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Course by: Siyavula Uploaders. E-mail the author

# Prime factors, square roots and cube roots

Module by: Siyavula Uploaders. E-mail the author

## PRIME FACTORS, SQUARE ROOTS AND CUBE ROOTS

CLASS ASSIGNMENT 1

1. Prime factors

• How do you write a number as the product of its prime factors?
• And how do you write it in exponent notation?

E.g. Question: Write 24 as the product of its prime factors(remember that prime factors are used as divisors only)

 2 24 2 12 2 6 3 3 1

Prime factors of 24 = {2; 3}

24 as product of its prime factors: 24 = 2 x 2 x 2 x 3

24 = 23 x 3 (exponential notation)

• Now express each of the following as the product of their prime factors(exponential notation) and also write the prime factors of each.

2. Square roots and cube roots

• How do you determine the square root ( size 12{ sqrt {} } {})or cube root ( 33 size 12{ nroot { size 8{3} } {} } {})of a number with the help of prime factors?
• Do you recall this?
• Determine: 324324 size 12{ sqrt {"324"} } {}Step 1: break down into prime factors Step 2: write as product of prime factors (in exponential notation)Step 3: 324324 size 12{ sqrt {"324"} } {} means (324)½ (obtain half of each exponent)
 2 324 2 162 3 81 3 27 3 9 3 3 1

Therefore: 324324 size 12{ sqrt {"324"} } {} = (22 x 34)½ = 21 x 32 = 2 x 9 = 18

(324 is a perfect square, because 18 x 18 = 324)

• Remember: size 12{ sqrt {} } {} means (......)½ and 33 size 12{ nroot { size 8{3} } {} } {} means (......)1/3

8x123=2x12÷3=48x123=2x12÷3=4 size 12{ nroot { size 8{3} } {8x rSup { size 8{"12"} } } =2x rSup { size 8{"12" div 3=4} } } {} therefore 2x42x4 size 12{2x rSup { size 8{4} } } {}

2.1 Calculate with the help of prime factors:

(i) 1 0241 024 size 12{ sqrt {"1 024"} } {}

 1024

(ii) 1000310003 size 12{ nroot { size 8{3} } {"1000"} } {}

 1000

2.2 Calculate:

a) (2 x 3)² =

b) 3 x 8² =

c) 1313 size 12{ nroot { size 8{3} } {1} } {} =

d) 11 size 12{ nroot {} {1} } {} =

e) 2222 size 12{ left ( sqrt {2} right ) rSup { size 8{2} } } {}=

f) then 172172 size 12{ left ( sqrt {"17"} right ) rSup { size 8{2} } } {} =

g) (3 + 4)3 + 14 =

h) 36+936+9 size 12{ sqrt {"36"} + sqrt {9} } {} =

i) 36+6436+64 size 12{ sqrt {"36"+"64"} } {} =

j) 273273 size 12{ nroot { size 8{3} } {"27"} } {} + 1313 size 12{ nroot { size 8{3} } {1} } {} =

k) (273)3(273)3 size 12{ $$nroot { size 8{3} } {"27"}$$ rSup { size 8{3} } } {} =

l) 64x1264x12 size 12{ sqrt {"64"x rSup { size 8{"12"} } } } {} =

HOMEWORK ASSIGNMENT 1

1. Determine the answers with the help of prime factors:

1.1 4096340963 size 12{ nroot { size 8{3} } {4"096"} } {} 1.2 1296412964 size 12{ nroot { size 8{4} } {1"296"} } {}

 4 096 1 296

2. Determine the answers without using a calculator.

2.1 3.3.3.3.3233.3.3.3.323 size 12{ nroot { size 8{3} } {3 "." 3 "." 3 "." 3 "." 3 rSup { size 8{2} } } } {} =

2.2 53a6b15353a6b153 size 12{ nroot { size 8{3} } {5 rSup { size 8{3} } a rSup { size 8{6} } b rSup { size 8{"15"} } } } {} =

2.3 8÷125×2738÷125×273 size 12{ nroot { size 8{3} } {8 div "125" times "27"} } {} =

2.4 643+(643)3643+(643)3 size 12{ nroot { size 8{3} } {"64"} + $$ nroot { size 8{3} } {"64"} $$ rSup { size 8{3} } } {} =

2.5 2(83)32(83)3 size 12{2 $$ nroot { size 8{3} } {8} $$ rSup { size 8{3} } } {} =

2.6 169169 size 12{ sqrt {"169"} } {} =

2.7 (6+4×12)2(6+4×12)2 size 12{ sqrt { $$6+4 times "12"$$ rSup { size 8{2} } } } {} =

2.8 6×18×126×18×12 size 12{ sqrt {6 times 18 times "12"} } {} =

2.9 2(9)22(9)2 size 12{2 $$sqrt {9}$$ rSup { size 8{2} } } {} =

2.10 (6+3)233(6+3)233 size 12{ sqrt { $$6+3$$ rSup { size 8{2} } }  - 3 rSup { size 8{3} } } {} =

CLASS ASSIGNMENT 2

1. Give the meaning of the following in your own words (discuss it in your group)

• LCM:

Explain it with the help of an example

• BCD:

Explain it with the help of an example

2. How would you determine the LCM and BCD of the following numbers?

8; 12; 20

Step 1: Write each number as the product of its prime factors.(Preferably not in exponential notation)

8 = 2 x 2 x 2

12 = 2 x 2 x 3

20 = 2 x 2 x 5

Step 2: First determine the BCD (the number/s occurring in each of the three)Suggestion: If the 2 occurs in each of the three, circle the 2 in each number and write it down once), etc.

BCD = 2 x 2 = 4

Step 3: Now determine the LCM. First write down the BCD and then find the number that occurs in two of the numbers and write it down, finally writing what is left over)

LCM = 4 x 2 x 3 x 5 = 120

3. Do the same and determine the BCD and LCM of the following:

38; 57; 95

Calculate it here:

38 = ....................................................................

57 = ....................................................................

95 = ....................................................................

BCD = .................................. and LCM = ..................................

Assessment

 Assessment of myself: by myself: Assessment by Teacher: I can…    1 2 3 4 Critical Outcomes 1 2 3 4 determine prime factors of a number; (Lo 1.2.6) Critical and creative thinking express a number as the product of its prime factors; (Lo 1.2.6; 1.2.3) Collaborating express prime factors in exponent notation; (Lo 1.2.7) Organising en managing determine the square root of a number; (Lo 1.2.7) Processing of information determine the cube root of a number. (Lo 1.2.7) Communication determine/define the smallest common factor (LCM); (Lo 1.2.6) Problem solving determine/define the biggest common divider (BCD). (Lo 1.2.6) Independence

good average not so good

 Comments by the learner: My plan of action: My marks: I am very satisfied with the standard of my work. < Date: I am satisfied with the steady progress I have made. Out of: I have worked hard, but my achievement is not satisfactory. Learner: I did not give my best. >
 Comments by parents: Comments by teacher: Signature: Date: Signature: Date:

Tutorial 1: (Number Systems)

Total: 30

1. Simplify:

1.1 1003610036 size 12{ sqrt {"100" - "36"} } {} [1]

1.2 25492549 size 12{ sqrt { { {"25"} over {"49"} } } } {} [1]

1.3 2631526315 size 12{ sqrt {2 rSup { size 8{6} } 3 rSup { size 8{"15"} } } } {} [2]

1.4 9(9+16)9(9+16) size 12{ sqrt {9}  $$sqrt {9} + sqrt {"16"}$$ } {} [3]

1.5 9² [1]

1.6 a=4,a=a=4,a= size 12{ sqrt {a} ="4,"~a=} {} [1]

1.7 a3=5,a=a3=5,a= size 12{ nroot { size 8{3} } {a} ="5,"~a={}} {} [1] [10]

2. Use the 324, and answer the following questions:

2.1 Is 324 divisible by 3? Give a reason for your answer. [2]

2.2 Write 324 as the product of its prime factors [3]

 324

2.3 Now determine 324324 size 12{ sqrt {"324"} } {} [2]

2.4 Is 324 a perfect square? Give a reason for your answer. [2] [9]

3. Determine each of the following without using your calculator.

3.1 8181 size 12{ sqrt {"81"} } {} [1]

3.2 364364 size 12{ sqrt { { {"36"} over {4} } } } {} [2]

3.3 32+4232+42 size 12{ sqrt {3 rSup { size 8{2} } +4 rSup { size 8{2} } } } {} [2]

3.4 16x1616x16 size 12{ sqrt {"16"x rSup { size 8{"16"} } } } {} [2]

4. If x = 3, determine:

4.1 4x4x size 12{4 rSup { size 8{x} } } {} [2]

4.2 27x27x size 12{ nroot { size 8{x} } {"27"} } {} [2] [11]

Tutorial

 I demonstrate knowledge and understanding of: Learning outcomes 0000 000 00 0 1. natural numbers (N) and whole numbers (N0) 1.1 2. the identification of the different types of numbers; 1.1 3. compound numbers; 1.2.6 4. divisibility rules; 1.2.6 5. the multiples of a number; 1.2.6 6. the factors of a number; 1.2.6 7. prime numbers; 1.1 8. prime factors; 1.2.6 9. expressing a number as the product of its prime factors; 1.2.6; 1.2.3 10. expressing prime factors in exponent notation; 1.2.3 11. even and odd numbers; 1.1 12. square roots of a number; 1.2.7 13. cube roots of a number; 1.2.7 14. the smallest common factor (LCM); 1.2.6 15. the biggest common divider (BCD). 1.2.6
 The learner’s … 1 2 3 4 work is… Not done.. Partially done. Mostly complete. Complete. layout of the work is… Not understandable. Difficult to follow. Sometimes easy to follow. Easy to follow. accuracy of calculations… Are mathematically incorrect. Contain major errors. Contain minor errors. Are correct.
 My BEST marks: Comments by teacher: Date: Out of: Learner: Signature: Date:

Parent signature: Date:

Test 1: (Number Systems)

Total: 30

1. Tabulate the following:

1.1 All the prime numbers between 20 and 30. [2]

1.2 All the factors of 12. [2]

1.3 All factors of 12 which are compound numbers [2] [6]

2. Determine the smallest natural number for * so that the following number is divisible by 3. (Give a reason for your answer)

1213156*3 [2]

3. Determine the following without using your calculator.

3.1 36+6436+64 size 12{ sqrt {"36"+"64"} } {} [2]

3.2 293293 size 12{ nroot { size 8{3} } {2 rSup { size 8{9} } } } {} [2]

3.3 279279 size 12{ sqrt {2 { { size 8{7} } over { size 8{9} } } } } {} [3]

3.4 0,040,04 size 12{ sqrt {"0,04"} } {} [2]

3.5 100 36100 36 size 12{ sqrt {"100"-" 36"} } {} [2]

3.6 8×2738×273 size 12{ nroot { size 8{3} } {8 times "27"} } {} [2]

3.7 (9)2(9)2 size 12{ $$sqrt {9}$$ rSup { size 8{2} } } {} [2]

3.8 64136413 size 12{ nroot { size 8{3} } {"64" - `1} } {} [2] [17]

4. Determine 1 72831 7283 size 12{ nroot { size 8{3} } {"1 728"} } {} using prime factors, without using a calculator.

[5]

5. Bonus question

If (n) means nn what is the value of ((2)) ? [2]

Enrichment Exercise for the quick learner

(Learning unit1)

Each question has five possible answers. Only one answer is correct. Place a cross (X) over the letter that indicates the correct answer.

1. If n and p are both odd, which of the following will be even?

a) np b) n²p + 2 c) n+p+1 d) 2n+3p+5 e) 2n+p

2. R 120 is divided amongst three men in the ratio 3 : 4: 9. The one with the smallest share will receive ...

a) R16 b) R20 c) R22,50 d) R24,50 e) R40

3. How many triangles are there in the figure?

a) 8 b) 12 c) 14 d) 16 e) 20

4. A decagon has 2 interior angles of 120° each. If all the remaining angles are of the same size, each angle will be equal to ...

a) 15° b) 30° c) 120° d) 150° e) 165°

5. The last digit of the number 31993 is ....

a) 1 b) 3 c) 6 d) 7 e) 9

6. The figure below has 5 squares. If AB = 6, the area of the figure is...

a) 12 b) 20 c) 24 d) 36 e) impossible

## Assessment

 Learning outcomes(LOs) LO 1 Numbers, Operations and RelationshipsThe learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems. Assessment standards(ASs) We know this when the learner : 1.1 describes and illustrates the historical and cultural development of numbers; 1.2 recognises, classifies and represents the following numbers in order to describe and compare them:1.2.3 numbers written in exponent form; including squares and cubes of natural numbers and their square roots and cube roots;1.2.6 multiples and factors;1.2.7 irrational numbers in the context of measurement (e.g. square and cube roots on non-perfect squares and cubes); 1.6 estimates and calculates by selecting suitable steps for solving problems that involve the following:1.6.2 multiple steps with rational numbers (including division with fractions and decimals);1.6.3 exponents.

## Memorandum

### CLASS ASSIGNMENT 2

1.1 48 = 24 × 3; 60 = 2² × 3 × 5; 450 = 2 x 3² x 5²;

P48 = {2, 3}; P60 = {2, 3, 5}; P450 = {2, 3, 5};

2.1 i) = = (210)

= 25

= 32

ii) = = (2³ x 5³)

= 2 x 5

= 10

2.2 a) 36

b) 192

c) 1

d) 1

e) 2

f) 17

g) 63

h) 9

i) 10

j) 4

k) 27

l) 8 x6

### HOMEWORK ASSIGNMENT 2

1.1 = (212)

= 24

= 16

1.2 = (24 x 34)

= 2 x 3

= 6

2.1 = 3² = 9

2.2 5a²b5

2.3 = x 3 =

= 1,2

2.4: 4 + 64 = 68

• :2(8) = 16
• :13

2.7 ()2 = 54

2.8 = 36

• :2(9) = 18
• :9 - 27 = -18

### CLASS ASSIGNMENT 3

21. LCM: Lowest common multiple

LCM of 2, 6, 12 :

24 HCF: Highest common factor

HCF of 24 and 48 :

2. 38 = 2 x 19

57 = 3 x 19

95 = 5 x 19

HCF = 19

LCM = 19 x 2 x 3 x 5

= 570

TUTORIAL 1

1.1 = 8

1.2

• 2³ . 37,5
• :3(3 + 4) = 21
• :81
• :16

1.7 :125

2.1 :3 + 2 + 4 = 9

9 ÷ 3 = 3 Yes!

2.2: 324 = 2² x 34

2.3: = (2² x 34)

= 2 x 3²

= 18

2.4: Yes! 18 x 18 = 324 /18² = 324

• :9
• : 6262 size 12{ { {6} over {2} } } {} = 3

3.3: 9+169+16 size 12{ sqrt {9+"16"} } {} = 2525 size 12{ sqrt {"25"} } {} = 5

3.4: 4 x8x8 size 12{x rSup { size 8{8} } } {}

• :43 = 64
• :3

### ENRICHMENT EXERCISE

1. d

2. c

3. d

4. 180(102)10180(102)10 size 12{ { {"180" $$"10" - 2$$ } over {"10"} } } {} = 144º (one angle) (1 440 – 240) ÷ 8 = 150 (d)

5. b 31992 ends on 1

6. d AB = 6

(2x)2 + x 2 = 36

4 x 2 + x 2 = 36

5 x 2 = 36

TEST 1

• :23, 29
• :1, 2, 3, 6, 12
• :4, 6, 12

2. :* 2 1 + 2 + 1 + 3 + 1 + 5 + 6 + 3 = 22

3.1 100100 size 12{ sqrt {"100"} } {} = 10

3.2 23 = 8

3.3 259259 size 12{ sqrt { { {"25"} over {9} } } } {} = 5353 size 12{ { {5} over {3} } } {} = 1 2323 size 12{ { {2} over {3} } } {}

3.4 41004100 size 12{ sqrt { { {4} over {"100"} } } } {} = 210210 size 12{ { {2} over {"10"} } } {} = 0,2 / 1515 size 12{ { {1} over {5} } } {}

3.5 6464 size 12{ sqrt {"64"} } {} = 8

• :2 x 3 = 6
• :9
• :4 – 1 = 3

4. 26x33326x333 size 12{ nroot { size 8{3} } {2 rSup { size 8{6} } x3 rSup { size 8{3} } } } {} = 22 x 3

= 4 x 3

= 12

5. (2) = 22 = 4

(4) = 44 = 256

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