Mathematics - Mathematics Pre-Diploma Studies SL (MYP 5 Plus) - First Edition

ALGEBRAIC SIMPLIFICATION AND EXPANSION (Chapter 9) 177

What to do:

1 Play the following ‘think of a Think of a number.
number’ game with a partner: Add 4.
Double the result.
Subtract 2.
Halve the result.
Subtract your original number.

Repeat the game choosing different numbers. You should find that the answer is
always 3. Why is this so?

2 Algebra can provide an insight into why the answer to this game is always 3.

Let x represent the starting number.

Copy and complete the following argument by writing down each step in terms of x:

Think of a number: x

Add 4 gives x+4

Double the result gives 2(x + 4) = :::: + ::::

Subtract 2 gives :::: + :::: ¡ 2 = :::: + ::::
Halve the result gives
Subtract your original number gives 1 (:::: + ::::) = :::: + ::::
2

:::: + :::: ¡ x = ::::

3 Try the following ‘think of a number’ game: Think of a number.
Double it.
Add 8.
Halve the result.
Subtract 4.

What is your answer? Repeat the game using different numbers.

4 For the game above, let x be the starting number. Use algebra to show how the game
works.

5 Make up your own ‘think of a number’ game. Test it with algebra before you try it
on others.

D THE EXPANSION OF (a¡+¡b)(c¡+¡d)

Products like (a + b)(c + d) can be expanded by repeated use of the distributive law.

For example, (x + 3)(x + 2) (1) Compare: 2(x + 2)
(2) =2£x + 2£2
= (x + 3)x + (x + 3)2 (3)
(4) Notice that the distributive law for
= x(x + 3) + 2(x + 3) bracket expansion was used three
= x2 + 3x + 2x + 6 times: once to get line (2) and twice
= x2 + 5x + 6 to get line (4).

178 ALGEBRAIC SIMPLIFICATION AND EXPANSION (Chapter 9)

Example 17 Self Tutor

Expand and simplify by repeated use of the distributive law:

a (x + 5)(x + 4) b (x ¡ 2)(2x ¡ 1)

a (x + 5)(x + 4) b (x ¡ 2)(2x ¡ 1)
= (x + 5)x + (x + 5)4 = (x ¡ 2)2x + (x ¡ 2)(¡1)
= x(x + 5) + 4(x + 5) = 2x(x ¡ 2) ¡ (x ¡ 2)
= x2 + 5x + 4x + 20 = 2x2 ¡ 4x ¡ x + 2
= x2 + 9x + 20 = 2x2 ¡ 5x + 2

EXERCISE 9D

1 Expand and simplify by repeated use of the distributive law:

a (x + 1)(x + 4) b (a + 3)(a + 2) c (c + 1)(c ¡ 4)
d (a ¡ 2)(a ¡ 5) e (w + x)(y + z) f (p + q)(a + b)
g (x ¡ 1)(3x + 2) h (1 ¡ x)(2x + 3) i (2x + 5)(x ¡ 3)
j (3x ¡ 2)(x + 4) k (4x ¡ 3)(3x ¡ 5) l (x ¡ 1)(x2 + 5)

Example 18 Self Tutor

Expand using the distributive law repeatedly: a (x + 5)2 b (x ¡ 5)2

a (x + 5)2 b (x ¡ 5)2
= (x + 5)(x + 5) = (x ¡ 5)(x ¡ 5)
= (x + 5)x + (x + 5)5 = (x ¡ 5)x + (x ¡ 5)(¡5)
= x(x + 5) + 5(x + 5) = x(x ¡ 5) ¡ 5(x ¡ 5)
= x2 + 5x + 5x + 25 = x2 ¡ 5x ¡ 5x + 25
= x2 + 10x + 25 = x2 ¡ 10x + 25

2 Expand using repeated use of the distributive law:

a (x + 2)2 b (x ¡ 2)2 c (5 + x)2 d (5 ¡ x)2
e (2x + 3)2 f (2x ¡ 3)2 g (a + b)2 h (a ¡ b)2
i (x ¡ 6)2 j (11 + z)2 k (3 ¡ 5x)2 l (2 + 7x)2

Example 19 Self Tutor

Use the distributive law (x + 2)(x2 + 2x ¡ 3)
to expand and simplify: = (x + 2)x2 + (x + 2)2x + (x + 2)(¡3)
(x + 2)(x2 + 2x ¡ 3) = x2(x + 2) + 2x(x + 2) ¡ 3(x + 2)
= x3 + 2x2 + 2x2 + 4x ¡ 3x ¡ 6
= x3 + 4x2 + x ¡ 6

ALGEBRAIC SIMPLIFICATION AND EXPANSION (Chapter 9) 179

3 Use the distributive law to expand and simplify:

a (x + 2)(x2 + 2x + 4) b (x + 1)(2x2 ¡ x + 3)

c (x ¡ 3)(x2 ¡ 2x + 1) d (x2 ¡ x + 3)(x + 5)

e (x2 ¡ 2x ¡ 4)(3x ¡ 5) f (3x2 + 2x + 1)(2x ¡ 7)

g (2x ¡ 5)(3 ¡ x2 + x) h (x + 8)(x2 + 1 ¡ 3x)

E THE EXPANSION RULES

Consider the following rectangle which is 8 units long and 6 units wide.

8 3 5 3
5 +4 +2
+2
4
6 =4

+
2

5 +3

Comparing the total number of squares on each side of the = sign, we notice that:
(4 + 2)(5 + 3) = 4 £ 5 + 4 £ 3 + 2 £ 5 + 2 £ 3.

We generalise this result by considering a rectangle with sides (a + b) and (c + d):

c d c d c d
a (a+b) = a +a +b

+b

b
(c+d )

The original rectangle has area = (a + b)(c + d). flength £ widthg
The sum of the areas of the smaller rectangles = ac + ad + bc + bd:

) (a + b)(c + d) = ac + ad + bc + bd:

This expansion rule is called the FOIL rule as:

inners

(a + b)(c + d) = ac + ad + bc + bd

outers Firsts Outers Inners Lasts