8 - Algebraic ManipulationBIDMAS - Brackets, Indices, Division, Multiplication, Addition, Subtraction

Negatives and Indices:-x² = -x²

Even Index: (-x)² = +x²-(x)² = -x²

-x³ = -x³Odd Index: (-x)³ = -x³

-(x)³ = -x³

-3² = - 3 x 3 = - 9 (square then add the -)(-3²) = -3 x -3 = 9

if a = -3, then a² = (-3)² not a = - 3², as the - belongs to the a

Expanding Brackets² =(x + 2)² = (x +2)(x +2)e.g. (3-x)(3-x) = 9 - 6x + x²when in brackets, — belongs to x, so (-x)²

Expanding Brackets:Claw Method = Sequence Method =

(x+4)² ⟶ x =1st term, 4 = 2nd term1. Square first term = x²2. Square second term = 4² = 163. Multiply the terms together then x by 2 = 8x

x² + 16 + 8x↗ ↑ ↖

1st term² 2nd term² (1stx2nd) x2

Factorising:

When coefficient of x = 1:e.g. x² -7x + 10

↗ ↖-7 = sum 10 = product

We need 2 numbers that ADD to get the sum and MULTIPLY to get the product= -5, -2

-5 + -2 = -7 (x - 5)(x -2)-5 x -2 = 10

When coefficient of x >1:3x² +10x + 81. Find product and sum: Product = 3 x 8 = 24, sum = 102. Find two numbers that x to 24 and + to 10 = 6, 43. Input factors into formula: 3x² + 6x + 4x + 84. Factorise expression, so that the brackets are repeating⟶ 3x(x+2)4(x+2)5. Cancel one of the repeating brackets = (3x+4)(x+2)(x+2)6. Answer = (3x+4)(x+2)

Factorising difference of two squares x² - y²:a² - b² = (√a + √b )(√a - √b ) x² - 64↑ ↑a b

e.g. x²-100 = (x+10)(x-10)must have √a and ± √b= (√x + √64 )(√x - √64)= (x +8)(x-8)

Notes by Ella Onyems and Revisely 1 See more at revisely.co.uk

Factorising completing the square x² + y² + 2yx:

e.g. x² + 14x + 49↑ ↑ ↑

x² + 2yx + y²

1. Divide second term by 2x to get y ⟶ y = 14x÷2x = 72. Square why to check y² ⟶ 7² = 49 = y²3. Put into brackets = (x + 7)²

Algebraic Manipulation• An equation is anything with an equal sign i.e. it is true for only certain value/s of xe.g. x² - 16 = 0 <— x has a particular value• An identity is true for all values of x i.e. like a rule or formulae.g. (x+1)² = x² + 2x + 1 <— x can be input as any valueTo prove it is an identity, the formula must be able to work when the value is not only an integer

Expansion of 2+ Binomials1. Expand 1st two brackets + simplify2. Put answer from expansion into brackets3. Expand 3rd bracket with new bracket4. Simplify - highest power first

e.g. (x+3)(x-4)(x-10)1. (x+3)(x-4) = x²-x-122. (x²-x-12)3. (x-10)(x²-x-12) = x³ -x² -12x -10x² + 10x

+1204. Answer = x³ -11x² - 2x +120

Notes by Ella Onyems and Revisely 2 See more at revisely.co.uk