# A1 - Expressions

A1 Expressions:

A1.1 - Simplifying using Index Laws

A1.2 - Expanding Single and Double Brackets

A1.3 - Factorisation

A1.4 - Further Factorisation

A1.5 - The Difference of Two Squares

A1.6 - Deriving and Using Formulae

## Simplifying using Index Laws - Grade B

An index is a power.

The base is the number which is raised to the power.

6³ = 6 is the base, and ³ is the index

You can simplify expressions using the three index laws:

When multiplying, add the indicies.  6³ x 6² = 6^5

When dividing, subtract the indicies. 6^6 ÷ 6³ = 6³

With brackets,multiply the indicies. (6³)² = 6^6

1 of 6

## Expanding single and double brackets - Grade B

Expanding brackets is where you remove brackets from an equation...

• To expand a single bracket, you multiply everything inside the bracket by whatever is outside the bracket:
• 2(9+7y) = 18+14y
• To expand double brackets, you times everything in one bracket by everything in the second bracket:
• (2x + 7) (3x-4) = 6x² - 8x +21x - 28 = 6x² + 13x -28
2 of 6

## Factorisation - Grade B/A

Factorising is the reverse of expanding...

• To factorise into single brackets, remove the common factors. Find the HCF of the coefficients, and then the HCF of the algebraic terms...

9xy + 6x² = 3x(3y = 2x)

9 and 6 are the coefficients, and the HCF is 3. X is the HCF of xy and x².

• For double brackets, look for two numbers that give the coefficent of x, and multiply to give the constant.

x² + 11x + 28 = (x + 4) (x + 7)

11 is the coefficent of x, and 28 is the constant.

Quadractic equations often factorise into double brackets

3 of 6

## Further factorisation - Grade A

To factorise quadratics with more than one x², you have to change your method...

Multiply the coefficient of x² and the constant (2x² + 11x + 12 --- 2 x 12 = 24)

Find two numbers that multiply to this, and add up to the coefficent of x (+3 and +8)

Write it again, with two numbers as x (2x² + 3x + 8x +12)

Factorise the pair of terms ( x(2x +3) + 4 (2x + 3))

Factorise again, with the brackets as the common factor (2x + 3) (x + 4)

(2x + 3)(x + 4) = 2x² + 11x + 12

4 of 6

## The difference of two squares - Grade A

Sometimes when you expand brackets, the 'x' terms cancel each other out...

(x+3) (x-3) = x² - 9

(x+5) (x-5) = x² -25

(2x-7)(2x+7) = 4x² - 49

You call quadratic expressions of the form x² - 16 = (x+4)(x-4) the 'difference of two squares'.

5 of 6

## Deriving and using Formulae - Grade B/A

You can subsitute numbers into formulae to work out the value of an unkown...

v² = u² + 2as

u= 0 a=6 s=300

0² +2(6x300)

0 + 3600 = 3600

Square root of 3600 = 60

60 = v

You can use given information to work out a formula

6 of 6