Module - Introduction to Algebra - Part...

2

LAST REVISED Oct., 2008

Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved.

Algebra Module A11

Introduction to Algebra - Part 2

Module A11 − Introduction to Algebra: Part II

1

Introduction to Algebra - Part 2 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? Algebra is that branch of mathematics that makes use of symbols to represent unknown quantities. An understanding of how to manipulate algebraic equations allows the student to solve for many unknown quantities. A solid grasp of algebra is essential if the student wishes to succeed in other areas of math such as trigonometry and calculus.

Learning Outcome When you complete this module you will be able to… Solve problems using algebra.

Learning Objectives 1. Use the laws of exponents to simplify algebraic expressions. 2. Perform algebraic division. 3. Work with zero and negative exponents. 4. Perform the operation of simplification on fractions having more than one term in the

numerator. 5. Perform algebraic long division.

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Module A11 − Introduction to Algebra: Part II 2

OBJECTIVE ONE When you complete this objective you will be able to… Use the laws of exponents to simplify algebraic expressions.

Exploration Activity

EXAMPLE 1 Simplify the following: a) (23)(24) = (2 · 2 · 2)(2 · 2 · 2 · 2) = 27

b) (y2)(y5) = (y · y)(y · y · y · y · y) = y7

This is a tedious process and a shorter method can be used which is shown in Example 2.

EXAMPLE 2 Simplify the following: a3 · a4 = (a · a · a)(a · a · a · a) = a7 however a3 · a4 = a3+4 = a7 This brings us to the rule for multiplying which is:

(am)(an) = am+n

that is, we keep the base a, and we add the exponents m and n. Another form of multiplication is to raise a power of a number to a power.


3

EXAMPLE 3 Simplify the following: a) (a3)4 = (a3)(a3)(a3)(a3)

= a · a · a · a · a · a · a · a · a · a · a · a = a12

b) (x2)3 = x2 · x2 · x2 = x6

However this can be done by the following rule for multiplication:

(am)n = amn that is, we multiply the powers, i.e.

(y2)4 = y8 NOTE: It is very important to remember the difference between the 2 rules:

RULE 1: (am)(an) = am+n ← multiplying like bases RULE 2: (am)n = amn ← raising a power to a power

EXAMPLE 4 Simplify:

a) (x3)(x2) = x5

b) (x7)2 = x14

c) x2 · x3 ≠ (x2)3 Why? (Symbol ≠ means not equal to)


EXAMPLE 5 Simplify the following: a) (a3b2)(a2b4)

regroup the terms to obtain = a3 · a2 · b2 · b4

apply rule 1 for multiplying exponents = a3+2 · b2+4 = a5b6

b) (2ab3)(3a2bc) regroup the terms to obtain = 2 · 3 · a · a2 · b3 · b · c apply rule 1 = 6a3b4c

c) (2x2y)(3xy2)

regroup terms to obtain = 2 · 3 · x2 · x · y · y2 apply rule 1 = 6x3y3 There is one more important rule we must consider, when raising a power to an exponent:

RULE 3 states:

(apbq)r = apqbqr


5

EXAMPLE 6 Simplify the following: a) (xy)4

expand = (xy)(xy)(xy)(xy) regroup x · x · x · x · y · y · y · y

= x4y4 OR apply Rule 3 above: (xy)4 = x4y4

b) (x2y2)3

expand

= (x2y2)(x2y2)(x2y2)

regroup

= x2 · x2 · x2 · y2 · y2 · y2

= x6y6

OR apply Rule 3

(x2y2)3

= x2·3 · y2·3 = x6y6 OR keep the brackets in the first step if you prefer (x2y2)3 = (x2)3(y2)3 = x6y6


c) (a2b3)4 apply Rule 3 = a2·4b3·4 = a8b12

d) (4c2d3)2

apply Rule 3 = (4)2 · c2·2 · d 3·2 note: the”4” must also be squared = 16c4d6

e) (2x2y3)5

apply Rule 3 = 25x10y15 = 32x10y15


7

Experiential Activity One Perform the indicated multiplications. 1. (n2)6 2. (y2)5 3. (3ax)(−2a2xy) 4. (−5act2)(9at5) 5. (−2axy3)(7ay4)(−ax4y) 6. (8dr3s2)(−4drs4) 7. (2st3)2 8. (3axt7)4 Show Me. 9. (x2y2)2 10. (7πR2)3

Experiential Activity One Answers 1. n12 2. y10 3. −6a3x2y 4. −45a2ct7 5. 14a3x5y8 6. −32d2r4s6 7. 4s2t6 8. 81a4x4t28

(TLM 81a4t28x4) 9. x4y4 10. 343π 3R 6


OBJECTIVE TWO When you complete this objective you will be able to… Perform algebraic division.

Exploration Activity In algebra, division of one quantity by another is most commonly written as a fraction:

ba

Only factors common to both numerator and denominator can be divided out.

EXAMPLE 1 Simplify the following:

23

32 a) =⋅

the 3 is common to both the numerator and the denominator so it is divided out

33 b) =xx

the x is common, so it divides out to leave the 3

yxxy 66 c) =

the common factor is x so it is divided out

xyxy

yxy 3

232

26 d) =

⋅⋅

=

the common factor is 2y, so when it is divided out we are left with 3x

As in multiplication, algebraic division uses rules of exponents to simplify expressions.

8



2

5 a)

aa

Expand:

aaaaaaa

⋅⋅⋅⋅⋅

=

Simplify by dividing the two a’s in the denominator into two a’s in the numerator, = a · a · a = a3

3

6 b)

xx

expand:

xxxxxxxxx

⋅⋅⋅⋅⋅⋅⋅

=

simplify by dividing the three x’s in the denominator into three x’s in the numerator, to be left with = x · x · x = x3

Again, this is a tedious process and a shorter method can be used. This shorter method is shown in Example 3.

9



a) 3

4

aa

aaaa

aaaa=

⋅⋅⋅⋅⋅

=

However, this can be done by the following rule for division:

RULE 4: nmn

ma

aa −= ← a very important rule!

Thus 343

4−= a

aa

= a1 or just a.

b) 2

63cc

apply rule 4.

= 3c6-2

= 3c4

c) 2

323

424

xyzzyx

expand:

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛= 2

323

424

zz

yy

xx

apply rule 4.

= 6x3-1 · y2-1 · z3-2

= 6x2yz

10


d) 2

64

816

abba

−−

expand:

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

−−

= 2

64

816

bb

aa

apply rule 4. = 2a4−1b6-2 = 2a3b4

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Experiential Activity Two Perform the indicated divisions.

1. a

xa2 2.

nn2

8 4

3. yx

yx2

34 4.

abcdscba

39 422

5. yt

uyt2

8 3 6.

zxyyzx2

22

93

7. ts

srt7

21 23 8. 2

2

662

wwx Show Me.

Experiential Activity Two Answers 1. ax 2. 4n3 3. x2y2 4. 3abc3ds 5. 4t2u 6.

yxz3

7. 3rt2s 8.

wx

331 2

12


OBJECTIVE THREE When you complete this objective you will be able to… Work with zero and negative exponents.

Exploration Activity RULE 5:

a0 = 1 here a ≠ 0

i.e. any quantity raised to the zero power is equal to 1. REASON:

1=b

b

xx because we are dividing something into itself, and 0xx

xx bb

b

b== − from the rule for

division,

therefore, x0 = 1 since both x0 and 1 are equal to b

b

xx .

EXAMPLE 1 Observe the “one” in each of the following expressions: a) 50 = 1 b) x0 = 1 c) 5x0

= 5(x0) = 5(1) =5

d) (4x−8)0 = 1 e) −x0 = −1 f) (−x)0 = 1 RULE 6:

nn

aa 1

=− a ≠ 0

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EXAMPLE 2

a) 22 1

xx =−

b) 22 1

bb =−

c) 22 44

xx =− since only x has the exponent of −2

d) 31−x

3

3

3

11

11

x

x

x

=

⋅=

=

e) 43−b

4

4

4

3

13

13

b

b

b

=

⋅=

=

f) 131

−a

3

31

131

1

a

a

a

=

⋅=

⋅=−

HINT: A negative exponent on a quantity moves that quantity from a numerator into a denominator, or from a denominator into a numerator.

eg. 22 1

xx =− , here x is moved from the numerator to the denominator.

eg. 22

1 xx

=−

, here x is moved from the denominator to the numerator.

Of course the exponent goes along with the quantity, only its sign changes to positive. 14



a) 7

4

bb

apply the rule for dividing with exponents

74−= b3−=

b

eliminate the negative exponent by applying rule 6:

31

b=

b) 4

22

yyx

expand:

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛= 4

22

1 yyx

apply the rule for dividing with exponents:

422 −⋅= yx

22 −⋅= yx

eliminate negative exponents by applying rule 6:

2

2

yx

=

c) 3

2

32

−

−

yx


2

3

32

xy

=

NOTE: The 2 and the 3 each have exponents of 1. The negative exponents only apply to the variables. Think of the question as the following:

31

21

32

−

−

yx

The 2 and 3 have positive exponents and stay where they are!!!

15


d) 44

32

zxyyzx

expand:

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛= 4

3

4

2

zz

yy

xx

apply the rule for dividing exponents

434112 −−− ⋅⋅= zyx 131 −− ⋅⋅= zyx

eliminate negative exponents by rewriting individual factors of the term and get:

11

33 1and1

zz

yy == −−

and then substituting to get:

zyx3=

e) 3

15

−

−−

xyyx

simplify

( )3115 −−−−− ⋅= yx

26 yx−=


6

2

xy

=

16


The Rules of Exponents

1. nmnm aaa +=))((

2. mnnm aa =)(

3. mnlnnml baba =)(

4. nmn

m

aaa −=

5. 10 =a

6. mm

aa 1

=−

17


Experiential Activity Three Express the following with only positive exponents. PART A 1. 3−b 2. 5−c

3. 41−a

4. 21−b

5. 12 −b 6. ( ) 12 −b

7. 121

−b 8.

( ) 121

−b

PART B Simplify the following and express your answers with only positive exponents.

1. 312 −− ⋅⋅ axa 2. 7

5

28

aa

3. 62

422yxyx 4.

mnnm 22 −−

5. ( )32

21

−

−−

yxxy 6.

( ) 321

2

−−

−

tsr

rst

7. bca

bca3

12

63

−

−−

8. ( )( ) 312

21

−−−

−−

yxa

axy Show Me.

9. ( )42525 zyx − 10. ( )3422 −ba

18


Experiential Activity Three Answers PART A

1. 31

b 2. 5

1c

3. 4a 4. 2b

5. b2 6.

b21

7. 2b 8. b2

PART B

1. ax1 2. 2

4a

3. 22y

4. 331nm

5. 4

5

xy 6. 2

7

rts

7. 22ca 8. 58

5

xay

9. 20

88625y

zx 10. 12

68ba

19


OBJECTIVE FOUR When you complete this objective you will be able to… Perform the operation of simplification on fractions having more than one term in the numerator.



a) n

nmn 32 2 −

This can be simplified, but we must be very careful, because there are two terms in the numerator and a single term in the denominator. Rewrite, putting each term of the numerator over the common denominator:

nn

nmn 32 2

−

simplify by dividing out the n in each fraction to obtain:

= 2mn − 3

b) 4

84 +

rewrite, putting each term over the common denominator.

48

44

+=

= 1 + 2 = 3

In general,

nb

na

nba

+=+ n ≠ 0

20


EXAMPLE 2

a) xy

xyyx2

24 32 +

rewrite the fraction as 2 separate fractions to get:

xyxy

xyyx

22

24 32

+=

simplify by dividing out the 2xy in each fraction to obtain:

22 yx +=

b) 2

223323

735287

abbababa ++

rewrite

2

22

2

33

2

23

735

728

77

abba

abba

abba

++=

simplify by dividing out the 7ab2 in each fraction to obtain:

abaa 54 22 ++=

c) 2 225 10

5x y xy

xy−

rewrite the fraction as 2 separate fractions to get:

2 225 105 5

x y xyxy x

= −y

simplify by dividing out the 5xy in each fraction to obtain:

5 2x y= −

21


Experiential Activity Four Simplify the following:

1. abc

bcacbacba 23643322 2−− 2. 2

34352

216188

rstrststrtrs

−

−−

3. ab

ababbaba−

−−− 24332 2 4. 2

82222

535305

mnnmmnynm

−

+− Show Me.

5. 22

2453

74914qp

qpqp−

+− 6. 32

384573

31294

yxyxyxyx ++

7. 33

343475

51520

bcacbacba − 8. 255

466357

153015wyx

wyxwyx −

9. 43

654467

3915

bcacbacba − 10. 244

345574

61218tyx

tyxtyx +

Experiential Activity Four Answers 1.

(TLM −a2b3c5 – 2a2c + abc2) cacbaabc 25322 2−− 2.

(TLM 9r2t2 – 4st3 +8t) ttrst 894 223 ++−

3. (TLM 2a2b3 – ab2 +b+1)

12 322 +++− bbaab 4. (TLM –7mn6 – my + 6)

676 mnmy −+−

5. (TLM –7p2 +2pq3)

23 72 ppq − 6. 634 4334 xyxxy ++

7. 362 34 bcba − 8. 22 2xywwx −

9. 2454 35 cabba −10.

(TLM 2xt + 3y3t3) xtty 23 33 +

Note: The answers for questions 1,2,3,4,5 and 10 had to be rearranged to be in the correct

TLM format. The other answers are were already in the correct format.

22


OBJECTIVE FIVE When you complete this objective you will be able to… Perform algebraic long division.


EXAMPLE 1

a) Perform the division: 12

352 2

−−+

xxx

NOTE: When the denominator consists of more than one term, in this case (2x – 1), we

CANNOT simplify by dividing out common factors. Therefore, we apply the rules of long division to simplify this expression.

Rewrite the original question in the following form:

35212 2 −+− xxx Here 2x – 1 is the divisor and 2x2 + 5x – 3 is the dividend.

Now in any long division of this style there are 3 basic steps:

1. Divide the first term of the divisor into the first term of the dividend and put what you get up top.

2. Multiply the whole divisor by the term you got in step 1, and put the product under the dividend.

3. Subtract what you got in step 1 from the dividend and write that answer down. 4. Repeat steps 1 to 3 starting by dividing the first term of the divisor into the first

term of your answer in step 3 and continue as many times as you have to until you get a remainder that is not divisible by the first term of the divisor!

Check out the following procedure:

In the division 35212 2 −+− xxx

the FIRST step is to divide 2x into 2x2 to get x. (Step 1)

multiply 2x − 1 by this x to get 2x2 − x:

x

xxxxx

−−+−

2

2

235212

23


Subtract to get 6x – 3, and start again by dividing 2x into 6x to get 3, then multiply this 3 by the divisor 2x – 1 to get 6x – 3, then subtract to get a remainder of 0. You are now finished. Thus:

3

03636

2

352122

2

+

−−

−

−+−

x

xx

xx

xxx Note: remember you’re subtracting here so: 5x – (–x) = 6x

Therefore:

312

352 2+=

−−+ x

xxx

Check:

(2x – 1)(x + 3) = 2x2 + 5x − 3

Notice in this example the key words: divisor, dividend, divide, multiply, and subtract. These 5 words state the essence of every long division problem.

EXAMPLE 2 Perform the division:

1723 4

−+−

xxx

Rewrite, only this time leave spaces for the missing x3 and x2 terms and also be sure the terms are in descending powers of x. Note in this example the x3 and the x2 terms are "missing" because their coefficients are 0, i.e. they can be written in as 0x3 and 0x2.

720031 234 +−++− xxxxx

Treat 0x3 and 0x2 as "ordinary" terms.

24


1333

817

33723

337203

33720031

23

2

2

23

23

34

234

+++

−+

−+−

−+−+

−+−++−

xxx

xxxxxx

xxxxx

xxxxxxx

The 8 at the end is the remainder.

Thus we say (x – 1) goes into (3x4 – 2x + 7), [3x3 + 3x2 + 3x + 1] and 1

8−x

times.

Compare this to saying 3 goes into 19, 6 and 1/3 times, they read the same way.

So just as 316

319

+= then1

813331

723 234

−++++=

−+−

xxxx

xxx .

• In most practical problems involving long division the remainder will be something

other than zero. • Also, the remainder is always “simpler” than the divisor, i.e. its degree is less than

the degree of the divisor. • Thus, when a term is reached which is of a smaller degree than that of the divisor, the

division is finished and this term is the remainder. The next example illustrates this.

EXAMPLE 3 Divide:

5091 232 −++− xxxx

The divisor is not of the 1st degree. Again proceed as usual.

9

49959

5091

2

2

3

232

+

+−−+

−−++−

x

xx

xxxx

xxxx 9

4909

519

10

50910

2

2

23

232

+

+−+

−+

−+

−++−+

x

xxx

xx

xxx

xxxxx or add in 0x

which fills in the terms and simplifies the

question.

The remainder is 1st degree, therefore simpler than 2nd degree divisor. The division is finished! .

So 149

159

22

23

−+

++=−

−+xxx

xxx

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Experiential Activity Five Perform the indicated division.

1. 12

352 2

−−+

xxx 2.

1+322 −−

xxx

3. 2−

1032 −+x

xx 4. 2+

352 2 −−x

xx

5. 2+

543 2 −+x

xx 6. 1−

453 23 ++−x

xxx

7. 3+

453 23 ++−x

xxx 8. 1−

532 24 ++x

xx

9. 1+

3743 253 +++−x

xxxx 10. 4+

71154 356 +−+−x

xxxx

11. 1−

1 23 +−x

xx 12. 4+

753 23 ++−t

ttt

13. 2−i

343 ++ ii 14. 1+

723 23 ++−M

MMM

15. 13 −a

256 2 +− aa 16. 32

4−

3434 23 −−−

x

xxx

17. 32 +278 3 +

xx 18.

62 −x123 24 −+ xx Show Me.

19. ( ) ( ) ( ) ( )742 224 −÷−− xxx 20. 32523 224 +−÷−+ xxxx

3+x 3

Experiential Activity Five Answers 1. 2. −x

5+x 15,92

3. 4. − =rx1, −=r 7,3 =+ rx

65,23 −=+ rx 10,5 =+ r

5,52 −=++ rx 7923,1979492123 2 =−+− rxx

5. 23 −x

2

6. 22 −x37. 6−x 8. 522 2 ++ xxx

9. 44 234 +− xxx 10. 328 345 +− xxx211. 1,222 =++ rxx

2

12. 125,337 −=+− rtt213. 19,82 =++ rii 14. 1,64 =+− rMM

15. 1,12 =− ra 16. 42

32 2 1++

xx

17. 964 2 +− xx 18. 42,92 =+ rx

19. 62 +x 20. 208 −−= x,563 2 ++ rxx

26


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Practical Application Activity Complete the Introduction to Algebra – Part 2 assignment in TLM.

Summary This module continued with the introductory concepts of a beginning algebra course.

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