1. Multiplying Pairs of Brackets and Simplifying
EXPANDING
Multiplying Pairs of Brackets and Simplifying

2. This means to multiply out a pair, or more, of brackets e. g
This means to multiply out a pair, or more, of brackets e.g.(ax + b) (cx + d). There are 2 ways of expanding brackets. They are:
The Box Method
The FOIL Method

3. The Box Method (2a + 3)(4a + 2) 2a +3 4a +2
Multiply the individuals components together
(2a + 3)(4a + 2) 2a +3 4a +2

4. The Box Method 8a2 +12a +4a +6 (2a + 3)(4a + 2) 2a +3 4a +2 2a +3 4a

5. We now have four separate values so
(2a+3) (4a+2) = 8a2 + 12a + 4a +6
This can be simplified to give us our final answer
(2a+3) (4a+2) = 8a2 + 16a+6

6. Exercise (x + 2)(x + 1) (x + 8)(x + 2) (x + 3)(x + 3) (x + 4)(x – 5)

7. The FOIL method FOIL stands for: FIRST, OUTER, INNER, LAST
And refers to the order in which the values are multiplied.
For Example : (x + 3) (x - 5)
FIRST, x x x = x2
OUTER, x x -5 = -5x
INNER, +3 x x = 3x
LAST +3 x -5= -15
SIMPLIFY (x + 3) (x - 5) = x2 -2x -15

8. Example 2 : (2x - 3) (4x - 5)
FIRST, 2x x 4x = 8x2
OUTER, 2x x -5 = -10x
INNER, -3 x 4x = -12x
LAST -3 x -5 = +15
SIMPLIFY (2x - 3) (4x - 5) = 8x2 -22x +15

9. Exercise (x + 5) (x + 5) (2x + 3) (5x – 4) (x – 3) (x – 3)

10. The Separation Method
Separating the brackets will often make life much easier.
For example (x + 5) (2x – 3) = x(2x – 3) +5(2x-3)
Now we can multiply out the separate brackets to obtain
x(2x – 3) = 2x2 -3x and 5(2x-3) = 10x -15
Adding these will give
2x2 -3x + 10x -15 = 2x2 + 7x -15

11. Example 2
(x2 + 2x - 5) (2x – 3) = x2(2x – 3) + 2x(2x-3) – 5(2x-3)
Now we can multiply out the separate brackets to obtain
x2(2x – 3) = 2x3 - 3x2
2x(2x – 3) = 4x2 – 6x
and -5(2x-3) = -10x +15
Adding these will give
2x3 -3x2 +4x2 -6x - 10x +15 = 2x3 + x2 - 16x +15

12. Exercise (x + 4) (x + 4) (x + 3) (5x – 4) (x + 3) (x – 3)