1. Expanding two brackets
With practice we can expand the product of two linear expressions in fewer steps. For example,
(x – 5)(x + 2) = x2
+ 2x
– 5x
– 10
= x2 – 3x – 10
Notice that –3 is the sum of –5 and 2 …
… and that –10 is the product of –5 and 2.
Point out that for any expression in the form (x + a)(x + b), where a and b are fixed numbers, the expanded expression will have an x with a coefficient of a + b and the number at the end will be a × b.

2. Matching quadratic expressions 1
Select a bracketed expression and ask a volunteer to find its corresponding expansion.

3. Matching quadratic expressions 2
Select a bracketed expression and ask a volunteer to find its corresponding expansion.

4. Squaring expressions Expand and simplify: (2 – 3a)2
We can write this as,
(2 – 3a)2 = (2 – 3a)(2 – 3a)
Expanding,
(2 – 3a)(2 – 3a) =
2(2 – 3a) – 3a(2 – 3a) = 4
– 6a
– 6a
+ 9a2
= 4 – 12a + 9a2

5. Squaring expressions In general, (a + b)2 = a2 + 2ab + b2
The first term squared …
… plus 2 × the product of the two terms …
… plus the second term squared.
For example,
(3m + 2n)2 = 9m2 + 12mn + 4n2

6. Squaring expressions
Any of the terms in the expansion can be hidden or revealed to practice squaring expressions.

7. The difference between two squares
Expand and simplify (2a + 7)(2a – 7)
Expanding,
(2a + 7)(2a – 7) =
2a(2a – 7) + 7(2a – 7) =
4a2
– 14a
+ 14a
– 49
= 4a2 – 49
When we simplify, the two middle terms cancel out.
This is the difference between two squares.
In general,
(a + b)(a – b) = a2 – b2

8. The difference between two squares
This animation demonstrate geometrically why a2 – b2 = (a – b)(a + b).
The first two steps in the animation demonstrate a square of area b2 being subtracted from a square of area a2 to give a shape of area a2 – b2.
The shape of area a2 – b2 is then rearranged in to a rectangle of width a + b and height a – b to show that its area is equal to (a – b)(a + b).
a2 – b2 is therefore equal to (a – b)(a + b).

9. Matching the difference between two squares
Select a bracketed expression and ask a volunteer to find its corresponding expansion.

10. A1 Algebraic manipulation
Contents
A1 Algebraic manipulation A
A1.1 Using index laws A
A1.2 Multiplying out brackets A
A1.3 Factorization A
A1.4 Factorizing quadratic expressions A
A1.5 Algebraic fractions

11. Factorizing expressions
Factorizing an expression is the opposite of expanding it.
Expanding or multiplying out
Factorizing
a(b + c)
ab + ac
Often:
When we expand an expression we remove the brackets.
When we factorize an expression we write it with brackets.

12. Factorizing expressions
Expressions can be factorized by dividing each term by a common factor and writing this outside a pair of brackets.
For example, in the expression
5x + 10
the terms 5x and 10 have a common factor, 5.
We can write the 5 outside of a set of brackets and mentally divide 5x + 10 by 5.
We can write the 5 outside of a set of brackets
Encourage pupils to check this by multiplying the expression out to 5x + 10.
(5x + 10) ÷ 5 = x + 2
This is written inside the bracket.
5(x + 2)
5(x + 2)

13. Factorizing expressions
Writing 5x + 10 as 5(x + 2) is called factorizing the expression.
Factorize 6a + 8
Factorize 12n – 9n2
The highest common factor of 6a and 8 is
The highest common factor of 12n and 9n2 is 2.
3n.
(6a + 8) ÷ 2 =
3a + 4
(12n – 9n2) ÷ 3n =
4 – 3n
Point out that we do not normally show the line involving division. This is done mentally. We can check the answer by multiplying out the bracket.
6a + 8 =
2(3a + 4)
12n – 9n2 =
3n(4 – 3n)

14. Factorizing expressions
Writing 5x + 10 as 5(x + 2) is called factorizing the expression.
Factorize 3x + x2
Factorize 2p + 6p2 – 4p3
The highest common factor of 3x and x2 is
The highest common factor of 2p, 6p2 and 4p3 is x.
2p.
(2p + 6p2 – 4p3) ÷ 2p =
(3x + x2) ÷ x =
3 + x
1 + 3p – 2p2
3x + x2 =
x(3 + x)
2p + 6p2 – 4p3 =
2p(1 + 3p – 2p2)

15. Factorization
Start by asking pupils to give you the value of the highest common factor of the two terms. Reveal this and then ask pupils to give you the values of the terms inside the brackets.

16. Factorization by pairing
Some expressions containing four terms can be factorized by regrouping the terms into pairs that share a common factor. For example,
Factorize 4a + ab b
Two terms share a common factor of 4 and the remaining two terms share a common factor of b.
4a + ab b = 4a ab + b
= 4(a + 1) + b(a + 1)
4(a + 1) and + b(a + 1) share a common factor of (a + 1) so we can write this as
(a + 1)(4 + b)

17. Factorization by pairing
Factorize xy – 6 + 2y – 3x
We can regroup the terms in this expression into two pairs of terms that share a common factor.
When we take out a factor of –3, – 6 becomes + 2
xy – 6 + 2y – 3x = xy + 2y – 3x – 6
= y(x + 2) – 3(x + 2)
This expression could also be written as xy – 3x + 2y – 6 to give x(y – 3) + 2(y – 3) = (y – 3)(x + 2)
y(x + 2) and – 3(x + 2) share a common factor of (x + 2) so we can write this as
(x + 2)(y – 3)

18. A1.4 Factorizing quadratic expressions
Contents
A1 Algebraic manipulation A
A1.1 Using index laws A
A1.2 Multiplying out brackets A
A1.3 Factorization A
A1.4 Factorizing quadratic expressions A
A1.5 Algebraic fractions

19. Quadratic expressions
A quadratic expression is an expression in which the highest power of the variable is 2. For example, t2 2
x2 – 2,
w2 + 3w + 1,
4 – 5g2 ,
The general form of a quadratic expression in x is:
ax2 + bx + c (where a = 0)
x is a variable.
As well as the highest power being two, no power in a quadratic expression can be negative or fractional.
Compare each of the quadratic expressions given with the general form.
In x2 – 2, a = 1, b = 0 and c = –2.
In w2 + 3w + 1, a = 1, b = 3 and c = 1. This is a quadratic in w.
In 4 – 5g2, a = –5, b = 0 and c = 4. This is a quadratic in g.
In t2/2, a = ½, b = 0 and c = 0. This is a quadratic in t.
a is a fixed number and is the coefficient of x2.
b is a fixed number and is the coefficient of x.
c is a fixed number and is a constant term.

20. Factorizing expressions
Remember: factorizing an expression is the opposite of expanding it.
Expanding or multiplying out
Factorizing
(a + 1)(a + 2)
a2 + 3a + 2
Often:
When we expand an expression we remove the brackets.
When we factorize an expression we write it with brackets.

21. Factorizing quadratic expressions
Quadratic expressions of the form x2 + bx + c can be factorized if they can be written using brackets as
(x + d)(x + e)
where d and e are integers.
If we expand (x + d)(x + e) we have,
(x + d)(x + e) = x2 + dx + ex + de
= x2 + (d + e)x + de
Pupils will require lots of practice to factorize quadratics effectively.
This slide explains why when we factorize an expression in the form x2 + bx + c to the form (x + d)(x + e) the values of d and e must be chosen so that d + e = b and de = c.
(x + d)(x + e) = x2 + (d + e)x + de is an identity. This means that the coefficients and constant on the left-hand side are equal to the coefficients and constant on the right-hand side.
Comparing this to x2 + bx + c we can see that:
The sum of d and e must be equal to b, the coefficient of x.
The product of d and e must be equal to c, the constant term.

22. Factorizing quadratic expressions 1
Factorize the given expression by finding two integers that add together to give the coefficient of x and multiply together to give the constant.
It may be a good idea to practice adding and multiplying negative numbers before attempting this activity. Use slide 31 in N1.2 Calculating with integers to do this if required.
The lower of the two hidden integers will be given first in each case.

23. Matching quadratic expressions 1
Select a quadratic expression and ask a volunteer to find its corresponding factorization.

24. Factorizing quadratic expressions
Quadratic expressions of the form ax2 + bx + c can be factorized if they can be written using brackets as
(dx + e)(fx + g)
where d, e, f and g are integers.
If we expand (dx + e)(fx + g)we have,
(dx + e)(fx + g)= dfx2 + dgx + efx + eg
= dfx2 + (dg + ef)x + eg
Discuss the factorization of quadratics where the coefficient of x2 is not 1.
Most examples at this level will have a as a prime number so that there are only two factors, 1 and the number itself.
Comparing this to ax2 + bx + c we can see that we must choose d, e, f and g such that: a = df,
b = (dg + ef)
c = eg

25. Factorizing quadratic expressions 2
Factorize each given expression using trial and improvement and the relationships shown on the previous slide.
For each expression in the form ax2 + bx + c, start by using the pen tool to write down pairs of integers that multiply together to make a and pairs of integers that multiply together to make c. Use these to complete the factorization.

26. Matching quadratic expressions 2
Select a quadratic expression and ask a volunteer to find its corresponding factorization.

27. Factorizing the difference between two squares
A quadratic expression in the form
x2 – a2
is called the difference between two squares.
The difference between two squares can be factorized as follows:
x2 – a2 = (x + a)(x – a)
For example,
See slide 40 to demonstrate the expansion of expressions of the form (x + a)(x – a).
Pupils should be encouraged to spot the difference between two squares whenever possible.
9x2 – 16 = (3x + 4)(3x – 4)
25a2 – 1 = (5a + 1)(5a – 1)
m4 – 49n2 = (m2 + 7n)(m2 – 7n)

28. Factorizing the difference between two squares

29. Matching the difference between two squares
Select an expression involving the difference between two squares and ask a volunteer to find the corresponding factorization.

30. A1 Algebraic manipulation
Contents
A1 Algebraic manipulation A
A1.1 Using index laws A
A1.2 Multiplying out brackets A
A1.3 Factorization A
A1.4 Factorizing quadratic expressions A
A1.5 Algebraic fractions

31. Algebraic fractions 3x 4x2 2a 3a + 2
and are examples of algebraic fractions.
The rules that apply to numerical fractions also apply to algebraic fractions.
For example, if we multiply or divide the numerator or the denominator of a fraction by the same number or term we produce an equivalent fraction.
It is important to realize that, like numerical fractions, multiplying or dividing the numerator and the denominator of an algebraic fraction by the same number, term or expression does not change the value of the fraction. This fact is used both when simplifying algebraic fractions and when writing algebraic fractions over a common denominator to add or subtract them.
For example, 3x
4x2 = 3 4x = 6 8x = 3y
4xy =
3(a + 2)
4x(a + 2)

32. Equivalent algebraic fractions
It is important to realize that, like numerical fractions, multiplying or dividing the numerator and the denominator of an algebraic fraction by the same number, term or expression does not change the value of the fraction. This fact is used both when simplifying algebraic fractions and when writing algebraic fractions over a common denominator to add or subtract them.

33. Simplifying algebraic fractions
We simplify or cancel algebraic fractions in the same way as numerical fractions, by dividing the numerator and the denominator by common factors. For example,
Simplify
6ab
3ab2 2
6ab
3ab2 =
6 × a × b
3 × a × b × b = 2 b

34. Simplifying algebraic fractions
Sometimes we need to factorize the numerator and the denominator before we can simplify an algebraic fraction. For example,
Simplify
2a + a2
8 + 4a
2a + a2
8 + 4a =
a (2 + a)
4(2 + a) = a 4

35. Simplifying algebraic fractions
b2 – 36 is the difference between two squares.
Simplify
b2 – 36
3b – 18
b2 – 36
3b – 18 =
(b + 6)(b – 6)
3(b – 6)
b + 6 3 =
Pupils should be encouraged to spot the difference between two squares whenever possible.
If required, we can write this as 6 3 = b + b 3
+ 2

36. Manipulating algebraic fractions
Remember, a fraction written in the form
a + b c
can be written as b a +
However, a fraction written in the form c
a + b
cannot be written as b a +
Stress that if two terms are added or subtracted in the numerator of a fraction, we can split the fraction into two fractions written over a common denominator. The converse is also true.
However, if two terms are added or subtracted in the denominator of a fraction, we cannot split the fraction into two.
Verify these rules using the numerical example.
For example,
1 + 2 3 = 2 1 + 3
1 + 2 = 2 1 +
but

37. Multiplying and dividing algebraic fractions
We can multiply and divide algebraic fractions using the same rules that we use for numerical fractions.
In general, a b
× = c d ac bd a b
÷ = c d
× = ad bc
and,
Point out to pupils that in the example we could multiply out the denominator. However, it is usually preferable to leave expressions in a factorized form.
For example, 3p 4
× = 2
(1 – p) 3 6p
4(1 – p) = 3p
2(1 – p) 2

38. Multiplying and dividing algebraic fractions
What is 2
3y – 6 ÷ 4
y – 2 ?
This is the reciprocal of 4
y – 2 2
3y – 6 ÷ = 4
y – 2 2
3y – 6 × 4
y – 2 2
3(y – 2) × = 4
y – 2 2 1 6 =

39. Adding algebraic fractions
We can add algebraic fractions using the same method that we use for numerical fractions. For example,
What is 1 a + 2 b ?
We need to write the fractions over a common denominator before we can add them. 1 a + 2 b = b ab + 2a =
b + 2a ab
If necessary review the method for adding numerical fractions.
In general, + = a b c d
ad + bc bd

40. Adding algebraic fractions3 y
What is + ? x 2
We need to write the fractions over a common denominator before we can add them. 3 x + y 2 = +
3 × 2
x × 2
y × x
2 × x + 6 2x xy = =
6 + xy 2x

41. Subtracting algebraic fractions
We can also subtract algebraic fractions using the same method as we use for numerical fractions. For example,
What is – ? p 3 q 2
We need to write the fractions over a common denominator before we can subtract them. – = p 3 q 2 – = 2p 6 3q
2p – 3q 6
In general, – = a b c d
ad – bc bd

42. Subtracting algebraic fractions
2 + p 4 3 2q
What is – ? = –
2 + p 4 3 2q –
(2 + p) × 2q
4 × 2q
3 × 4
2q × 4 = –
2q(2 + p) 8q 12 =
2q(2 + p) – 12 8q 6
The denominators in this example share a common factor. That means that we will either have to cancel at the end of the calculation. Alternatively, we could use a common denominator of 4q in the first step. 4 =
q(2 + p) – 6 4q

43. Addition pyramid – algebraic fractions
Start by revealing the three fractions on the bottom row of the wall. Add the fractions together to find the missing values in the bocks above. Each block is the sum of the two fractions below it.
Ask pupils if it is true to say that the fraction in the top row is the sum of the three fractions in the bottom row. Conclude that if the three factions on the bottom row are a, b and c then the fraction on the top row is a + 2b + c.
The activity can be varied by revealing one fraction in each row and using subtraction to find those that are missing.