- Created by: Charlotte Posner
- Created on: 07-09-14 19:12
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
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
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
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)
Expand to check your answer
(2x + 3)(x + 4) = 2x² + 11x + 12
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'.
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
0 + 3600 = 3600
Square root of 3600 = 60
60 = v
You can use given information to work out a formula