2 3 xy 5 y 2 is ***. ____ ____ ____ ____ Mathematics A30 1 3 xa 3 xa 1 xa 1 The middle term of the expanded form of 7 a 3 b is ***. ____ ____ ____ ____ 10. a. b. c. d. One factored form of 4 x 2 4 ax a 2 is ***. ____ ____ ____ ____ 9. 7 xc 2bc 7 x 2b 7 xc 2b 7 x 2bc 7 x 2bc 7 xc 2b 7 xc 2bc 7 x 2b One of the factors of 9 x 2 a 2 1 is ***. ____ ____ ____ ____ 7. a. b. c. d. a. b. c. d. 8 x 2 18 xy 15 xy 5 y 2 8 x 2 4 xy xy 5 y 2 8 x 2 8 xy 3 xy 5 y 2 8 x 2 2 xy xy 5 y 2 110 Assignment 2 11. 12. Another way of writing x 2 6 x 11 is ***. ____ ____ ____ a. b. c. x 9 2 2 x 10 x 1 ____ d. x 3 14. a. b. c. d. ____ ____ ____ a. b. c. ____ d. 2 2 9 x5 27 x 2 27 x 6 9x2 5ay 2 25 ay 2 5ay 3 125 a 9 y 18 A factor of x 1 2 3 is ***. 3 a. b. c. d. x 1 x 3 x2 4 x 1 x2 5 A factor of 8 x 3 27 y 6 is ***. ____ ____ ____ ____ Mathematics A30 6x 2 1 The cube root of 125 a 3 y 6 is ***. ____ ____ ____ ____ 15. 2 The cube of 3 x 2 is ***. ____ ____ ____ ____ 13. x a. b. c. d. 4 x 2 6 xy 2 9 y 2 2x 3 y2 2x 3 y 4 x 2 6 xy 2 9 y 4 111 Assignment 2 16. A factor of (2 x)3 ( x 2)3 is ***. ____ ____ ____ ____ 17. 3x2 3x 1 3x2 3x 1 1 1 a. b. c. d. 1, 3, 1 2, 3, 1 2, 3, 1 2 x 2 , 3x, 1 a. b. c. d. 1 + x3 1 + x + x2 1 x x2 1+x One of the factors of 12 x 2 8 x 15 is ***. ____ ____ ____ ____ Mathematics A30 x3 x2 x3 x3 One of the factors of x 2 x 5 is ***. ____ ____ ____ ____ 20. a. b. c. d. If ax2 + 3x + c = 2x2 + bx 1 , then a, b, c are equal to ***. ____ ____ ____ ____ 19. 3x 2 4 x 2 2 x( x 2) ( x 2) 2 x2 x 2 The expanded form of ( x 1)3 is ***. ____ ____ ____ ____ 18. a. b. c. d. a. b. c. d. 6x 5 2x 3 2x + 3 3x + 5 112 Assignment 2 Answer Part B and Part C in the space provided. Evaluation of your solution to each problem will be based on the following. • A correct mathematical method for solving the problem is shown. • The final answer is accurate and a check of the answer is shown where asked for by the question. • The solution is written in a style that is clear, logical, well organized, uses proper terms, and states a conclusion. B. 1. Completely factor each of the following: (2) a. a 3 216 (2) b. 8 a 3 b3 c6 (2) c. 16 x 4 y 2 Mathematics A30 113 Assignment 2 (2) d. 2 y3 5 y2 3 y (2) e. x 3 2 2 x 1 2 (2) f. 36 x 4 r 42 x 3 r 2 18 x 2 r 3 Mathematics A30 114 Assignment 2 2. Completely factor each of the trinomials and show a check of the factors by multiplying. (3) a. 12x 2 2x 14 (3) b. x 2 35 xy 300 y 2 (3) c. 6 x 2 30 x 900 (3) d. 8 x 2 24 x 1 440 Mathematics A30 115 Assignment 2 (3) e. 2 x 4 10 x 3 72 x 2 (3) f. 16 x 2 34 xy 15 y 2 3. Find the greatest common factor from each of the following expressions. (3) a. 48 y 2 ( y 2)( y 3) 2 ; 32 y 3 ( y 3) (3) b. 6 x 2 6 x; 4 x 2 24 x 20 Mathematics A30 116 Assignment 2 (4) (6) c. C. 16 ab 2 a 5 b 2 ; a 3 b 8 b 1. R r Cross section of pipe The formula for the area of a circle is A r 2 , where r represents the radius of the circle. Show that the cross sectional area of the pipe shown in the diagram is times the product of the sum of the two radii and the difference of the two radii. Mathematics A30 117 Assignment 2 (6) 2. The volume of a sphere of radius R is given by V 4 R 3 3 A hollow spherical ball is made so that the outside radius is R and the inside radius is r. Write an expression for the volume of material used for the ball using the thickness T of the shell in the expression. (8) 3. (STUDENT JOURNAL) For this lesson, write no more than a one page summary of the material in this lesson. This summary is to be written in such a way that it will be useful to you for studying for examinations. (100 ) Mathematics A30 118 Assignment 2