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Form 4 Algebra Part 1

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Chapter 2 – Algebra Part 1

Section 2.1 – Expansion (Revision)

In Mathematics EXPANSION really means MULTIPLY. For example 3(2x + 4) can be expanded by

multiplying them out.

Remember: There is an invisible multiplication sign between the outside number and the opening

bracket. Therefore 3(2x + 4) is really 3 ╳ (2x+4)

You expand by multiplying everything inside the bracket by what is outside the bracket.

Example

1) 3(2x + 4) = 3 ╳ (2x+4) = (3 ╳ 2x) + (3 ╳ 4) = 6x + 12

2) 4y2(2y + 3) = 4y2 ╳ (2y + 3) = (4y2 ╳ 2y) + (4y2 ╳ 3) = 8y3 + 12y2

3) −3(2 + 3x) = −3 ╳ (2 + 3x) = (−3 ╳ 2) + (−3 ╳ 3x) = −6 – 9x [Note: The sign changes when a minus is

outside the brackets]

Consolidation

1) 2(3 + m) _____________________________

2) t (t + 4) _____________________________

3) 5h(3h – 2) _____________________________

4) 3d (5d2 – d3) _____________________________

5) 2m2 (4m + m2) _____________________________


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Expand and Simplify

When two brackets are expanded there are often like terms that can be collected together. Algebraic expressions should always be simplified as much as possible.

Example

1) 3(4 + m) + 2(5 + 2m) = 12 + 3m + 10 + 4m = 22 + 6m

2) 3t(5t + 4) – 2t(3t – 5) = 15t2 + 12t – 6t2 + 10t = 9t2 + 22t

Consolidation: Expand and Simplify the following:-

1) 4a(2b + 3c) + 3b(3a + 2c)

2) 3y(4w + 2t) + 2w(3y – 4t)

3) 5m(2n – 3p) – 2n(3p – 2m)

4) 2r(3r + r2) – 3r2(4 – 2r)

5) 4e(3e – 5) – 2e(e – 7)

6) 3k(2k + p) – 2k(3p – 2m)

7) 2y(3 + 4y) + y(5y – 1)


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Quadratic Expansion

A quadratic expression is one which the highest power of the terms is 2.

For example:

y2 2d2 + 4d 5m2 + 3m – 2

In the expansion method, split the terms in the first set of brackets, make each of them multiply both terms in the second set of brackets, and then simplify the outcome.

Example

(x + 3)(x + 4) = x ╳ (x + 4) + 3 ╳ (x + 4)

= x2 + 4x + 3x + 12

= x2 + 7x + 12

Example

1) (y -‐ 2)(y + 5) = y ╳ (y + 5) – 2 ╳ (y + 5)

= y2 + 5y – 2y – 10

= y2 + 3y – 10

2) (2t + 3)(3t + 1) = 2t ╳ (3t + 1) + 3 ╳ (3t + 1)

= 6t2 + 2t + 9t + 3

= 6t2 + 11t + 3


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3) (x + 3)2 = (x + 3)(x + 3)

= x ╳ (x+ 3) + 3 ╳ (x+ 3)

= x2 + 3x + 3x + 9

= x2 + 6x + 9

Consolidation: Expand and Simplify the following:-

1) (w + 3)(w -‐ 1)

2) (p -‐ 2)(p -‐ 1)

3) (7 + g)(7 -‐ g)

4) (4 + 3p)(2p + 1)

5) (3g -‐ 2)(5g -‐ 2)

6) (3 – 2q)(4 + 5q)


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7) (1 – 3p)(3 + 2p)

8) (m + 4)2

9) (4t + 3)2

10) (m -‐ n)2

11) (x -‐ 2)2 – 4

Support Exercise Pg 107 Exercise 8A No 1 – 4

Pg 110 Exercise 8C No 1 – 4

Section 2.2 – Factorisation by taking out the common factor

Factorisation is the process of putting mathematical expressions into brackets. It is the opposite of expansion. If we write the very first expression that you saw backwards, then we have factorised it:

5 (x + 2) = 5x + 10


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In this case, we look at the terms (two of them in this case, although they could be more) and we find something that divides into BOTH of them. This is written outside the brackets, and the rest of each term (with the appropriate + or -‐ sign) is written inside. In order to do this we must find the HCF of the terms.

Example

1) 6m + 12t = 6(m + 2t)

2) 5g2 + 3g = g(5g + 3)

3) 8abc + 6bed = 2b(4ac + 3ed) [We sometimes have both a letter and number which are common]

4) 6mt2 – 3mt + 9m2t = 3mt(2t – 1 + 3m)

Consolidation: Factorise the following:-

1) 9t + 3p

2) mn + 3n

3) 3m2 – 3mp

4) 5b2c – 10bc

5) 6ab + 9bc + 3bd

6) 5t2 + 4t + at


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7) 8ab2 + 2ab – 4a2b

8) 10pt2 + 15pt + 5p2t

Support Exercise Pg 108 Exercise 8B No 1 – 2

Section 2.3 – Factorising by Grouping Like Terms

In the previous section, whilst factorizing, the common factor was always a single term (e.g. 3, 4a, ab, etc…)

The common factor does not always have to be a single term, it can be a sum or difference of terms (e.g. x + 2, 3x – 4)

Example

1. 2(x – 4) + x(x – 4) [(x -‐ 4) can be considered as a common term]

(x – 4)(2 + x)

We can have an expression which has both a number and a sum or difference which are common.

2. 12(x + 2)2 -‐ 9(x + 2) [(x + 2) can be considered as a common term]

3 ╳ 4 ╳ (x + 2)(x + 2) – 3 ╳ 3 ╳ (x + 2) [3 and (x + 2) are both factors]

3(x + 2)[4(x + 2) – 3] [So write 3(x + 2) outside the square brackets]

3(x + 2)[4x + 8 – 3] [Simplify the terms inside the square brackets]

3(x + 2)(4x + 5)


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3. 10x(x – 5) – 5(x – 5)2

5 ╳ 2 ╳ x ╳ (x – 5) – 5 ╳ (x – 5) ╳ (x – 5)

5(x – 5)[2x – (x – 5)]

5(x – 5)(2x – x + 5)

5(x – 5)(x + 5)

Consolidation: Factorise the following completely:-

1. a(b + c) – d(b + c)

2. y(x – 6) + 2(x – 6)

3. 6(x +3)2 – 3(x + 3)

4. (y + 2)2 – 4(y + 2)


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When four or more terms come together to form an expression, you always look for a greatest common factor first.

If you can’t find a factor common to all the terms at the same time, your other option is grouping. To group, you take the terms two at a time and look for common factors for each of the pairs on an individual basis.

After factoring, you see if the new groupings have a common factor.

The best way to explain this is to demonstrate the factoring by grouping on a few examples.

Example:

1. The four terms

don’t have a common factor. However, the first two terms have a common factor of

and the last two terms have a common factor of 3:

Notice that you now have two terms, not four, and they both have the factor (x – 4). Now, factoring (x – 4) out of each term, you have

Factoring by grouping only works if a new common factor appears — the exact same one in each term.

2. Now, consider the expression 7x + 14y + bx + 2by. Clearly, there is no factor common to every term.

However, it is clear that 7 is a common factor of the first two terms and b is a common factor of the last two terms. So, the expression can be grouped into two pairs of two terms as shown.


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3. The six terms

don’t have a common factor, but, taking them two at a time, you can pull out the factors

Factoring by grouping, you get the following:

The three new terms have a common factor of (x – 2), so the factorization becomes

Consolidation: Factorise the following completely:-

1. 6x + 9 + 2ax + 3a



2. x2 – 6x + 5x – 30

3. 5x + 10y – ax – 2ay

4. a2 – 2a – ax + 2x

Support Exercise Pg 111 Exercise 8D Nos 1 – 2

Section 2.4 – Factorising a Trinomial of the form of x2 + bx + c

Expanding (x + 4)(x + 2) gives x2 + 2x + 4x + 8

x2 + 6x + 8

Since factorization is the opposite of expanding the factorization of the expression x2 + 6x + 8 gives (x + 4)(x + 2)



Sometimes it is easy to put a quadratic expression back into its brackets, other times it seems hard. However, there are some simple rules that will help you to factorise.

• The expression inside each set of brackets will start with an x, and the signs in the quadratic expression show which signs to put after the xs.

• When the second sign in the expression is a plus, the signs in both sets of brackets are the same as the first sign.

x2 + ax + b = (x + ?)(x + ?) Since everything is positive.

x2 – ax + b = (x -‐ ?)(x -‐ ?) Since negative ╳ negative = positive

• Next, look at the last number, b, in the expression. When multiplied together, the two numbers in the brackets must give b.

• Finally, look at the coefficient of x, which is a. The sum of the two numbers in the brackets will give a.

Example

1. Factorise x2 + 5x + 6

Because of the signs we know that the signs must be of the form (x + ?)(x + ?).

Two numbers that have a product of 6 and a sum of 5 are 3 and 2.

Therefore, (x + 2)(x + 3)

2. Factorise x2 – 9x + 20

Because of the signs the brackets must be of the form (x -‐ ?)(x -‐ ?)

Two numbers that have a product which gives 20 and a sum of 9 are 4 and 5.

Therefore, (x – 4)(x – 5)

3. Factorise x2 – 7x + 10 Because of the signs the brackets must be of the form (x -‐ ?)(x -‐ ?)

Two numbers that have a product which gives 10 and a sum of -‐7 are -‐5 and -‐2.

Therefore, (x – 5)(x – 2)



Consolidation: Factorise the following expressions:-

1. x2 + 5x + 6

2. k2 + 10k + 24

3. w2 + 11w + 18

4. t2 – 5t + 6

5. y2 – 16y + 48

6. y2 + 6y + 8

7. x2 + 16y + 39



8. x2 – 11x + 30

9. x2 – 9x + 14

10. x2 + 15x + 56

Support Exercise Pg 113 Exercise 8E No 2 (a – g, m – o)

Section 2.5 – Factorising a Trinomial of the form of x2 + bx – c

Expanding (x – 3)(x + 2) gives x2 + 2x – 3x – 6

x2 – x – 6

Since factorization is the opposite of expanding the factorization of the expression x2 – x – 6 gives (x – 3)(x + 2)

• When the second sign is a minus, the signs in the brackets are different.

x2 + ax – b = (x + ?)(x -‐ ?) Since positive ╳ negative = negative

x2 – ax – b = (x + ?)(x -‐ ?)

The larger factor will have the minus sign before it.

• Next, look at the last number, b, in the expression. When multiplied together, the two numbers in the brackets must give b.

• Finally, look at the coefficient of x, which is a. The sum of the two numbers in the brackets will give a.



1. Factorise x2 – x – 6

Because of the signs we know that the signs must be of the form (x + ?)(x -‐ ?).

Two numbers that have a product of -‐6 and a sum of -‐1 are 3 and 2.

The larger factor of these two factors is 3, therefore the minus must go with it.

Therefore, (x + 2)(x – 3)

2. Factorise x2 + 3x – 18

Because of the signs we know that the signs must be of the form (x + ?)(x -‐ ?).

Two numbers that have a product of -‐18 and a sum of 3 are 6 and 3.

The larger factor of these two factors is 6, therefore the plus must go with it.

Therefore, (x + 6)(x – 3)

Consolidation: Factorise the following expressions:-

1. y2 + 5y – 6

2. m2 – 4m – 12

3. h2 – h – 72



4. x2 + 4x – 21

5. x2 – 4x – 12

6. r2 – 12r – 28

7. x2 + 2x – 24

8. x2 – x – 20

9. x2 -‐ 4x – 21

10. h2 + h -‐ 72

Support Exercise Pg 113 Exercise 8E No 2 (h – l, p, q)



Section 2.6 Factorising Mixed Examples

Mixed Consolidation Examples

1. x2 -‐ 10x + 9

2. x2 + x – 12

3. x2 – 6x – 16

4. x2 – 5x – 14

5. x2 – x – 2

6. x2 – 12x + 20



7. x2 – 14x + 24

8. x2 + 6x + 8

9. x2 – 9x + 20

10. x2 + 4x + 3

Support Exercise Handout

Section 2.7 : Factorising ax2 + bx + c

We can adapt the method for factorizing x2 + ax + b to take into account the factors of the coefficient of

x2.

Example

1. Factorise 3x2 + 8x + 4

• First, note that both signs are positive. So the signs in the brackets must be (?x + ?)(?x + ?)

• As 2 has only 3 ╳ 1 as factors, the brackets must be (3x + ?)(x + ?)



• Next, notes that the factors of 4 are 4 ╳ 1 and 2 ╳ 2

• Now find which pair of factors of 4 combined with the factors 3 give 8

3 4 2

1 1 2

You can see that the combination 3 ╳ 2 and 1 ╳ 2 adds up to 8

• So the complete factorization becomes (3x + 2)(x + 2)

2. Factorise 6x2 – 7x – 10

• Note that both signs are negative. So the signs in the brackets must be (?x + ?)(?x -‐ ?)

• As 6 has 6 ╳ 1 and 3 ╳ 2 as fctors, the brackets could be (6x ± ?)(x ± ?) or (3x ± ?)(2x ± ?)

• Note that the factors of 10 are 5 ╳ 2 and 10 ╳ 1

• Now find which pair of factors of 10 combined with the factors of 6 give – 7 .

3 6 ±1 ±2

2 1 ±10 ±5

You can see that the combination 6 ╳ -‐ 2 and 1 ╳ 5 adds up to -‐7.

• So , the complete factorization becomes (6x + 5)(x – 2)




1. 2x2 + 5x + 2

2. 7x2 + 8x + 1

3. 4x2 + 3x – 7

4. 24t2 + 19t + 2

Support Exercise Pg 458 Exercise 28A No 1 – 26

Harder Trinomial Factorisation Handout

Section 2.8 Factorisation of Harder Trinomials –ax2 +bx + c

It is not always possible to have a positive x2 in the trinomial which we will be factorizing. In order to

factorise polynomials with a negative x2 we must follow the following steps.



Example 1

Factorise –x2 + 5x – 6

• Factorise by making the leading term POSITIVE. We do this by taking out a -‐1. [Remember to

change the signs throughout the trinomial].This will give:

-‐x2 + 5x – 6 = -‐1(x2 – 5x + 6) = -‐ (x2 – 5x + 6)

• Factorise the bracket normally (Remember not to forget the minus sign outside the brackets)

-‐ (x2 – 5x + 6) = -‐ (x – 3)(x – 2)

• Once the bracket is factorised you may multiply the -‐1 with the first bracket

-‐ (x – 3)(x – 2) = (-‐x – 3)(x – 2)

Consolidation

1. –x2 – 2x + 3

2. –x2 +x + 6

3. – 2x2 – 5x + 3



4. – m2 – 10m – 16

5. -‐6x2 – x + 7

Support Exercise Harder Trinomial Factorisation Handout

Section 2.9 Factorising a Difference of Two Squares

In Section 2.1 we multiplied out, for example (a + b)(a – b) and obtained a2 – b2. This type of quadratic expression, with only two terms, both of which are perfect squares separated by a minus sign, is called the difference of two squares.

The following are examples of differences of two squares.

x2 – 9 x2 – 25 x2 – 4 x2 – 100

There are three conditions that must be met for difference of two squares to work.

• There must be two terms • They must be separated by a minus sign • Each term must be a perfect square, say x2 and n2

When these three conditions are met the factorization is:

x2 – n2 = (x + n)(x – n)



Example

1. Factorise x2 – 36

Recognise the difference of two squares x2 and 62

So it factorises to (x + 6)(x – 6)

To check your answer, expand the brackets once again.

2. Factorise 9x2 – 169

Recognise the difference of two squares (3x)2 and 132

So it factorises to (3x + 13)(3x – 13)

To check your answer, expand the brackets once again.


1. x2 – 9

2. m2 – 16

3. 9 – x2

4. x2 – 64

5. t2 – 81



6. x2 – y2

7. 9x2 – 1

8. 4x2 – 9y2

9. 16y2 – 25x2

Support Exercise Pg 115 Exercise 8F Nos 1 – 4

Section 2.10 : Simplifying Algebraic Fractions (Rational Expressions)

Algebraic expressions in the form of fractions are called Rational Expressions.

Each of these rational expressions can be simplified by factorizing the numerator and denominator and then cancelling any expression which is common.

For this section we must keep in mind all the factorizing methods which we have learnt till now.

The following rules are used to work out the value of fractions:

• Multiplication 𝑎𝑏×𝑐𝑑=𝑎𝑐𝑏𝑑

• Division 𝑎𝑏÷𝑐𝑑=𝑎𝑏×𝑑𝑐=𝑎𝑑𝑏𝑐

Note that a, b, c and d can be numbers, other letters or algebraic expressions. Remember:



• use brackets, if necessary

• factorise if you can

• cancel if you can

Example

1. !!!

!!!= !

! [We just cancel out top and bottom]

2. !(!!!)!!! ! =

!(!!!)!!! (!!!)

= !(!!!)

[(x – 3) is common in the numerator and denominator

and therefore we can cancel ]

3. Simplify fully !!!!!!!!!!

2x2 + 4x = 2x(x + 2) [Factorise the numerator]

x2 – 4 = (x + 2)(x – 2) [Factorise the denominator with difference of two squares]

!!!!!!!!!!

= !!(!!!)!!! (!!!)

[Write !!!!!!!!!!

as a fully factorised term]

= !!(!!!)!!! (!!!)

[Cancel the common factor (x + 2)]

!!!!!!!!!!

= !!(!!!)

[ !!(!!!)

is usually written as !!!!!

It is not possible to simplify !!!!!

any further.]

4. Simplify fully !!!!!!!!!!!

3x+ 3 = 3(x + 1) [Factorise the numerator]

x2 + 3x + 2 = (x + 2)(x + 1) [Factorise the denominator]

!!!!!!!!!!!

= !(!!!)!!! (!!!)

3𝑥 + 3𝑥! + 3𝑥 + 2

=3

𝑥 + 2



Consolidation: Simplify fully:-

1. !(!!!)!

(!!!)

2. !"!!!

!"!!!

3. !!!!!!!!!!!




Section 2.11 : Simplifying Rational Expressions

Example

1. Simplify fully !!!!!!!!!!!

x2 – 9 = (x + 3)(x – 3) [Factorise the numerator]

x2 – 2x – 3 = (x – 3)(x + 1) [Factorise the denominator]

!!!!!!!!!!!

= !!! (!!!)!!! (!!!)

𝑥! − 9𝑥! − 2𝑥 − 3

=𝑥 + 3𝑥 + 1

2. Simplify fully !!!!

!!!!!!!

4 – x2 = (2 + x)(2 – x) [Factorise the numerator]

x2 – 3x + 2 = (x – 2)(x – 1) [Factorise the denominator]

4 − 𝑥!

𝑥! − 3𝑥 + 2=

2 − 𝑥 (2 + 𝑥)𝑥 − 2 (𝑥 − 1)

!!!!

!!!!!!!= !!! (!!!)

!! !!! (!!!) [(x – 2) = -‐1 (2 – x)]

4 − 𝑥!

𝑥! − 3𝑥 + 2=−1(2 + 𝑥)(𝑥 − 1)

Consolidation: Simplify fully:-

1. !!!!!!!!!!!"



2. !!!!!!!!!!!!!

3. !!!!!!!!!!

4. !!!!!"!!!!

!!!!!!!




Section 2.12 Adding and Subtracting Rational Expressions

The following rules are used to work out the value of fractions:

Addition 𝑎𝑏+𝑐𝑑=𝑎𝑑 + 𝑏𝑐𝑏𝑑

Subtraction 𝑎𝑏−𝑐𝑑=𝑎𝑑 − 𝑏𝑐𝑏𝑑

Example

1. Simplify !!+ !

!!

!!+ !

!!= ! !! ! ! (!)

! (!!) [Find the LCM by multiplying the denominators

and arrange the numerators accordingly]

1𝑥+𝑥2𝑦

=2𝑦 + 𝑥!

2𝑥𝑦

2. Simplify !!− !

!!

!!− !

!!= ! !! !!(!)

!!! [Find LCM and arrange numerator]

!!− !

!!= !!!!"

!!!

!!− !

!!= !!!!"

!!!= ! !!!

!!! [Take out the common if possible and cancel]

2𝑏−𝑎2𝑏

=4 − 𝑎2𝑏

Consolidation: Simplify the following:-

1. !!+ !

!



2. !"!− !

!

3. !!!!+ !!!!

!

4. !!!!!

− !!!!!

Example

1. Write !!!!

− !!!!

as a single fraction as simply as possible.

• To find the LCM we have to multiply the denominator

3𝑥 − 1

−2

𝑥 + 1=3 𝑥 + 1 − 2(𝑥 − 1)

𝑥 − 1 (𝑥 + 1)

• Expand the numerator

3𝑥 − 1

−2

𝑥 + 1=3𝑥 + 3 − 2𝑥 + 2𝑥 − 1 (𝑥 + 1)



• Collect like terms

3𝑥 − 1

−2

𝑥 + 1=

𝑥 + 5𝑥 − 1 (𝑥 + 1)

2. Write !!− !

!!!!! as a single fraction in its simplest form

• First factorise all the denominators where it is possible

1𝑥−

2𝑥! + 2𝑥

=1𝑥−

2𝑥(𝑥 + 2)

• The LCM is x(x+2)

1𝑥=

𝑥 + 2𝑥(𝑥 + 2)

1𝑥−

2𝑥(𝑥 + 2)

=𝑥 + 2

𝑥(𝑥 + 2)−

2𝑥(𝑥 + 2)

• Since we now have the same denominator we just have to subtract the numerators

1𝑥−

2𝑥(𝑥 + 2)

=𝑥 + 2 − 2𝑥(𝑥 + 2)

1𝑥−

2𝑥(𝑥 + 2)

=𝑥

𝑥(𝑥 + 2)

• Cancel if possible

1𝑥−

2𝑥(𝑥 + 2)

=1

𝑥 + 2

3. Simplify !!!!!

+ !!!!!"

− !!"!!!"

• Factorise the denominators

13𝑥 + 6

+1

5𝑥 + 10−

215𝑥 + 30

=1

3(𝑥 + 2)+

15(𝑥 + 2)

−2

15(𝑥 + 2)

╳ (x+2)



• To find the LCM we need a common factor for the number (15) and we can notice that (x+2) is in each fraction.

LCM = 15(x + 2)

• We must arrange each fraction with denominator 15(x + 2)

13(𝑥 + 2)

=3

15(𝑥 + 2)

15(𝑥 + 2)

=3

15(𝑥 + 2)

13(𝑥 + 2)

+1

5(𝑥 + 2)−

215(𝑥 + 2)

=5

15(𝑥 + 2)+

315(𝑥 + 2)

−2

15(𝑥 + 2)

• Since the denominators are the same just combine the numerators

515(𝑥 + 2)

+3

15(𝑥 + 2)−

215(𝑥 + 2)

=5 + 3 − 215(𝑥 + 2)

=6

15(𝑥 + 2)

• Cancel top and bottom by 3

• = !!(!!!)

Consolidation: Simplify the following

1. !!!!!

− !!!!!



2. !!!!

+ !!!!!

3. !!!!!!!!

+ !!!!!

4. !!!!!!!!

+ !!!!

Support Exercise Pg 463 Exercise 28C No 1-‐20

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