Algebra Part 1 Notes - coremaths.webs.com 4/Algebra Part 1 Notes.pdf · In Mathematics EXPANSION...
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Form 4 Algebra Part 1
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) = 4y
2 ╳ (2y + 3) = (4y
2 ╳ 2y) + (4y
2 ╳ 3) = 8y
3 + 12y
2
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 – d
3) _____________________________
5) 2m2 (4m + m
2) _____________________________
Form 4 Algebra Part 1
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 – 6t
2 + 10t = 9t
2 + 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) – 3r
2(4 – 2r)
5) 4e(3e – 5) – 2e(e – 7)
6) 3k(2k + p) – 2k(3p – 2m)
7) 2y(3 + 4y) + y(5y – 1)
Form 4 Algebra Part 1
Quadratic Expansion
A quadratic expression is one which the highest power of the terms is 2.
For example:
y2 2d
2 + 4d 5m
2 + 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
Form 4 Algebra Part 1
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)
Form 4 Algebra Part 1
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
Form 4 Algebra Part 1
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 + 9m
2t = 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
Form 4 Algebra Part 1
7) 8ab2 + 2ab – 4a
2b
8) 10pt2 + 15pt + 5p
2t
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)
Form 4 Algebra Part 1
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)
Form 4 Algebra Part 1
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.
Form 4 Algebra Part 1
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
Form 4 Algebra Part 1
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)
Form 4 Algebra Part 1
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)
Form 4 Algebra Part 1
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
Form 4 Algebra Part 1
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.
Form 4 Algebra Part 1
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
Form 4 Algebra Part 1
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)
Form 4 Algebra Part 1
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
Form 4 Algebra Part 1
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 + ?)
Form 4 Algebra Part 1
• 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)
Form 4 Algebra Part 1
Consolidation: Factorise the following:-
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.
Form 4 Algebra Part 1
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(x
2 – 5x + 6) = - (x
2 – 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. –x2 + x + 6
Form 4 Algebra Part 1
5. – m2 – 10m – 16
6. -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 – b
2. 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 x
2 – 25 x
2 – 4 x
2 – 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 n
2
When these three conditions are met the factorization is:
x2 – n
2 = (x + n)(x – n)
Form 4 Algebra Part 1
Example
1. Factorise x2 – 36
Recognise the difference of two squares x2 and 6
2
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 13
2
So it factorises to (3x + 13)(3x – 13)
To check your answer, expand the brackets once again.
Consolidation: Factorise the following:-
1. x2 – 9
2. m2 – 16
3. 9 – x2
4. x2 – 64
5. t2 – 81
Form 4 Algebra Part 1
6. x2 – y
2
7. 9x2 – 1
8. 4x2
– 9y2
9. 16y2 – 25x
2
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
• Division
Note that a, b, c and d can be numbers, other letters or algebraic expressions. Remember:
Form 4 Algebra Part 1
• 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]
Form 4 Algebra Part 1
Support Exercise Pg 460 Exercise 28B No 1 – 11
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]
2. Simplify fully
4 – x2 = (2 + x)(2 – x) [Factorise the numerator]
x2 – 3x + 2 = (x – 2)(x – 1) [Factorise the denominator]
[(x – 2) = -1 (2 – x)]
Form 4 Algebra Part 1
4.
Support Exercise Pg 460 Exercise 28B No 12 – 25
Section 2.12 Adding and Subtracting Rational Expressions
The following rules are used to work out the value of fractions:
Addition
Subtraction
Example
1. Simplify
[Find the LCM by multiplying the denominators
and arrange the numerators accordingly]
Form 4 Algebra Part 1
2. Simplify
[Find LCM and arrange numerator]
[Take out the common if possible and cancel]
Consolidation: Simplify the following:-
1.
2.
3.
Form 4 Algebra Part 1
4.
Example
1. Write as a single fraction as simply as possible.
• To find the LCM we have to multiply the denominator
• Expand the numerator
• Collect like terms
2. Write as a single fraction in its simplest form
• First factorise all the denominators where it is possible
Form 4 Algebra Part 1
• The LCM is x(x+2)
• Since we now have the same denominator we just have to subtract the numerators
• Cancel if possible
3. Simplify
• Factorise the denominators
• 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)
╳ (x+2)
Form 4 Algebra Part 1
• Since the denominators are the same just combine the numerators
• Cancel top and bottom by 3
•
Consolidation: Simplify the following
1.
2.