#### Algebra

__Factorization__

__Factorization__

- Common Factors:
- 3x
^{2}+ 6x = 3x(x + 2)

- 3x
- Difference of Two Squares:
- 25 – x
^{2}= (5 + x)(5 – x)

- 25 – x
- Group Factorization:
- 4d + ac + ad + 4c
- = 4d + 4c + ac + ad
- = 4(d + c) + a(c + d)
- = (4 + a) (c + d)

- Trinomial:
- x
^{2}+ 14x + 24 - = x
^{2}+ 12x + 2x + 24 - = x(x + 12) + 2(x + 12)
- = (x + 2)(x + 12)

- x

__Quadratic Factorization__

__Quadratic Factorization__

- General Equation:
- ax
^{2}+ bx + c = 0

- ax
- Solve quadratics by:
- Trinomial factorization
- Quadratic formula

**NOTE**: When the question asks to round off, use the quadratic formula to solve.

__Simultaneous Equations__

__Simultaneous Equations__

- Simultaneous linear equations can be solved either by substitution or elimination.
- Simultaneous linear and non-linear equations are generally solved by substitution as follows:
- Step 1: Obtain an equation in one unknown and solve this equation
- Step 2: Substitute the result from step 1 into the equation to find the other unknown.

- The points of intersection of two graphs are given by the solution of their simultaneous equations.

__Inequalities__

__Inequalities__

- You would solve these like you would equations.
- Multiplying or dividing by negative would cause a switch of the inequality sign. For example:
^{y}/_{-3}≥ 7- Multiply through by -3: y ≤ -7 × -3
- ∴ y ≤ 21

- When two inequalities are present, split into two. For example:
- x < 3x – 1 < 2x + 7 ⇒ Split into two groups:
- x < 3x – 1; x >
^{1}/_{2} - 3x – 1 < 2x + 7; x < 8

- x < 3x – 1; x >

- x < 3x – 1 < 2x + 7 ⇒ Split into two groups:

**Linear Programming**- For strict inequalities (‘<‘, ‘>’), use broken lines.
- For non-strict inequalities (‘≤’, ‘≥’), use solid lines.
- Steps to solve:
- Draw straight line graphs
- Shade
- Solve

__Sequences__

__Sequences__

- Linear Sequence: Find the common, e.g. 3, then multiply it by n and work out what needs to be added.
- Quadratic Sequence: Using 2, 6, 12, 20, 30, 42 as an example.
- The general formula is u
_{n}= an^{2}+ bn + c. - To find the nth term of the sequence:
- Find the differences between the term sin the sequence
- Find the difference between the differences in the above step.
**Note**: The second difference is common (the same throughout). - The coefficient of n
^{2}is gotten by dividing the second difference by 2. Therefore, a = 1. - Then, find the first few terms of the first term (an
^{2}); in this case, 1, 4, 9, 16, 25 (we found for the first 5 terms) - Then, create a 3-rowed table; the first column would have the first number of terms from the original sequence (according to the number of terms you found above), the second column would have the terms you found above, and the third term would have the subtractions of the first column and the second column.
- Treat d
_{n}like a linear sequence and find the formula for its nth term; in this case d_{n}= n (the common difference is 1, and you need not add anything to get the terms) - Therefore, u
_{n}= n^{2}+ n. **NOTE**: If d_{n}= something else, say 4n + 3; u_{n}= n^{2}+ 4n + 3.

- The general formula is u
**Geometric Sequence**- This is a sequence where the terms have been multiplied by a constant to form the next term.
- The n
^{th}term of G.P. = ar^{(n – 1)}- a = 1
^{st}term - r = common difference

- a = 1

__Distance-Time Graphs__

__Distance-Time Graphs__

- From O to A: Uniform speed.
- From B to C: Uniform speed (return journey)
- From A to B: Stationary (speed = 0)

- On the graph, the gradient is equal to the speed of the journey.

__Speed-Time Graphs__

__Speed-Time Graphs__

- From O to A: Uniform speed.
- From A to B: Constant speed (acceleration = 0)
- From B to C: Uniform deceleration/retardation

- Area under a graph = distance traveled
- Gradient = acceleration
- If the acceleration is negative, it is called deceleration or retardation. (Moving body is slowing down)

__Functions__

__Functions__

- Function Notation:
- f: x → 2x – 1
- Function f such that x maps onto 2x -1

- Composite Function:
- Given two functions f(x) and g(x), the composite function of f and g is the function which maps x onto f(g(x))

- f(2)
- Substitute x = 2 and solve for f(2)

- f(g(x))
- Substitute x = g(x)

- f
^{-1}(x)- Let y = f(x) and make x the subject.