# Connexions

You are here: Home » Content » Algebra » The algebra of the four basic operations

### Lenses

What is a lens?

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

#### In these lenses

• GETSenPhaseMaths

This module is included inLens: Siyavula: Mathematics (Gr. 7-9)
By: SiyavulaAs a part of collection: "Mathematics Grade 9"

Collection Review Status: In Review

Click the "GETSenPhaseMaths" link to see all content selected in this lens.

Click the tag icon to display tags associated with this content.

### Recently Viewed

This feature requires Javascript to be enabled.

### Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.

Inside Collection (Textbook):

Textbook by: Jason Haap. E-mail the author

# The algebra of the four basic operations

Module by: Siyavula Uploaders. E-mail the author

MATHEMATICS

ALGEBRA AND GEOMETRY

Module 6

THE ALGEBRA OF THE FOUR BASIC OPERATIONS

Activity 1

To refresh understanding of conventions in algebra as applied to addition and subtraction

[LO 1.2, 1.6]

A We will first have a look at terms.

Remember, terms are separated by + or –. In each of the following, say how many terms there are:

1. a + 5

2. 2a2

3. 5a (a+1)

4. 3a1a2+5a3a1a2+5a size 12{ { {3 size 11{a} - 1} over { size 11{a rSup { size 8{2} } }} } size 12{+5a}} {}

In the next exercise you must collect like terms to simplify the expression:

1. 5a + 2a

2. 2a2 + 3a – a2

3. 3x – 6 + x + 11

4. 2a(a–1) – 2a2

Example:

Add 3x + 4 by x + 5.

(x + 5) + (3x + 4) Write, with brackets, as sum.

x + 5 + 3x + 4 Remove brackets, with care.

4x + 9 Collect like terms.

In this exercise, add the two given expressions:

1. 7a + 3 and a + 2

2. 5x – 2 and 6 – 3x

3. x + ½ and 4x – 3½

4. a2 + 2a + 6 and a – 3 + a2

5. 4a2 – a – 3 and 1 + 3a – 5a2

C Subtracting expressions

Study the following examples very carefully:

Subtract 3x – 5 from 7x + 2.

(7x + 2) – (3x – 5)

Notice that 3x – 5 comes second, after the minus.

7x + 2 – 3x + 5

The minus in front of the bracket makes a difference!

4x + 7

Collecting like terms.

Calculate 5a – 1 minus 7a + 12: (5a – 1) – (7a + 12)

5a – 1 – 7a – 12

–2a – 13

D Mixed problems

Do the following exercise (remember to simplify your answer as far as possible):

1. Add 2a – 1 to 5a + 2.

2. Find the sum of 6x + 5 and 2 – 3x.

3. What is 3a – 2a2 plus a2 – 6a?

4. (x2 + x) + (x + x2) = . . .

5. Calculate (3a – 5) – (a – 2).

6. Subtract 12a + 2 from 1 + 7a.

7. How much is 4x2 + 4x less than 6x2 – 13x?

8. How much is 4x2 + 4x more than 6x2 – 13x?

9. What is the difference between 8x + 3 and 2x +1?

Use appropriate techniques to simplify the following expressions:

1. x2 + 5x2 – 3x + 7x – 2 + 8

2. 7a2 – 12a + 2a2 – 5 + a – 3

3. (a2 – 4) + (5a + 3) + (7a2 + 4a)

4. (2x – x2) – (4x2 – 12) – (3x – 5)

5. (x2 + 5x2 – 3x) + (7x – 2 + 8)

6. 7a2 – (12a + 2a2 – 5) + a – 3

7. (a2 – 4) + 5a + 3 + (7a2 + 4a)

8. (2x – x2) – 4x2 – 12 – (3x – 5)

9. x2 + 5x2 – 3x + (7x – 2 + 8)

10. 7a2 – 12a + 2a2 – (5 + a – 3)

11. a2 – 4 + 5a + 3 + 7a2 + 4a

12. (2x – x2) – [(4x2 – 12) – (3x – 5)]

Here are the answers for the last 12 problems:

1. 6x2 + 4x + 6

2. 9a2 – 11a – 8

3. 8a2 + 9a – 1

4. – 5x2 – x + 17

5. 6x2 + 4x + 6

6. 5a2 – 11a + 2

7. 8a2 + 9a – 1

8. – 5x2 – x – 7

9. 6x2 + 4x + 6

10. 9a2 – 13a – 2

11. 8a2 + 9a – 1

12. – 5x2 + 5x + 7

Activity 2

To multiply certain polynomials by using brackets and the distributive principle

[LO 1.2, 1.6, 2.7]

A monomial has one term; a binomial has two terms; a trinomial has three terms.

A Multiplying monomials.

Brackets are often used.

Examples:

2a × 5a = 10a2

3a3 × 2a × 4a2 = 24 a6

4ab × 9a2 × (–2a) × b = –36a4b2

a × 2a × 4 × (3a2)3 = a × 2a × 4 × 3a2 × 3a2 × 3a2 = 126a8

(2ab2)3 × (a2bc)2 × (2bc)2 = (2ab2) (2ab2) (2ab2) × (a2bc) (a2bc) × (2bc) (2bc) = 32a7b10c4

Exercise:

1. (3x) (5x2)

(x3) (–2x)

(2x)2 (4)

(ax)2 (bx2) (cx2)2

B Monomial × binomial

Brackets are essential.

Examples:

5(2a + 1) means multiply 5 by 2a as well as by 1. 5 (2a + 1) = 10a + 5

Make sure that you work correctly with your signs.

4a(2a + 1) = 8a + 4a

–5a(2a + 1) = –10a2 – 5a

a2(–3a2 – 2a) = –3a4 – 2a3

–7a(2a – 3) = –14a2 + 21a

Note: We have turned an expression in factors into an expression in terms. Another way of saying the same thing is: A product expression has been turned into a sum expression.

Exercise:

1. 3x (2x + 4)

x2 (5x – 2)

–4x (x2 – 3x)

(3a + 3a2) (3a)

C Monomial × trinomial

Examples:

5a(5 + 2a – a2) = 25a + 10a2 – 5a3

– ½ (10x5 + 2a4 – 8a3) = – 5x5 – a4 +4a3

Exercise:

3x (2x2 – x + 2)

–ab2 (–bc + 3abc – a2c)

12a ( ¼ + 2a + ½ a2)

Also try: 4. 4x (5 – 2x + 4x2 – 3x3 + x4)

D Binomial × binomial

Each term of the first binomial must be multiplied by each term of the second binomial.

(3x + 2) (5x + 4) = (3x)(5x) + (3x)(4) + (2)(5x) + (2)(4) = 15x2 + 12x + 10x + 8

= 15x2 + 22x + 8

Here is a cat–face picture to help you remember how to multiply two binomials:

The left ear says multiply the first term of the first binomial with the first term of the second binomial.

The chin says multiply the first term of the first binomial with the second term of the second binomial.

The mouth says multiply the second term of the first binomial with the first term of the second binomial.

The right ear says multiply the second term of the first binomial with the second term of the second binomial.

There are some very important patterns in the following exercise – think about them.

Exercise:

(a + b) (c + d)

(2a – 3b) (–c + 2d)

(a2 + 2a) (b 2 –3b)

(a + b) (a + b)

(x2 + 2x) (x2 + 2x)

(3x – 1) (3x – 1)

(a + b) (a – b)

(2y + 3) (2y – 3)

(2a2 + 3b) (2a2 – 3b)

(a + 2) (a + 3)

(5x2 + 2x) (x2 – x)

(–2a + 4b) (5a – 3b)

E Binomial × polynomial

Example:

(2a + 3) (a3 – 3a2 + 2a – 3) = 2a4 – 6a3 + 4a2 – 6a + 3a3 – 9a2 + 6a – 9

= 2a4 – 3a3 – 5a2 – 9 (simplified)

Exercise:

(x2 – 3x) (x2 + 5x – 3)

(b + 1) (3b2 – b + 11)

(a – 4) (5 + 2a – b + 2c)

(–a + 2) (a + b + c – 3d)

Activity 3

To find factors of certain algebraic expressions

[LO 1.6, 2.1, 2.7]

A Understanding what factors are

This table shows the factors of some monomials

 Expression Smallest factors 42 2 × 3 × 7 6ab 2 × 3 × a × b 21a2b 3 × 7 × a × a × b (5abc2)2 5 × a × b × c × c × 5 × a × b × c × c –8y4 –2 × 2 × 2 × y × y × y × y (–8y4)2 –2 × 2 × 2 × y × y × y × y × –2 × 2 × 2 × y × y × y × y

You can write the factors in any order, but if you stick to the usual order your work will be easier. Two lists of factors in the table are not in the usual order – rewrite them in order.

B Finding common factors of binomials

Take the binomial 6ab + 3ac.

The factors of 6ab are 2 × 3 × a × b and the factors of 3ac are 3 × a × c.

The factors that appear in both 6ab and 3ac are 3 and a – they are called common factors.

We can now use brackets to group the factors into the part that is common and the rest, as follows:

6ab = 3a × 2b and 3ac = 3a × c

Now we can factorise 6ab + 3ac. This is how to set it out:

6ab + 3ac = 3a (2b + c).

An expression in terms has been written as an expression in factors.

Or: A sum expression has been turned into a product expression.

Here are some more examples:

6x2 + 12x = 3x (2x + 4)

5x3 – 2x2 = x2 (5x – 2)

–4x3 + 12x2 = –4x (x2 – 3x)

9a2 + 9a3 = (3a + 3a2) (3a)

Look back at the exercise in section B of the previous activity – did you recognise the problems?

C Finding common factors of polynomials

In exactly the same way we can find the common factors of more than two terms. Here are some examples:

Examples:

6x3 – 3x2 + 6x = 3x (2x2 – x + 2)

ab3c – 3a2b3c + a3b2c = ab2c (b – 3ab + a2)

3a + 24a2 + 6a3 = 3a ( 1 + 8a + 2a2)

20x – 8x2 + 16x3 – 12x4 +4x5 = 4x (5 – 2x + 4x2 – 3x3 + x4)

You will notice that the terms remaining in the brackets don’t have any more common factors left. This is because they have been fully factorised. You must always take out the highest common factor from all the terms.

Exercise:

Fully factorise the following expressions by taking out the highest common factor:

12abc + 24ac

15xy – 21y

3abc + 18ab2c3

8x2y2 – 2x

2a2bc2 + 4ab2c – 7abc

12a(bc)2 – 8(abc)3 + 4(ab)2c3 – 20bc + 4a

Pair activity:

Did you notice that in each case the number of terms in the brackets after factorising was the same as the number of terms in the original expression?

Explain to your partner why you think this will always happen.

D Factorising difference of squares

In section D of the previous activity you had to multiply these three pairs of binomials:

(a + b) (a – b) ,

(2y + 3) (2y – 3) and

(2a2 + 3b) (2a2 – 3b)

Here are the solutions:

(a + b) (a – b) = a2 – b2

(2y + 3) (2y – 3) = 4y2 – 9

(2a2 + 3b) (2a2 – 3b) = 4a4 – 9b2

You will have noticed that the answers have a very special pattern: square minus square.

This is called a difference of squares and this is how it is factorised:

First–square minus second–square

= ( firstsquarefirstsquare size 12{ sqrt { ital "first" - ital "square"} } {}plus secondsquaresecondsquare size 12{ sqrt {"sec" ital "ond" - ital "square"} } {}) ( firstsquarefirstsquare size 12{ sqrt { ital "first" - ital "square"} } {}minus secondsquaresecondsquare size 12{ sqrt {"sec" ital "ond" - ital "square"} } {})

Examples:

x2 – 25 = (x + 5) (x – 5)

4 – b2 = (2 + b) (2 – b)

9a2 – 1 = (3a + 1) (3a – 1)

YOU HAVE TO BE VERY SURE OF THE MOST COMMON SQUARES AND THEIR ROOTS.

These are a few important ones – you must add to this list.

22 = 4 32 = 9 (a2)2 = a4

(a3)2 = a6

(½)2 = ¼ 12 = 1

Exercise:

Factorise fully:

1. a2 – b 2

4y2 – 9

4a4 – 9b2

1 – x2

25 – a6

a8 – ¼

4a2b 2 – 81

0,25 – x2y6

9. 2a2 – 2b2 (take care!)

E Combining common factors with differences of squares

As you saw in the last exercise (number 9), it is essential to check for common factors first and then to factorise the bracketed polynomial if possible.

Another example:

Factorise 12ax2 – 3ay2

First recognise that there is a common factor of 3a, before saying that this can’t be a difference of squares.

12ax2 – 3ay2 = 3a (4x2 – y2) Now we recognise 4x2 – y2 as the difference of two squares.

12ax2 – 3ay2 = 3a (4x2 – y2) = 3a(2x + y)(2x – y).

Exercise:

Factorise completely:

1. ax2 – ay4

2. a3 – ab2

3. 0,5a2x – 4,5b2x

4. a5b3c – abc

F Successive differences of squares

Try factorising this binomial completely by keeping your eyes open: a4 – b4

Now do this exercise – as usual factorise as far as possible.

1. x6 – 64

2. 1 – m8

3. 3a4 – 24b8

4. x – x9

G Factorising trinomials

If you study the answers to the following four problems (they are from a previous activity), you will notice that the answers, after simplifying, sometimes have two terms, sometimes three terms and sometimes four terms. Discuss what you see happening (with a partner) and decide why they are different.

1. (a + b) (a – b) = a2 – ab + ab – b2 = a2 – b2 (when simplified)

2. (a + 2) (a + 3) = a2 + 3a + 2a + 6 = a2 + 5a + 6

3. (a + b) (a + b) = a×a +ab + ba + b×b = a2 + ab + ab + b2 = a2 + 2ab + b2 (in the simplest form)

4. (a + b) (c + d) = ac + ad + bc + bd (and this answer can not be simplified)

The answer to the type of problem represented by number 1 above is a difference of squares.

The answers to numbers 2 and 3 are trinomials, and we will now see how to factorise them

The first fact that you must always remember is that not all trinomials can be factorised.

Work backwards through problem 2:

a2 + 5a + 6 = a2 + 3a + 2a + 6 = (a + 2) (a + 3).

So that you can see clearly where the a2 came from, and the 5a and the 6.

Now try to factorise a2 + 7a + 12 = ( …………. ) ( …………… ) by filling in two correct binomials in the brackets.

You can check your answer by multiplying the binomials in your answer as you were taught in activity 2. Keep trying and checking your answer until you have it right. Do the same in the following three exercises:

Match up the two columns:

A. a2 – 5a – 6 1. (x + 2)(x + 3)

B. a2 – a – 6 2. (x – 2)(x + 3)

C. a2 – 5a + 6 3. (x + 1)(x – 6)

D. a2 + 7a + 6 4. (x – 2)(x – 3)

E. a2 + 5a + 6 5. (x + 1)(x + 6)

F. a2 + 5a – 6 6. (x – 1)(x + 6)

G. a2 + a – 6 7. (x + 2)(x – 3)

H. a2 – 7a + 6 8. (x – 1)(x – 6)

Now factorise the following trinomials by using the same techniques you have just learnt. The last two are more difficult than the first four!

a2 + 3a + 2

a2 + a – 12

a2 – 4a + 3

a2 – 9a + 20

a2 + ab – 12b2

2a2 – 18a + 40

 Problem One f a ctor correct Both f a ctors correct A nswer checked by multiplic a tion 1 2 3 1 2 3
 4 5 6

Finally, in groups of 3, 4 or 5, work out exactly how one should go about factorising these trinomials, and write down a strategy that will get you to the answer accurately and quickly.

Activity 4

To use factorising in simplifying fractions, and in multiplying, dividing and adding fractions

[LO 1.2, 1.6, 2.9]

A. Simplifying algebraic fractions

Two of the following four fractions can be simplified and the others can’t. Which is which?

2 + a 2 a 2 + a 2 a size 12{ { {2+a} over {2 - a} } } {}

3 a + b a + b 3 a + b a + b size 12{ { {3 left (a+b right )} over {a+b} } } {}

4 + x x + 4 4 + x x + 4 size 12{ { {4+x} over {x+4} } } {}

a b c 2 b + c a b c 2 b + c size 12{ { {a left (b - c right )} over {2 left (b+c right )} } } {}
(1)

As you have seen in the previous activity, factorising is a lot of trouble. So, why do we do it?

The following expression cannot be simplified as it is 6a2b6b2a26a2b6b2a2 size 12{ { {6a rSup { size 8{2} } b - 6b} over {2a - 2} } } {} because we are not allowed to cancel terms. If we can change the sum expressions into product expressions (by factorising) then we will be able to cancel the factors, and simplify.

6a2b – 6b = 6b (a2 – 1) = 6b (a + 1) (a – 1) and 2a – 2 = 2(a – 1)

So, the reason we factorise is that it allows us to simplify expressions better.

Now: 6a2b6b2a26a2b6b2a2 size 12{ { {6a rSup { size 8{2} } b - 6b} over {2a - 2} } } {} = 6ba+1a12a16ba+1a12a1 size 12{ { {6 size 11{b left (a+1 right ) left (a - 1 right )}} over {2 left (a - 1 right )} } } {} = 3ba+113ba+11 size 12{ { {3b left (a+1 right )} over {1} } } {} = 3b(a + 1) .

It is very important that you factorise completely.

Exercise:

Factorise the numerator and denominator, cancel factors and write in the simplest form:

1 12a+6b2a+b12a+6b2a+b size 12{ { { size 11{"12"}a+6b} over {2a+b} } } {}

2 x29x+3x29x+3 size 12{ { {x rSup { size 8{2} } - 9} over {x+3} } } {}

3 2a+1a16a+122a+1a16a+12 size 12{ { {2 left (a+1 right ) left (a - 1 right )} over {6 left (a+1 right ) rSup { size 8{2} } } } } {}

4 5a255a+55a255a+5 size 12{ { {5a rSup { size 8{2} } - 5} over {5a+5} } } {}

B. Multiplying and dividing fractions

Our normal rules for multiplication and division of fractions are still the same. Study the following examples, taking special note of the factorising and cancelling.

4x3y6y2÷xy3x2×2xy23x4x3y6y2÷xy3x2×2xy23x size 12{ { {4x rSup { size 8{3} } y} over {6y rSup { size 8{2} } } } div { { ital "xy"} over {3x rSup { size 8{2} } } } times { {2 ital "xy" rSup { size 8{2} } } over {3x} } } {} = 4x3y6y2×3x2xy×2xy23x4x3y6y2×3x2xy×2xy23x size 12{ { {4x rSup { size 8{3} } y} over {6y rSup { size 8{2} } } } times { {3x rSup { size 8{2} } } over { ital "xy"} } times { {2 ital "xy" rSup { size 8{2} } } over {3x} } } {} = 4x434x43 size 12{ { {4x rSup { size 8{4} } } over {3} } } {}

a292×14a212aa292×14a212a size 12{ { {a rSup { size 8{2} } - 9} over {2} } times { {1} over {4a rSup { size 8{2} } - "12"a} } } {} = a+3a32×14aa3a+3a32×14aa3 size 12{ { { left (a+3 right ) left (a - 3 right )} over {2} } times { {1} over {4a left (a - 3 right )} } } {} = a+38aa+38a size 12{ { { left (a+3 right )} over {8a} } } {}

3a+65÷a24103a+65÷a2410 size 12{ { {3a+6} over {5} } div { {a rSup { size 8{2} } - 4} over {"10"} } } {} = 3a+65×10a243a+65×10a24 size 12{ { {3a+6} over {5} } times { {"10"} over {a rSup { size 8{2} } - 4} } } {} = 3a+25×10a+2a23a+25×10a+2a2 size 12{ { {3 left (a+2 right )} over {5} } times { {"10"} over { left (a+2 right ) left (a - 2 right )} } } {} = 6a26a2 size 12{ { {6} over {a - 2} } } {}

Exercise:

Simplify:

1. 2ab2b3c×9ac24b÷3ac2b22ab2b3c×9ac24b÷3ac2b2 size 12{ { {2 ital "ab" rSup { size 8{2} } } over {b rSup { size 8{3} } c} } times { {9 ital "ac" rSup { size 8{2} } } over {4b} } div { {3 ital "ac"} over {2b rSup { size 8{2} } } } } {}

2. 2a+1a22a23a+3×9a+1a+324a2÷3a+1a+32a222a+1a22a23a+3×9a+1a+324a2÷3a+1a+32a22 size 12{ { {2 left (a+1 right ) left (a - 2 right ) rSup { size 8{2} } } over { left (a - 2 right ) rSup { size 8{3} } left (a+3 right )} } times { {9 left (a+1 right ) left (a+3 right ) rSup { size 8{2} } } over {4 left (a - 2 right )} } div { {3 left (a+1 right ) left (a+3 right )} over {2 left (a - 2 right ) rSup { size 8{2} } } } } {}

3. 4a2+8a2b+4×3b2+23a2+6a4a2+8a2b+4×3b2+23a2+6a size 12{ { {4a rSup { size 8{2} } +8a} over {2b+4} } times { {3 left (b rSup { size 8{2} } +2 right )} over {3a rSup { size 8{2} } +6a} } } {}

4. x215x5÷x+1215x+15x215x5÷x+1215x+15 size 12{ { {x rSup { size 8{2} } - 1} over {5x - 5} } div { { left (x+1 right ) rSup { size 8{2} } } over {"15"x+"15"} } } {}

5. 7x3xy÷3x+65x2y÷5x103x2127x3xy÷3x+65x2y÷5x103x212 size 12{ { {7x} over {3 ital "xy"} } div { {3x+6} over {5x rSup { size 8{2} } y} } div { {5x - "10"} over {3x rSup { size 8{2} } - "12"} } } {}

6. 5x2+5xx2x5x+5x215x2+5xx2x5x+5x21 size 12{ { { { {5x rSup { size 8{2} } +5x} over {x rSup { size 8{2} } - x} } } over { { {5x+5} over {x rSup { size 8{2} } - 1} } } } } {} (here we have a fraction divided by a fraction – first rewrite it like number 4)

You already know quite well that adding and subtracting fractions is a lot more difficult than multiplying and dividing them. The reason is that we can only add and subtract the same kind of fractions, namely fractions with identical denominators. If the denominators are different, find the lowest common multiple of the denominators (LCD), rewrite all the terms with this as denominator, and then simplify by gathering like terms together. Finally, the answer has to be simplified by cancelling factors occurring in both numerator and denominator. Here are some examples – all the steps have been shown:

Simplify:

1. 5abx2cx+4ac3x+cx2a5abx2cx+4ac3x+cx2a size 12{ { {5 ital "abx"} over {2 ital "cx"} } + { {4 ital "ac"} over {3x} } + { { ital "cx"} over {2a} } } {} (LCD = 6acx)

5abx2cx×3a3a+4ac3x×2ac2ac+cx2a×3cx3cx5abx2cx×3a3a+4ac3x×2ac2ac+cx2a×3cx3cx size 12{ left ( { {5 ital "abx"} over {2 ital "cx"} } times { {3a} over {3a} } right )+ left ( { {4 ital "ac"} over {3x} } times { {2 ital "ac"} over {2 ital "ac"} } right )+ left ( { { ital "cx"} over {2a} } times { {3 ital "cx"} over {3 ital "cx"} } right )} {}= 15a2bx6acx+8a2c26acx+3c2x26acx15a2bx6acx+8a2c26acx+3c2x26acx size 12{ { {"15"a rSup { size 8{2} } ital "bx"} over {6 ital "acx"} } + { {8a rSup { size 8{2} } c rSup { size 8{2} } } over {6 ital "acx"} } + { {3c rSup { size 8{2} } x rSup { size 8{2} } } over {6 ital "acx"} } } {} = 15a2bx+8a2c2+3c2x26acx15a2bx+8a2c2+3c2x26acx size 12{ { {"15"a rSup { size 8{2} } ital "bx"+8a rSup { size 8{2} } c rSup { size 8{2} } +3c rSup { size 8{2} } x rSup { size 8{2} } } over {6 ital "acx"} } } {}

2. a+b2+b+c3a+c6a+b2+b+c3a+c6 size 12{ { {a+b} over {2} } + { {b+c} over {3} } - { {a+c} over {6} } } {} (LCD = 6) Take careful note of how the signs are handled below!

3a+b6+2b+c6a+c63a+b6+2b+c6a+c6 size 12{ { {3 left (a+b right )} over {6} } + { {2 left (b+c right )} over {6} } - { {a+c} over {6} } } {} = 3a+b+2b+ca+c63a+b+2b+ca+c6 size 12{ { {3 left (a+b right )+2 left (b+c right ) - left (a+c right )} over {6} } } {} = 3a+3b+2b+2cac63a+3b+2b+2cac6 size 12{ { {3a+3b+2b+2c - a - c} over {6} } } {} = 2a+5b+c62a+5b+c6 size 12{ { {2a+5b+c} over {6} } } {}

3. a+3a24+13a+6+25a10a+3a24+13a+6+25a10 size 12{ { {a+3} over {a rSup { size 8{2} } - 4} } + { {1} over {3a+6} } + { {2} over {5a - "10"} } } {}

To find Lowest Common Denominator first factorise denominators!

a + 3 a + 2 a 2 + 1 3 a + 2 + 2 5 a 2 a + 3 a + 2 a 2 + 1 3 a + 2 + 2 5 a 2 size 12{ { {a+3} over { left (a+2 right ) left (a - 2 right )} } + { {1} over {3 left (a+2 right )} } + { {2} over {5 left (a - 2 right )} } } {}

Do you see that the LCD is 3×5×(a+2)(a–2)?

a + 3 a + 2 a 2 × 15 15 + 1 3 a + 2 × 5 a 2 5 a 2 + 2 5 a 2 × 3 a + 2 3 a + 2 a + 3 a + 2 a 2 × 15 15 + 1 3 a + 2 × 5 a 2 5 a 2 + 2 5 a 2 × 3 a + 2 3 a + 2 size 12{ left ( { {a+3} over { left (a+2 right ) left (a - 2 right )} } times { {"15"} over {"15"} } right )+ left ( { {1} over {3 left (a+2 right )} } times { {5 left (a - 2 right )} over {5 left (a - 2 right )} } right )+ left ( { {2} over {5 left (a - 2 right )} } times { {3 left (a+2 right )} over {3 left (a+2 right )} } right )} {}

= 15a+3+5a2+6a+215a+2a215a+3+5a2+6a+215a+2a2 size 12{ { {"15" left (a+3 right )+5 left (a - 2 right )+6 left (a+2 right )} over {"15" left (a+2 right ) left (a - 2 right )} } } {} = 15a+45+5a10+6a+1215a+2a215a+45+5a10+6a+1215a+2a2 size 12{ { {"15"a+"45"+5a - "10"+6a+"12"} over {"15" left (a+2 right ) left (a - 2 right )} } } {} = 26a+4715a+2a226a+4715a+2a2 size 12{ { {"26"a+"47"} over {"15" left (a+2 right ) left (a - 2 right )} } } {}

Exercise:

Simplify the following expressions by using factorising:

1. ax2ax+5a2xax2ax+5a2x size 12{ { {a} over {x rSup { size 8{2} } } } - { {a} over {x} } + { {5a} over {2x} } } {}

2. 13+2x+12xx13x13+2x+12xx13x size 12{ { {1} over {3} } + { {2x+1} over {2x} } - { {x - 1} over {3x} } } {}

3. 4a4b2a22b232a2b4a4b2a22b232a2b size 12{ { {4a - 4b} over {2a rSup { size 8{2} } - 2b rSup { size 8{2} } } } - { {3} over {2a - 2b} } } {}

4. 12a2+23a+134a312a2+23a+134a3 size 12{ { {1} over {2} } left (a - 2 right )+ { {2} over {3} } left (a+1 right ) - { {3} over {4} } left (a - 3 right )} {}

Assessment: Assess the 4 problems in the exercise above.

Here is one final trick. We could simplify 2x13x+3×9x+31x2x13x+3×9x+31x size 12{ { {2 left (x - 1 right )} over {3 left (x+3 right )} } times { {9 left (x+3 right )} over { left (1 - x right )} } } {} better if (1–x) had been: (x–1).

So, we make the change we want by changing the sign of the whole binomial as well:

(1–x) = –(x–1) because –(x–1) = –x + 1, which is 1–x. Finish the problem yourself.

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 development of number systems in a variety of historical and cultural contexts (including local); 1.2 recognises, uses and represent rational numbers (including very small numbers written in scientific notation), moving flexibly between equivalent forms in appropriate contexts; 1.3 solves problems in context, including contexts that may be used to build awareness of other Learning Areas, as well as human rights, social, economic and environmental issues such as: 1.3.1 financial (including profit and loss, budgets, accounts, loans, simple and compound interest, hire purchase, exchange rates, commission, rentals and banking); 1.3.2 measurements in Natural Sciences and Technology contexts; 1.4 solves problems that involve ratio, rate and proportion (direct and indirect);

 1.5 estimates and calculates by selecting and using operations appropriate to solving problems and judging the reasonableness of results (including measurement problems that involve rational approximations of irrational numbers); 1.6 uses a range of techniques and tools (including technology) to perform calculations efficiently and to the required degree of accuracy, including the following laws and meanings of exponents (the expectation being that learners should be able to use these laws and meanings in calculations only): 1.6.1 x n × x m = xn + m 1.6.2 x n  x m = xn – m 1.6.3 x 0 = 1 1.6.4 x –n = 1xn1xn size 12{ { {1} over {x rSup { size 8{n} } } } } {} 1.7 recognises, describes and uses the properties of rational numbers. LO 2 Patterns, Functions and AlgebraThe learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills. We know this when the learner: 2.1 investigates, in different ways, a variety of numeric and geometric patterns and relationships by representing and generalising them, and by explaining and justifying the rules that generate them (including patterns found in natural and cultural forms and patterns of the learner’s own creation); 2.7 uses the distributive law and manipulative skills developed in Grade 8 to: find the product of two binomials; factorise algebraic expressions (limited to common factors and difference of squares). 2.8 uses the laws of exponents to simplify expressions and solve equations; 2.9 uses factorisation to simplify algebraic expressions and solve equations.

Memorandum

TEST 1

1. Simplify the following expressions by collecting like terms:

1.1 3a2 + 3a2 – 6a + 3a – 4 + 1

1.2 2y2 – 1y + 2y2 – 6 + 2y – 9

1.3 8x2 – (5x + 12x2 – 1) + x – 4

1.4 (3a – a2) – [(2a2 – 11) – (5a – 3)]

2. Give the answers to the following problems in the simplest form:

2.1 Add 3x2 + 5x – 1 to x2 – 3x

2.2 Find the sum of 2a + 3b – 5 and 3 + 2b – 7a

2.3 Subtract 6a + 7 from 5a2 + 2a + 2

2.4 How much is 3a – 8b + 3 less than a + b + 2?

3.1 (3x2) × (2x3)

3.2 (abc) (a2c) (2b2)

3.3 abc(a2c + 2b2)

3.4 –3a(2a2 – 5a)

3.5 (a – 2b) (a + 2b)

3.6 (3 – x2) (2x2 + 5)

3.7 (x – 5y)2

3.8 (2 – b) (3a + c)

TEST 1 – Memorandum

1.1 6a2 – 3a – 3

1.2 4y2 + y – 15

1.3 – 4x2 – 4x – 3

1.4 – 3a2 + 8a + 8

2.1 4x2 + 2x – 1

2.2 – 5a + 5b – 2

2.3 5a2 – 4a – 5

2.4 – 2a + 9b – 1

3.1 6x5

3.2 2a3b3c2

3.3 a3bc2 + 2ab3c

3.4 – 6a3 + 15a2

3.5 a2 – 4b2

3.6 –2x4 + x2 + 15

3.7 x2 – 10xy + 25y2

3.8 6a + 2c – 3ab – bc

TEST 2

1. Find the Highest Common Factor of these three expressions: 6a2c2 and 2ac2 and 10ab2c3.

2. Completely factorise these expressions by finding common factors:

2.1 12a3 + 3a4

2.2 –5xy – 15x2y2 – 20y

2.3 6a2c2 – 2ac2 + 10ab2c3

3. Factorise these differences of squares completely:

3.1 a2 – 4

3.2 19a29b219a29b2 size 12{ { {1} over {9} } a rSup { size 8{2} } - 9b rSup { size 8{2} } } {}

3.3 x4 – 16y4

3.4 1 – a4b4

4. Factorise these expressions as far as possible:

4.1 3x2 – 27

4.2 2a – 8ab2

4.3 a2 – 5a – 6

4.4 a2 + 7a + 6

5. Simplify the following fractions by making use of factorising:

5.1 3a236a+63a236a+6 size 12{ { {3a rSup { size 8{2} } - 3} over {6a+6} } } {}

5.2 6x2y6y2x26x2y6y2x2 size 12{ { {6x rSup { size 8{2} } y - 6y} over {2x - 2} } } {}

5.3 a292×14a212aa292×14a212a size 12{ { {a rSup { size 8{2} } - 9} over {2} } times { {1} over {4a rSup { size 8{2} } - "12"a} } } {}

5.4 3x+65÷x24153x+65÷x2415 size 12{ { {3x+6} over {5} } div { {x rSup { size 8{2} } - 4} over {"15"} } } {}

5.5 abx2cx+2ac3x+3cx2aabx2cx+2ac3x+3cx2a size 12{ { { ital "abx"} over {2 ital "cx"} } + { {2 ital "ac"} over {3x} } + { {3 ital "cx"} over {2a} } } {}

5.6 2ax23ax+a2x2ax23ax+a2x size 12{ { {2a} over {x rSup { size 8{2} } } } - { {3a} over {x} } + { {a} over {2x} } } {}

5.7 4a4b2a22b232a2b4a4b2a22b232a2b size 12{ { {4a - 4b} over {2a rSup { size 8{2} } - 2b rSup { size 8{2} } } } - { {3} over {2a - 2b} } } {}

5.8 23a+2+13a114a523a+2+13a114a5 size 12{ { {2} over {3} } left (a+2 right )+ { {1} over {3} } left (a - 1 right ) - { {1} over {4} } left (a - 5 right )} {}

TEST 2 – Memorandum

1. 2ac2

2.1 3a3 (4 + a2)

2.2 –5y (x + 3x2y + 4)

2.3 2ac2 (3a – 1 + 5b2c)

3.1 (a + 2) (a – 2)

3.2 13a+3b13a3b13a+3b13a3b size 12{ left ( { {1} over {3} } a+3b right ) left ( { {1} over {3} } a - 3b right )} {}

3.3 (x2 + 4y2) (x + 2y) (x – 2y)

3.4 (1 + a2b2) (1 + ab) (1 – ab)

4.1 3 (x + 3) (x – 3)

4.2 2a (1 + 4b) (1 – 4b)

4.3 (a + 1) (a – 6)

4.4 (a + 1) (a + 6)

5.1 a12a12 size 12{ { {a - 1} over {2} } } {}

5.2 3y (x + 1)

5.3 a+38aa+38a size 12{ { {a+3} over {8a} } } {}

5.4 9x29x2 size 12{ { {9} over {x - 2} } } {}

5.5 3a2bx+4a2c2+9c2x26acx3a2bx+4a2c2+9c2x26acx size 12{ { {3a rSup { size 8{2} } ital "bx"+4a rSup { size 8{2} } c rSup { size 8{2} } +9c rSup { size 8{2} } x rSup { size 8{2} } } over {6 ital "acx"} } } {}

5.6 4a5ax2x24a5ax2x2 size 12{ { {4a - 5 ital "ax"} over {2x rSup { size 8{2} } } } } {}

5.7 a7b2a+baba7b2a+bab size 12{ { {a - 7b} over {2 left (a+b right ) left (a - b right )} } } {}

5.8 3a+943a+94 size 12{ { {3a+9} over {4} } } {}

## Content actions

EPUB (?)

### What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

PDF | EPUB (?)

### What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

#### Collection to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

#### Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

### Reuse / Edit:

Reuse or edit collection (?)

#### Check out and edit

If you have permission to edit this content, using the "Reuse / Edit" action will allow you to check the content out into your Personal Workspace or a shared Workgroup and then make your edits.

#### Derive a copy

If you don't have permission to edit the content, you can still use "Reuse / Edit" to adapt the content by creating a derived copy of it and then editing and publishing the copy.

| Reuse or edit module (?)

#### Check out and edit

If you have permission to edit this content, using the "Reuse / Edit" action will allow you to check the content out into your Personal Workspace or a shared Workgroup and then make your edits.

#### Derive a copy

If you don't have permission to edit the content, you can still use "Reuse / Edit" to adapt the content by creating a derived copy of it and then editing and publishing the copy.