Module - Factoringtlment.nait.ca/tlm40FileStore/Math/Asset/25000/index_htm.25029/A32... ·...

LAST REVISED November, 2008

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Algebra Module A32

Factoring - 1

Module A32 – Factoring - 1

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Factoring - 1 Statement of Prerequisite Skills Complete all previous TLM modules before beginning this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale Why is it important for you to learn this material? Factoring is the reverse of finding a product. Some factors are so common it is useful to recognize them by sight. This module will introduce you to strategies for approaching common factoring problems.

Learning Outcome When you complete this module you will be able to… Factor a variety of algebraic expressions.

Learning Objectives 1. Identify the factors of a polynomial that has a common monomial factor. 2. Identify the factors of ac + bc + ad + bd. 3. Identify the factors of a2 − b2. 4. Identify the factors of a2 − b2 where a and/or b is a binomial. 5. Identify the factors of a2 + 2ab + b2. 6. Identify the factors of a2 − 2ab + b2.

Connection Activity Recall from the products module that (a – b)(a + b) = a2 – b2. Knowing this it is easy to recognize the factors of a2 – b2. They are (a – b) and (a + b). Can you think of the factors of other common products that you have encountered?

http://www.microsoft.com/windows/ie/default.asp

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Module A32 − Factoring - 1 2

OBJECTIVE ONE When you complete this objective you will be able to… Identify the factors of a polynomial that has a common monomial factor.

Exploration Activity The greatest common factor of a set of terms is the largest term that will divide evenly into every term in the set. Given an algebraic expression such as ax − bx, x is a factor common to both terms, therefore:

ax − bx = x(a − b)

EXAMPLE 1 Factor the following. 2x2 + 4y = _____________ SOLUTION: The number 2 will divide evenly into each term 2x2 and 4y. Therefore 2 is a common factor and can be factored. 2x2 + 4y = 2(x2 + 2y) Checking your answer you get: 2(x2 + 2y) = 2x2 + 4y

EXAMPLE 2 Factor the following. 2x2

- 6x = ______________ SOLUTION: The term 2x will divide evenly into each term 2x2 and 6x. Therefore 2x is a common factor and can be factored. 2x2 − 6x = 2x(x − 3) Checking your answer you get: 2x(x − 3) = 2x2 − 6x


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EXAMPLE 3 Factor the following. 4x3y + 4xy2 + 4xy = ____________ SOLUTION: The term 4xy will divide evenly into each of the terms 4x3y , 4xy2 and 4xy. Therefore 4xy is a common factor and can be factored. 4x3y + 4xy2 + 4xy = 4xy(x2 + y + 1) Checking your answer you get: 4xy(x2 + y + 1) = 4x3y + 4xy2 + 4xy


Experiential Activity One Factor the following expressions completely and check your answers. 1. 3x + 9 2. 8y − 6 3. x2 + xy 4. ax2 − ay 5. 21x2 + 14x + 7 6. 4x3y − 8x2y2 − 4xy3 7. 3x3y3 + 6x4y3 + 3x2y2 Show Me. 8. l00x3y2 − l0x2y2 − 5xy3

Experiential Activity One Answers 1. 3(x + 3) 2. 2(4y − 3) 3. x(x + y) 4. a(x2 − y) 5. 7(3x2 + 2x + 1) 6. 4xy(x2 − 2xy − y2) 7. 3x2y2(2x2y + xy + 1) 8. 5xy2(20x2 − 2x − y)


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OBJECTIVE TWO When you complete this objective you will be able to… Identify the factors of ac + bc + ad + bd.

Exploration Activity Factoring expressions of the type ac + bc + ad + bd is done by grouping the first two terms and the last two terms. We get the following expression:

(ac + bc) + (ad + bd) From the first grouping we can remove a c as a common factor, while from the second grouping we can remove a d as a common factor. The expression now takes the following form:

c(a + b) + d(a + b) Note: (a + b) is a common factor found in both groupings. If we remove it as a common factor we complete the factoring as follows:

ac + bc + ad + bd = c(a + b) + d(a + b) = (a + b)(c + d)

EXAMPLE 1 Factor the following. 3cx + 6dx + 4cy + 8dy = ____________ SOLUTION: Step 1: Grouping the terms we get:

= (3cx + 6dx) + (4cy + 8dy) Step 2: Find a common factor in each grouping:

= 3x(c + 2d) + 4y(c + 2d) Step 3: Look for a common binomial factor:

= (c + 2d)(3x + 4y)


Step 4: You should always check using FOIL:

= (c + 2d)(3x + 4y) = 3cx + 4cy + 6dx + 8dy

Step 5 Therefore the factors are:

= (c + 2d)(3x + 4y)

EXAMPLE 2 Factor the following. l0rs −15rt + 8ps − 12pt = ____________ SOLUTION: Step 1: Grouping the terms we get:

= (10rs − 15rt) + (8ps − 12pt) Step 2: Find a common factor in each grouping:

= 5r(2s − 3t) + 4p(2s − 3t) Step 3: Look for a common binomial factor:

= (2s − 3t)(5r + 4p) Step 4: You should always check using FOIL:

= (2s − 3t)(5r + 4p) = l0rs + 8ps − 15rt − 12pt


= (2s − 3t)(5r + 4p)


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EXAMPLE 3 Factor the following. 6x2 − 4ax − 9bx + 6ab = ____________ SOLUTION: Step 1: Grouping the terms we get:

= (6x2 − 4ax) − (9bx − 6ab) ← watch the signs Step 2: Find a common factor in each grouping:

= 2x(3x − 2a) − 3b(3x − 2a) Step 3: Look for a common binomial factor:

= (3x − 2a)(2x − 3b) Step 4: You should always check using FOIL:

= (3x − 2a)(2x − 3b) = 6x2 − 4ax − 9bx + 6ab


= (3x − 2a)(2x − 3b)


Experiential Activity Two Factor the following expressions completely and check your answers.

1. ay + 4y + 2a + 8 2. 5pr − ps + 20qr − 4qs 3. 4x2 + 9x − 4xy − 9y 4. 15rs − 40rt + 9sx − 24tx 5. 2ax − 9bx − 6ay + 27by 6. 3x2 − 4x + 3xy − 4y 7. a3 − a2b + ab2 − b3 Show Me.

Experiential Activity Two Answers 1. (a + 4)(y + 2) 2. (5r − s)(p + 4q) 3. (4x + 9)(x − y) 4. (3s − 8t)(5r + 3x) 5. (2a − 9b)(x − 3y) 6. (3x − 4)(x + y) 7. (a − b)(a2 + b2)


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OBJECTIVE THREE When you complete this objective you will be able to… Identify the factors of a2 − b2.

Exploration Activity The expression (a2 − b2), is a difference of squares. The square root of the first term a2 is a while the square root of the second term b2 is b. Therefore the factors are the sum of the square roots times the difference of the square roots. The factors of (a2 − b2) are (a + b) and (a − b) and so: a2 − b2 = (a + b)(a − b)

EXAMPLE 1 Factor the following. 4x2 − 25y2 = ____________ SOLUTION: Step 1: Recognizing the expression as a difference of squares you get:

4x2 − 25y2 = (2x + 5y)(2x − 5y) Step 2: Checking the answer using FOIL you get:

= 4x2 + l0xy − l0xy − 25y2 = 4x2 − 25y2

EXAMPLE 2 Factor the following.

16r2 − 49s2 = ____________

SOLUTION: Step 1: Recognizing the expression as a difference of squares you get:

16r2 − 49s2 = (4r + 7s)(4r − 7s) Step 2: Checking the answer using FOIL you get:

= 16r2 + 28rs − 28rs − 49s2 = 16r2 − 49s2


EXAMPLE 3 Factor the following. 9c2 − 100d2 = __________ SOLUTION: Step 1: Recognizing the expression as a difference of squares you get:

9c2 − 100d2 = (3c + l0d)(3c − l0d)

Step 2: Checking the answer using FOIL you get:

= 9c2 + 30cd − 30cd − 100d2 = 9c2 − 100d2

Sometimes it is necessary to factor out a common factor in order to identify the expression as a difference of squares. Study the following example to see how this is done.

EXAMPLE 4 Factor the following. 3x4 − 75 = ___________ SOLUTION: Step 1: Factoring a 3 from each term you get:

3x4 − 75 = 3(x4 − 25) Step 2: Recognizing the second factor as a difference of squares you get:

x4 − 25 = (x2 + 5)(x2 − 5) don’t forget to include the 3 that you factored out in your final answer = 3(x2 + 5)(x2 − 5)

Step 3:Checking the answer using FOIL you get:

= 3(x4 + 5x2 − 5x2 − 25) = 3x4 − 75


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Experiential Activity Three Factor the following expressions completely and check your answers. 1. 121x2 − 9 2. 25a2 − 81b2 3. 9x2 − y2 4. 4a2b2 − 49x2y2 5. 144x2 − 25 6. 49p2 − 121r2s4 7. m4n4 − 81 8. 2x2 − 50 9. 20x3 − 45x 10. 16x4 − 16 Show Me.

Experiential Activity Three Answers 1. (11x + 3)(11x − 3) 2. (5a + 9b)(5a − 9b) 3. (3x + y)(3x − y) 4. (2ab + 7xy)(2ab − 7xy) 5. (12x + 5)(12x − 5) 6. (7p + 11rs2)(7p − 11rs2) 7. (m2n2 + 9)(mn − 3)(mn + 3) 8. 2(x + 5)(x − 5) 9. 5x(2x + 3)(2x − 3) 10. 16(x2 + 1)(x + 1)(x − 1)

Module A32 − Factoring - 1

OBJECTIVE FOUR When you complete this objective you will be able to… Identify the factors of a2 − b2 where a and/or b is a binomial.

Exploration Activity The expression (w + x)2 − (y + z)2 is a difference of squares. The first factor (w + x)2 is a

perfect square, taking the square root we get: ( ) ( xwxw +=+ 2 ) . The second factor

(y + z)2 is also a perfect square, taking the square root we get: ( ) ( zyzy +=+ 2 ) . The two factors are separated by a minus sign therefore the rule for factoring difference of squares can be used. Therefore:

a2 − b2 = (a + b)(a − b) is a model for: (w + x)2 − (y + z)2 = [(w + x) + (y + z)][(w + x) − (y + z)] =(w + x + y + z)(w + x − y – z)

EXAMPLE 1 Factor the following. (x + 2)2 − (x + 3)2 SOLUTION: The first term (x + 2)2 is a perfect square with a square root of (x + 2). The second term (x + 3)2 is a perfect square with a square root of (x + 3). Therefore when we factor this expression we get:

(x + 2)2 − (x + 3)2 = [(x + 2) + (x + 3)][(x + 2) − (x + 3)] This can be simplified, by collecting like terms. Watch the − sign in the second factor.

[(x + 2) + (x + 3)][(x + 2) − (x + 3)]

= (x + 2 + x + 3)(x + 2 − x − 3)

= (2x + 5)(−1)

= −1(2x + 5) Unfortunately, the solution cannot be checked by foil.

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EXAMPLE 2 Factor the following. (x2 + 1)2 − x2 = __________ SOLUTION:

The first term (x2 + 1)2 is a perfect square with a square root of (x2 + 1). The second term x2 is a perfect square with a square root of x. Therefore when we factor this expression we get: (x2 + 1)2 − x2 = [(x2 + 1) − x][(x2 + 1) + x] This can be rewritten as: [(x2 + 1) − x][(x2 + 1) + x] = (x2 − x + 1)(x2 + x + 1)

EXAMPLE 3 Factor the following.

( ) ___________9

242

22 =+

−xba

SOLUTION:

The first term 4a2b2 is a perfect square with a square root of 2ab.

The second term ( )9

2 2+x is a perfect square with a square root of ( )3

2+x .

Therefore when we factor this expression we get:

( ) ( ) ( )⎟⎠⎞

⎜⎝⎛ +

−⎟⎠⎞

⎜⎝⎛ +

+=+

−3

223

229

242

22 xabxabxba

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Experiential Activity Four Factor the following expressions completely and check your answers. 1. (3a − 5b)2 − 4c2 2. 25r2 − (4s + 3t)2 3. (5x + 2y)2 − z2 4. 16p2 − (5q − r)2 5. (x + 2)2 − 81y2 6. (2x − 1)2 − (3y + 1)2

7. 251

94 2 −x

8. 6(a − b)2 − 24(c + d)2 Show Me.

Experiential Activity Four Answers 1. (3a − 5b − 2c)(3a − 5b + 2c) 2. (5r − 4s − 3t)(5r + 4s + 3t) 3. (5x + 2y − z)(5x + 2y + z) 4. (4p − 5q + r)(4p + 5q − r) 5. (x + 9y + 2)(x − 9y + 2) 6. (2x − 3y − 2)(2x + 3y)

7. ⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ −

51

32

51

32 xx

8. 6(a − b + 2c + 2d)(a − b − 2c − 2d)

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OBJECTIVE FIVE When you complete this objective you will be able to… Identify the factors of a2 + 2ab + b2.

Exploration Activity The expression (a2 + 2ab + b2), is a perfect square. That is, it has two identical factors. Therefore the factors of (a2 + 2ab + b2) are (a + b) and (a + b). You must identify the expression as a perfect square using the following procedure: Step 1: Take the square root of the first term and the square root of the last term. Step 2: Does 2 × square root of the first term × square root of the last term = middle

term? Yes: the expression is a perfect square. No: the expression is not a perfect square. If it does not, another factoring technique must be used.

Step 3: Therefore once you identify the expression as a perfect square you can use the

following formula to factor the expression. a2 + 2ab + b2 = (a + b)(a + b) or (a + b)2

EXAMPLE 1 Factor the following. 9x2 + 30x + 25 = __________ SOLUTION: Step 1: Check to see if the expression is a perfect square.

9x2 + 30x + 25 = 29 3 and 25x x= = 5

Step 2: Does 2 · 3x · 5 = middle term?

Yes, so it is a perfect square. Step 3: Writing the factors you get:

= (3x + 5)(3x + 5) Checking the answer using FOIL you get:

= 9x2 + 15x + 15x + 25 = 9x2 + 30x + 25

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EXAMPLE 2 Factor the following. 16x2 + 56xy + 49y2 = ________________ SOLUTION: Step 1: Check to see if the expression is a perfect square.

16x2 + 56xy + 49y2 yyxx 749 and 416 22 ===

Step 2: Does 2 · 4x · 7y = middle term Yes, so it is a perfect square. Step 3: Writing the factors you get:

= (4x + 7y)(4x + 7y) Checking the answer using FOIL you get:

= 16x2 + 28xy + 28xy + 49y2 = 16x2 + 56xy + 49y2

EXAMPLE 3 Factor the following. 36p2q2 + 132pqrs + 121r2s2 = ____________ SOLUTION: Step 1: Check to see if the expression is a perfect square.

36p2q2 + 132pqrs + 121r2s2 rssrpqqp 11121and 636 2222 ==

Step 2: Does 2 · 6pq · 11rs = middle term Yes, so it is a perfect square. Step 3: Writing the factors you get:

= (6pq + 11rs)(6pq + 11rs)

Checking the answer using FOIL you get:

= 36p2q2 + 66pqrs + 66pqrs + 121r2s2 = 36p2q2 + 132pqrs + 121r2s2

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Experiential Activity Five Factor the following expressions completely and check your answers. 1. 9x2 + 6x + 1 2. 4x2 + 12x + 9 3. 25x2 + 20xy + 4y2 4. 9y2 + 48y + 64 5. 25x2 + 60x + 36 6. 16x4 + 72x2 + 81 7. a2x2 + 2abxy + b2y2 8. 2x3y3 + 16x2y4 + 32xy5 Show Me.

Experiential Activity Five Answers 1. (3x + 1)2 2. (2x + 3)2 3. (5x + 2y)2 4. (3y + 8)2 5. (5x + 6)2 6. (4x2 + 9)2 7. (ax + by)2 8. 2xy3(x + 4y)2


OBJECTIVE SIX When you complete this objective you will be able to… Identify the factors of a2 − 2ab + b2.

Exploration Activity The trinomial a2 − 2ab + b2 is also a perfect square because it has two identical factors. The only difference between this trinomial and the one in the previous objective is the sign of the second term is negative. Therefore the sign of the factors changes to a negative. Therefore once you identify the expression as a perfect square you can use the following formula to factor the expression.

a2 − 2ab + b2 = (a − b)(a − b) or (a − b)2

EXAMPLE 1 Factor the following. 16x2 − 24x + 9 = ___________ SOLUTION: Step 1: Check to see if the expression is a perfect square.

16x2 − 24x + 9 39and 416 2 == xx

Step 2: Does 2 · 4x · 3 = middle term Yes, so it is a perfect square. Step 3: Writing the factors you get:

= (4x − 3)(4x − 3) Checking the answer using FOIL you get:

= 16x2 − 12x − 12x + 9 = 16x2 − 24x + 9

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EXAMPLE 2 Factor the following. 49x2 − 70xy + 25y2 = ___________ SOLUTION: Step l: Check to see if the expression is a perfect square. Step 2: Writing the factors you get:

= (7x − 5y)(7x − 5y) Checking the answer using FOIL you get:

= 49x2 − 35xy − 35xy + 25y2 = 49x2 − 70xy + 25y2

EXAMPLE 3 Factor the following. 4p2 − 100pq + 625q2 = __________ SOLUTION: Step l: Check to see if the expression is a perfect square. Step 2: Writing the factors you get:

= (2p − 25q)(2p − 25q)

Checking the answer using FOIL you get:

= 4p2 − 50pq − 50pq + 625q2 = 4p2 − 100pq + 625q2


Experiential Activity Six Factor the following expressions completely and check your answers.

1. 16x2 − 8x + 1 2. 25x2 − 40x + 16 3. x2 − 18x + 81 4. 100x2 − 60x + 9 5. 36x2 − 60xy + 25y2 6. 4a2x2 − 12a2x + 9a2 7. 98x2y2 − 168xy2 +72y2 Show Me. 8. 16x4 − 72x2 + 81

Experiential Activity Six Answers 1. (4x − 1)2 2. (5x − 4)2 3. (x − 9)2 4. (10x − 3)2 5. (6x − 5y)2 6. a2(2x − 3)2 7. 2y2(7x − 6)2 8. (2x + 3)2(2x − 3)2

Practical Application Activity Complete the factoring – 1 module assignment in TLM.

Summary Students were introduced to factoring a variety of algebraic expressions.

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