Algebra

1. Variables and Expressions

What Are Variables?

A variable is a letter (like x, y, or n) that stands in for an unknown number. Think of it as a labeled box that can hold any value. In the expression x + 5, the variable x could be 1, 7, -3, or any other number -- we just don't know which one yet.

Variables let us write general rules that work for any number, not just a specific one. Instead of saying "3 plus 5 is 8," we can say "x + 5" and it works for every possible value of x.

Variables in Daily Life

You use variables constantly without realizing it:

"I'll be there in t minutes" -- t is a variable for time
"Speed limit is s mph" -- s varies by road
"The bill comes to $x" -- x depends on what you ordered
"There are n people coming" -- n is unknown until people RSVP

Why This Matters for CS

Variables in algebra are exactly the same idea as variables in programming. When you write let x = 10; in JavaScript or x = 10 in Python, you're creating a named container that holds a value -- just like in algebra. Understanding algebraic variables makes programming variables intuitive.

Algebraic Expressions

An algebraic expression is a mathematical phrase that contains numbers, variables, and operations. Here are the key vocabulary words:

Term -- a single chunk separated by + or - signs. In 3x + 7, the terms are 3x and 7.
Coefficient -- the number multiplied by the variable. In 3x, the coefficient is 3.
Constant -- a term with no variable. In 3x + 7, the constant is 7.
Like terms -- terms with the same variable and exponent. 3x and 5x are like terms. 3x and 3x² are NOT like terms.

Simplifying Expressions

To simplify an expression, combine like terms by adding or subtracting their coefficients.

Example -- Combining Like Terms

Simplify: 4x + 3 + 2x - 1

Step 1: Group the like terms: (4x + 2x) + (3 - 1)

Step 2: Combine: 6x + 2

Answer: 6x + 2

Visual Method for Beginners

If you're struggling with like terms, try color-coding or underlining. Use one color for all x terms, another for all x² terms, and another for constants. Only combine terms of the same color!

More Practice with Like Terms

1) Simplify: 5a + 2b + 3a - b
Group: (5a + 3a) + (2b - b) = 8a + b

2) Simplify: 2x² + 4x + x² - 2x + 7
Group: (2x² + x²) + (4x - 2x) + 7 = 3x² + 2x + 7

3) Simplify: -3y + 5 + 7y - 2 + y
Group: (-3y + 7y + y) + (5 - 2) = 5y + 3

Key insight: Only combine terms with identical variable parts. 3x and 3x² cannot be combined!

The Distributive Property

a(b + c) = ab + ac

The distributive property says that multiplying a number by a group of numbers added together is the same as multiplying the number by each one individually and then adding the results. You "distribute" the multiplication across each term inside the parentheses.

Example -- Distributive Property

Expand: 3(2x + 4)

Step 1: Multiply 3 by each term inside: 3 * 2x + 3 * 4

Step 2: Simplify: 6x + 12

Answer: 6x + 12

Example -- Distribute then Combine

Simplify: 2(x + 3) + 4x

Step 1: Distribute: 2x + 6 + 4x

Step 2: Combine like terms: 6x + 6

Answer: 6x + 6

More Distribution Practice

1) Expand: 4(2x - 3)
4 × 2x + 4 × (-3) = 8x - 12

2) Expand: -3(x + 5)
-3 × x + (-3) × 5 = -3x - 15

3) Expand: 0.5(4y - 6)
0.5 × 4y + 0.5 × (-6) = 2y - 3

4) Expand and simplify: 3(2x + 1) - 2(x - 4)
= 6x + 3 - 2x + 8
= (6x - 2x) + (3 + 8) = 4x + 11

Common Mistake

Don't forget to distribute to every term inside the parentheses. A common error: writing 3(2x + 4) = 6x + 4 instead of the correct 6x + 12. The 3 must multiply both the 2x AND the 4.

Distributing with Fractions

Distribution works the exact same way when fractions are involved. Your brain might panic, but the rule hasn't changed: multiply by each term inside the parentheses.

The rule is always: a × (b + c) = a×b + a×c

It doesn't matter if a, b, or c are fractions, negatives, or have exponents. Same rule.

Example -- 3n × (40 + n³/2)

This is the kind of expression that shows up in Big-O analysis. Let's go step by step.

Step 1: Identify what you're distributing

a = 3n, b = 40, c = n³/2

Step 2: Distribute -- multiply 3n by each term

= 3n × 40 + 3n × (n³/2)

Step 3: Simplify each part

First part: 3n × 40 = 120n

Second part: 3n × n³ = 3n⁴, then ÷ 2 = 3n⁴/2

Result: 120n + 3n⁴/2

Example -- n/2 × (n + 1)

This is n(n+1)/2 written differently. Let's distribute:

= (n/2) × n + (n/2) × 1

= n²/2 + n/2

= (n² + n) / 2

This is the expanded form of n(n+1)/2 -- the sum of 1 to n.

Example -- Distributing a Negative with Fractions

-(x/3 + 2/5)

Distribute the -1:

= (-1) × x/3 + (-1) × 2/5

= -x/3 - 2/5

Splitting Fractions (When You Can and Can't)

This is the source of so much confusion. Here's the exact rule:

Splitting the NUMERATOR -- ALWAYS valid:
(a + b) / c = a/c + b/c ✓

Splitting the DENOMINATOR -- NEVER valid:
c / (a + b) ≠ c/a + c/b ✗

Valid Split: (40 + n³) / 2

The denominator is a single term (just 2), so we can split the numerator:

(40 + n³) / 2 = 40/2 + n³/2 = 20 + n³/2

Why does this work? Division distributes over addition in the numerator. Think of dividing a pizza: if you split 8 slices among 2 people, each person gets 4. If you split (5 + 3) slices among 2 people, each gets 5/2 + 3/2 = 2.5 + 1.5 = 4. Same result.

Invalid Split: DON'T Do This

12 / (x + 4) ≠ 12/x + 12/4

If x = 2: Left side = 12/6 = 2. Right side = 6 + 3 = 9. Not even close!

When the denominator has addition/subtraction, you cannot split. The fraction must stay as one piece, or you need to find another approach.

Common Denominator: Making Fractions Combine

When you need to add a whole number to a fraction, convert the whole number to a fraction first:

Example -- Combining 40 + n³/2 Into One Fraction

Step 1: Write 40 as 40/1

Step 2: Common denominator is 2, so: 40/1 = 80/2

Step 3: Add: 80/2 + n³/2 = (80 + n³) / 2

This is useful when you want to simplify an expression into a single fraction.

Example -- Combining 3n with a Fraction

You have 3n × 40/2. How to compute this?

Method 1 (just multiply): 3n × 40/2 = (3n × 40) / 2 = 120n/2 = 60n

Method 2 (simplify first): 40/2 = 20, so 3n × 20 = 60n

Both methods give the same answer. Use whichever feels simpler.

2. Solving Linear Equations

An equation is a statement that two expressions are equal, connected by an equals sign. Solving an equation means finding the value of the variable that makes the statement true.

The Golden Rule: Whatever you do to one side, you must do to the other.

This is the single most important idea in equation solving. An equation is like a perfectly balanced scale. If you add 5 to the left side, you must add 5 to the right side too, or the scale tips over and the equation becomes false.

One-Step Equations

These require a single operation to isolate the variable.

Example 1 -- Addition

Solve: x + 5 = 12

Goal: Get x by itself.

Subtract 5 from both sides: x + 5 - 5 = 12 - 5

Result: x = 7

Check: 7 + 5 = 12. Correct!

Why Do We "Move" Numbers to the Other Side?

We don't actually move numbers. We add the same amount to both sides to keep the equation balanced. Saying "move 5 to the other side" is shorthand for "subtract 5 from both sides." The goal is always to isolate the variable.

Example 2 -- Multiplication

Solve: 3x = 21

Divide both sides by 3: 3x / 3 = 21 / 3

Result: x = 7

Check: 3(7) = 21. Correct!

Two-Step Equations

Strategy: undo addition/subtraction first, then undo multiplication/division.

Example 3 -- Two Steps

Solve: 2x + 3 = 11

Step 1: Subtract 3 from both sides: 2x = 8

Step 2: Divide both sides by 2: x = 4

Check: 2(4) + 3 = 8 + 3 = 11. Correct!

More Two-Step Practice

1) Solve: 3x - 4 = 14
Add 4: 3x = 18
Divide by 3: x = 6
Check: 3(6) - 4 = 18 - 4 = 14 ✓

2) Solve: -2x + 7 = 1
Subtract 7: -2x = -6
Divide by -2: x = 3
Check: -2(3) + 7 = -6 + 7 = 1 ✓

3) Solve: 5 + 4y = 21
Subtract 5: 4y = 16
Divide by 4: y = 4
Check: 5 + 4(4) = 5 + 16 = 21 ✓

Strategy: Work Backwards from Order of Operations

To solve 2x + 3 = 11, ask yourself: "What operations were done to x?" Answer: "First multiplied by 2, then added 3." To undo this, work backwards: first subtract 3, then divide by 2. Always undo addition/subtraction before multiplication/division.

Example 4 -- Negative Coefficient

Solve: -5x + 10 = -15

Step 1: Subtract 10 from both sides: -5x = -25

Step 2: Divide both sides by -5: x = 5

Check: -5(5) + 10 = -25 + 10 = -15. Correct!

Multi-Step Equations (Variables on Both Sides)

When the variable appears on both sides, move all variable terms to one side and all constants to the other.

When You Get Negative Answers

Don't worry if x comes out negative! Many students think negative answers are "wrong," but they're often correct. For example, if x - 3 = -8, then x = -5. Negative numbers are perfectly valid solutions. Always check your answer by substituting back.

Example 5 -- Variables on Both Sides

Solve: 5x + 3 = 2x + 18

Step 1: Subtract 2x from both sides: 3x + 3 = 18

Step 2: Subtract 3 from both sides: 3x = 15

Step 3: Divide both sides by 3: x = 5

Check: 5(5) + 3 = 28, and 2(5) + 18 = 28. Correct!

More Variables on Both Sides Practice

1) Solve: 4x + 1 = x + 10
Subtract x: 3x + 1 = 10
Subtract 1: 3x = 9
Divide by 3: x = 3

2) Solve: 6y - 5 = 4y + 7
Subtract 4y: 2y - 5 = 7
Add 5: 2y = 12
Divide by 2: y = 6

3) Solve: 3a + 8 = 7a - 4
Subtract 3a: 8 = 4a - 4
Add 4: 12 = 4a
Divide by 4: a = 3

Example 6 -- Distributive Property First

Solve: 3(x - 2) = 2x + 1

Step 1: Distribute: 3x - 6 = 2x + 1

Step 2: Subtract 2x from both sides: x - 6 = 1

Step 3: Add 6 to both sides: x = 7

Check: 3(7 - 2) = 3(5) = 15, and 2(7) + 1 = 15. Correct!

Equations with Fractions

The trick: multiply every term by the least common denominator (LCD) to clear all the fractions first. Then solve normally.

Example 7 -- Clearing Fractions

Solve: (x/2) + (x/3) = 5

The LCD of 2 and 3 is 6. Multiply every term by 6:

6(x/2) + 6(x/3) = 6(5)

3x + 2x = 30

5x = 30

x = 6

Check: 6/2 + 6/3 = 3 + 2 = 5. Correct!

Common Mistake

When clearing fractions, you must multiply every single term by the LCD -- including terms that aren't fractions. A very common error is only multiplying the fraction terms and forgetting the constant on the other side.

Why This Matters for CS

Solving equations is essentially what computers do when they evaluate expressions and solve algorithms. When you write code that calculates a value from a formula -- say, converting Celsius to Fahrenheit with f = (9/5) * c + 32 -- you're applying the same equation-solving logic. Understanding how to rearrange formulas is also key for deriving algorithms, analyzing complexity, and optimizing code.

3. Inequalities

An inequality is like an equation, but instead of saying two things are equal, it says one is bigger or smaller. The four inequality symbols are:

Symbol	Meaning	Example
<	Less than	x < 5 (x is less than 5)
>	Greater than	x > 3 (x is greater than 3)
≤	Less than or equal to	x ≤ 10 (x is at most 10)
≥	Greater than or equal to	x ≥ 0 (x is non-negative)

Solving Inequalities

You solve inequalities almost exactly like equations, with one critical exception: when you multiply or divide both sides by a negative number, you must flip the inequality sign.

Example -- Standard Inequality

Solve: 2x + 1 < 9

Step 1: Subtract 1 from both sides: 2x < 8

Step 2: Divide both sides by 2: x < 4

Solution: x can be any number less than 4.

Example -- Flipping the Sign

Solve: -3x + 6 ≤ 15

Step 1: Subtract 6 from both sides: -3x ≤ 9

Step 2: Divide both sides by -3 (flip the sign!): x ≥ -3

Solution: x can be -3 or any number greater than -3.

Common Mistake

Forgetting to flip the inequality sign when multiplying or dividing by a negative number. This is the number one mistake students make with inequalities. Ask yourself: "Am I dividing by a negative?" If yes, flip the sign.

Graphing Inequalities on a Number Line

You can visualize inequality solutions on a number line:

x < 4 -- Draw an open circle at 4 (meaning 4 is NOT included) and shade everything to the left.
x ≥ -3 -- Draw a filled circle at -3 (meaning -3 IS included) and shade everything to the right.
Open circle = strict inequality (< or >), filled circle = includes the endpoint (≤ or ≥).

Why This Matters for CS

Inequalities show up constantly in programming: loop conditions (while i < n), array bounds checking (if index >= 0 && index < length), and algorithm analysis ("this algorithm runs in O(n) when n ≥ 1"). Understanding how inequalities work helps you reason about when loops terminate and whether your array accesses are safe.

4. Functions

What Is a Function?

A function is an input-output machine: you feed it an input, it follows a rule, and it produces exactly one output. The key idea is that each input gives exactly one output. If you put the same number in twice, you get the same result both times.

We usually write functions using the notation f(x), which reads as "f of x." The letter f is the name of the function, and x is the input. For example:

f(x) = 2x + 3

This function takes any number x, doubles it, then adds 3.

Example -- Evaluating a Function

If f(x) = 2x + 3, find f(4).

Replace x with 4: f(4) = 2(4) + 3 = 8 + 3 = 11

Answer: f(4) = 11

Domain and Range

Domain -- the set of all valid inputs (x-values). Ask: "What values of x can I plug in without breaking anything?"
Range -- the set of all possible outputs (y-values). Ask: "What values can actually come out?"

For example, the function f(x) = 1/x has a domain of all real numbers except 0 (because dividing by zero is undefined). Its range is also all real numbers except 0 (because 1/x never equals zero).

Linear Functions: y = mx + b

The most important type of function in beginning algebra is the linear function. Its graph is a straight line.

y = mx + b

m = slope (steepness and direction of the line)
b = y-intercept (where the line crosses the y-axis)

Slope (m) tells you how much y changes for each 1-unit increase in x. A slope of 2 means "go up 2 for every 1 you go right." A negative slope means the line goes downhill.
Y-intercept (b) is the y-value when x = 0. It's where the line crosses the vertical axis.

Example -- Identifying Slope and Intercept

For the line y = -3x + 7:

Slope: m = -3 (the line goes down 3 units for every 1 unit to the right)

Y-intercept: b = 7 (the line crosses the y-axis at the point (0, 7))

Slope Formula

Given two points on a line, (x₁, y₁) and (x₂, y₂), the slope is:

m = (y₂ - y₁) / (x₂ - x₁)

This is often described as "rise over run" -- the vertical change divided by the horizontal change.

Example -- Finding Slope from Two Points

Find the slope of the line through (1, 2) and (4, 11).

m = (11 - 2) / (4 - 1) = 9 / 3 = 3

The slope is 3. The line rises 3 units for every 1 unit to the right.

Understanding Slope Intuitively

Slope in Different Contexts

Positive slope: Line goes upward left-to-right (like climbing a hill)

• m = 2: For every 1 step right, go up 2 steps

• m = 0.5: For every 1 step right, go up 0.5 steps (gentle incline)

Negative slope: Line goes downward left-to-right (like descending a hill)

• m = -1: For each 1 step right, go down 1 step (45° decline)

• m = -3: For each 1 step right, go down 3 steps (steep decline)

Zero slope: m = 0 means horizontal line (no rise or fall)

Undefined slope: Vertical line (infinite steepness)

Finding the Equation of a Line

Given Slope and Y-Intercept

Write the equation of a line with slope -2 and y-intercept 5:

Use y = mx + b directly: y = -2x + 5

Given Two Points

Find the equation passing through (3, 7) and (5, 13):

Step 1: Find slope: m = (13 - 7)/(5 - 3) = 6/2 = 3

Step 2: Use point-slope form with either point. Using (3, 7):

y - 7 = 3(x - 3)

y - 7 = 3x - 9

y = 3x - 2

Answer: y = 3x - 2

Check with other point: 3(5) - 2 = 15 - 2 = 13 ✓

Given Slope and One Point

Find the equation with slope 4 passing through (-1, 3):

Point-slope form: y - y₁ = m(x - x₁)

y - 3 = 4(x - (-1))

y - 3 = 4(x + 1)

y - 3 = 4x + 4

y = 4x + 7

Graphing Lines Step-by-Step

Method 1: Using Slope and Y-Intercept

For y = 2x - 3:

1. Plot the y-intercept: (0, -3)

2. Use slope 2 = 2/1 to find next point: from (0, -3), go right 1 and up 2 to reach (1, -1)

3. Continue: from (1, -1) go right 1 and up 2 to reach (2, 1)

4. Draw line through these points

Method 2: Using Two Points

For y = -x + 4:

1. Choose any x-values, say x = 0 and x = 4

2. When x = 0: y = -(0) + 4 = 4, so (0, 4)

3. When x = 4: y = -(4) + 4 = 0, so (4, 0)

4. Plot these points and draw the line

Special Forms of Linear Equations

Standard Form: Ax + By = C
Point-Slope Form: y - y₁ = m(x - x₁)
Slope-Intercept Form: y = mx + b
Intercept Form: x/a + y/b = 1

Converting Between Forms

Convert 3x + 2y = 6 to slope-intercept form:

Solve for y:

2y = -3x + 6

y = (-3x + 6)/2

y = -1.5x + 3

So slope = -1.5, y-intercept = 3

Why This Matters for CS

Functions in algebra map directly to functions in programming. The concept of domain is like input validation -- what inputs are acceptable? In graphics programming, linear functions define lines on screen. Slope appears in machine learning as the "weight" in linear regression models. And the idea that a function always produces the same output for the same input is the foundation of "pure functions" in functional programming.

5. Systems of Equations

A system of equations is a set of two (or more) equations with the same variables. The solution is the values of the variables that satisfy ALL equations simultaneously. Geometrically, for two linear equations, the solution is where the two lines intersect.

Method 1: Substitution

Idea: Solve one equation for one variable, then substitute that expression into the other equation.

Example -- Substitution Method

Solve the system:

y = 2x + 1 ... (Equation 1)

3x + y = 16 ... (Equation 2)

Step 1: Equation 1 already has y isolated: y = 2x + 1

Step 2: Substitute into Equation 2:

3x + (2x + 1) = 16

5x + 1 = 16

5x = 15

x = 3

Step 3: Plug x = 3 back into Equation 1:

y = 2(3) + 1 = 7

Solution: (x, y) = (3, 7)

Check in Equation 2: 3(3) + 7 = 9 + 7 = 16. Correct!

Method 2: Elimination

Idea: Add or subtract the equations to eliminate one variable.

Example -- Elimination Method

Solve the system:

2x + 3y = 12 ... (Equation 1)

4x - 3y = 6 ... (Equation 2)

Step 1: Notice that 3y and -3y will cancel if we add the equations.

Step 2: Add Equation 1 + Equation 2:

(2x + 4x) + (3y - 3y) = 12 + 6

6x = 18

x = 3

Step 3: Plug x = 3 into Equation 1:

2(3) + 3y = 12

6 + 3y = 12

3y = 6

y = 2

Solution: (x, y) = (3, 2)

Check in Equation 2: 4(3) - 3(2) = 12 - 6 = 6. Correct!

Example -- Elimination with Multiplication

Solve the system:

3x + 2y = 16 ... (Equation 1)

x + 4y = 22 ... (Equation 2)

Step 1: The variables don't cancel yet. Multiply Equation 2 by -3 so the x terms cancel:

3x + 2y = 16

-3x - 12y = -66

Step 2: Add the equations:

-10y = -50

y = 5

Step 3: Plug y = 5 into Equation 2:

x + 4(5) = 22

x + 20 = 22

x = 2

Solution: (x, y) = (2, 5)

Which Method Should You Use?

Substitution is easiest when one variable is already isolated (like y = ...) or has a coefficient of 1. Elimination is easiest when the coefficients line up nicely or can be made to cancel with simple multiplication.

Why This Matters for CS

Systems of equations are everywhere in CS. In computer graphics, finding where two lines intersect is solving a 2-variable system. Linear programming (used for optimization problems like scheduling and resource allocation) involves solving systems of equations and inequalities. Machine learning's linear regression solves systems of equations to find the best-fit line. At scale, these become matrix equations -- which is why linear algebra matters so much for CS.

Understanding Solutions Graphically

What Does the Solution Mean?

The solution to a system of two linear equations is the intersection point of the two lines. This point satisfies both equations simultaneously.

Types of Solutions

One Solution (Most Common): Lines intersect at exactly one point

• Example: y = 2x + 1 and y = -x + 4 intersect at (1, 3)

No Solution: Lines are parallel (never intersect)

• Example: y = 2x + 1 and y = 2x + 5 (same slope, different y-intercepts)

Infinite Solutions: Lines are identical (overlap completely)

• Example: y = 2x + 1 and 2y = 4x + 2 (same line written differently)

More Practice Examples

Substitution Method Practice

Solve: x + y = 5 and 2x - y = 1

From equation 1: y = 5 - x

Substitute into equation 2: 2x - (5 - x) = 1

2x - 5 + x = 1

3x = 6

x = 2

So y = 5 - 2 = 3

Solution: (2, 3)

Check: 2 + 3 = 5 ✓ and 2(2) - 3 = 1 ✓

Elimination Method Practice

Solve: 2x + 3y = 7 and 4x - 3y = 5

Notice the y-coefficients are opposites (3 and -3), so add directly:

(2x + 4x) + (3y - 3y) = 7 + 5

6x = 12

x = 2

Substitute back: 2(2) + 3y = 7 → 4 + 3y = 7 → y = 1

Solution: (2, 1)

Real-World Application

Concert tickets cost $15 for adults and $8 for children. If 200 tickets were sold for $2,250 total, how many of each type were sold?

Let a = adult tickets, c = child tickets

Total tickets: a + c = 200

Total revenue: 15a + 8c = 2250

From equation 1: c = 200 - a

Substitute: 15a + 8(200 - a) = 2250

15a + 1600 - 8a = 2250

7a = 650

a = ~93 adult tickets, c = ~107 child tickets

6. Polynomials

What Are Polynomials?

A polynomial is an expression made up of terms where variables are raised to non-negative integer powers. Each term has a coefficient and a variable part. Here are some examples:

5x + 3 -- a polynomial with 2 terms (a binomial)
x² + 4x - 7 -- a polynomial with 3 terms (a trinomial)
2x³ - x² + 5x - 1 -- a polynomial with 4 terms

The degree of a polynomial is the highest exponent. The degree of x² + 4x - 7 is 2. The degree of 2x³ - x² + 5x - 1 is 3.

Adding and Subtracting Polynomials

Simply combine like terms (terms with the same variable and exponent).

Example -- Adding Polynomials

(3x² + 2x + 1) + (x² - 5x + 4)

Combine like terms:

x² terms: 3x² + x² = 4x²

x terms: 2x + (-5x) = -3x

Constants: 1 + 4 = 5

Answer: 4x² - 3x + 5

Example -- Subtracting Polynomials

(5x² + 3x - 2) - (2x² - x + 6)

Distribute the negative sign: 5x² + 3x - 2 - 2x² + x - 6

Combine like terms: 3x² + 4x - 8

Answer: 3x² + 4x - 8

Common Mistake

When subtracting polynomials, remember to distribute the negative sign to every term in the second polynomial. A common error: (5x² + 3x - 2) - (2x² - x + 6) = 3x² + 2x + 4. The mistake is not flipping -x to +x and +6 to -6.

Multiplying Polynomials (FOIL)

To multiply two binomials, use FOIL: First, Outer, Inner, Last.

Example -- FOIL Method

Multiply: (x + 3)(x + 5)

First: x * x = x²

Outer: x * 5 = 5x

Inner: 3 * x = 3x

Last: 3 * 5 = 15

Combine: x² + 5x + 3x + 15 = x² + 8x + 15

Answer: x² + 8x + 15

Example -- FOIL with Negatives

Multiply: (2x - 1)(x + 4)

First: 2x * x = 2x²

Outer: 2x * 4 = 8x

Inner: -1 * x = -x

Last: -1 * 4 = -4

Combine: 2x² + 8x - x - 4 = 2x² + 7x - 4

Answer: 2x² + 7x - 4

Beyond FOIL

FOIL only works for two binomials. For larger polynomials, use the general rule: multiply every term in the first polynomial by every term in the second, then combine like terms. FOIL is just a shortcut for this process when each polynomial has exactly two terms.

Special Products (Memorize These!)

These patterns appear so often that memorizing them will save you time and reduce errors.

Perfect Square Trinomials

(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²

Example: Expanding (x + 5)²

(x + 5)² = x² + 2(x)(5) + 5²

= x² + 10x + 25

Example: Expanding (3y - 2)²

(3y - 2)² = (3y)² - 2(3y)(2) + 2²

= 9y² - 12y + 4

Common Mistake

(x + 5)² ≠ x² + 25!

You MUST include the middle term (2ab). This is one of the most common algebra mistakes.

Difference of Squares

(a + b)(a - b) = a² - b²

Example: (x + 7)(x - 7)

= x² - 7² = x² - 49

The middle terms cancel out because one is +7x and one is -7x.

Sum and Difference of Cubes

a³ + b³ = (a + b)(a² - ab + b²)
a³ - b³ = (a - b)(a² + ab + b²)

Memory Trick: SOAP

For a³ ± b³, use SOAP: Same sign, Opposite sign, Always Positive

a³ + b³ = (a + b)(a² - ab + b²)

Multiplying Polynomials with More Than Two Terms

Example: (x + 2)(x² + 3x - 1)

Distribute each term in the first polynomial:

= x(x² + 3x - 1) + 2(x² + 3x - 1)

= x³ + 3x² - x + 2x² + 6x - 2

= x³ + 5x² + 5x - 2

Example: (x² + x + 1)(x - 1)

= x²(x - 1) + x(x - 1) + 1(x - 1)

= x³ - x² + x² - x + x - 1

= x³ - 1 (difference of cubes!)

Polynomial Vocabulary

Term	Definition	Example
Monomial	1 term	5x²
Binomial	2 terms	x + 3
Trinomial	3 terms	x² + 2x + 1
Degree	Highest exponent	x³ + x has degree 3
Leading term	Term with highest degree	In 2x³ + x, it's 2x³
Leading coefficient	Coefficient of leading term	In 2x³ + x, it's 2
Constant term	Term with no variable	In x² + 5, it's 5

Polynomial Division (Optional Advanced Topic)

You can divide polynomials using long division, similar to numerical long division.

Example: (x² + 5x + 6) ÷ (x + 2)

Step 1: x² ÷ x = x (first term of quotient)

Step 2: x(x + 2) = x² + 2x

Step 3: (x² + 5x + 6) - (x² + 2x) = 3x + 6

Step 4: 3x ÷ x = 3 (second term of quotient)

Step 5: 3(x + 2) = 3x + 6

Step 6: (3x + 6) - (3x + 6) = 0 (no remainder)

Answer: (x² + 5x + 6) ÷ (x + 2) = x + 3

Check: (x + 2)(x + 3) = x² + 5x + 6 ✓

7. Factoring

What is Factoring, Really?

You already know how to multiply: take (x + 3)(x + 5) and expand it into x² + 8x + 15. Factoring is the reverse. You start with x² + 8x + 15 and figure out that it came from (x + 3)(x + 5).

Think of it like this: multiplication is baking a cake (mixing ingredients together). Factoring is looking at a finished cake and figuring out the recipe. You are "un-multiplying."

Multiplication (forward): Factoring (reverse): (x + 3)(x + 5) x² + 8x + 15 | | | FOIL / expand | un-multiply v v x² + 8x + 15 (x + 3)(x + 5)

Why Do We Factor?

Factoring is how you solve equations. If you need to solve x² + 8x + 15 = 0, there is no way to "undo" the x² directly. But if you factor it into (x + 3)(x + 5) = 0, then you know either x + 3 = 0 or x + 5 = 0, giving you x = -3 or x = -5. That is the entire point: turn a hard problem into two easy ones.

First, Understand FOIL (So You Can Reverse It)

Before you can undo multiplication, you need to truly understand how multiplication works. FOIL stands for First, Outer, Inner, Last -- it is a way to multiply two binomials.

(x + 3)(x + 5) F: First x * x = x² O: Outer x * 5 = 5x I: Inner 3 * x = 3x L: Last 3 * 5 = 15 Add them up: x² + 5x + 3x + 15 = x² + 8x + 15 Notice: the middle term (8x) comes from adding the two numbers (3 + 5) the last term (15) comes from multiplying them (3 × 5)

This is the key insight for factoring: the middle coefficient is the SUM of two numbers, and the last term is the PRODUCT of those same two numbers. When you factor, you are hunting for those two numbers.

The Area Model -- A Visual Way to See It

There is a beautiful visual way to understand what FOIL is really doing. Think of multiplying (x + 3)(x + 5) as finding the area of a rectangle with sides (x + 3) and (x + 5):

x 5 +----------+------+ | | | x | x² | 5x | | | | +----------+------+ | | | 3 | 3x | 15 | | | | +----------+------+ Total area = x² + 5x + 3x + 15 = x² + 8x + 15 When you FACTOR, you are looking at the area (x² + 8x + 15) and trying to figure out the dimensions of the rectangle.

Factoring Strategy -- Follow This Every Time

The Factoring Flowchart

Step 1 -- GCF First: Always check if all terms share a common factor. If they do, pull it out. This is the single most important step and the one most people skip.
Step 2 -- Count the terms:
- 2 terms? Check for difference of squares: a² - b² = (a + b)(a - b)
- 3 terms? Find two numbers that multiply to give the last term and add to give the middle. This is the core technique.
- 4 terms? Try factoring by grouping (pair them up).
Step 3 -- Always verify: Multiply your answer back out. If it does not match the original, something went wrong.

Technique 1: Greatest Common Factor (GCF)

The GCF is the biggest thing that divides evenly into every term. Always do this first -- it makes everything else simpler.

How to find the GCF: look at the numbers (what is the biggest number that divides all of them?) and the variables (what is the lowest power of each variable that appears in every term?).

Example -- Simple GCF

Factor: 6x² + 9x

Numbers: GCF of 6 and 9 is 3.
Variables: both terms have at least one x, so the GCF includes x.

GCF = 3x. Divide each term by 3x:

6x² / 3x = 2x and 9x / 3x = 3

Answer: 3x(2x + 3)

Verify: 3x * 2x = 6x², and 3x * 3 = 9x. That is 6x² + 9x. Correct.

Example -- Bigger GCF

Factor: 12x³y² - 8x²y³ + 4xy

Numbers: GCF of 12, 8, 4 is 4.
Variables: smallest power of x across all terms is x¹, smallest power of y is y¹.

GCF = 4xy. Divide each term:

12x³y² / 4xy = 3x²y, 8x²y³ / 4xy = 2xy², 4xy / 4xy = 1

Answer: 4xy(3x²y - 2xy² + 1)

Technique 2: Factoring Trinomials (x² + bx + c)

This is the most common type. You have a trinomial where the x² has no coefficient (or a coefficient of 1). You need to find two numbers p and q where:

x² + bx + c = (x + p)(x + q)

Rule: p × q = c (they multiply to the last number)
p + q = b (they add to the middle number)

This is like a number puzzle: "I'm thinking of two numbers. They multiply to give 12, and they add to give 7. What are they?" The answer is 3 and 4.

Example -- Step by Step

Factor: x² + 7x + 12

We need two numbers that multiply to 12 and add to 7.

List the pairs that multiply to 12:

Pairs that multiply to 12: Do they add to 7? 1 × 12 = 12 1 + 12 = 13 ✗ 2 × 6 = 12 2 + 6 = 8 ✗ 3 × 4 = 12 3 + 4 = 7 ✓ FOUND IT So p = 3, q = 4 Answer: (x + 3)(x + 4) Verify with FOIL: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓

Example -- When Signs Are Negative

Factor: x² - 2x - 15

Need two numbers that multiply to -15 and add to -2.

Since the product is negative, one number must be positive and one must be negative. Since the sum is negative, the negative number must be bigger.

Pairs that multiply to -15: Do they add to -2? 1 × (-15) = -15 1 + (-15) = -14 ✗ -1 × 15 = -15 -1 + 15 = 14 ✗ 3 × (-5) = -15 3 + (-5) = -2 ✓ FOUND IT -3 × 5 = -15 -3 + 5 = 2 ✗ So p = 3, q = -5 Answer: (x + 3)(x - 5) Verify: (x + 3)(x - 5) = x² - 5x + 3x - 15 = x² - 2x - 15 ✓

Sign Cheat Sheet for Trinomials

The signs in x² + bx + c tell you a lot about p and q:

c is positive, b is positive: both p and q are positive. (e.g., x² + 7x + 12 = (x+3)(x+4))
c is positive, b is negative: both p and q are negative. (e.g., x² - 7x + 12 = (x-3)(x-4))
c is negative: one is positive, one is negative. The bigger one has the same sign as b. (e.g., x² - 2x - 15 = (x+3)(x-5))

Example -- Both Numbers Negative

Factor: x² - 9x + 20

c = +20 (positive) and b = -9 (negative), so both numbers are negative.

Need: multiply to 20, add to -9. Pairs: (-4)(-5) = 20 and -4 + -5 = -9.

Answer: (x - 4)(x - 5)

Technique 3: Factoring Harder Trinomials (ax² + bx + c, where a ≠ 1)

When the x² term has a coefficient other than 1 (like 2x² or 3x²), the basic "find two numbers" method needs a twist. Use the AC method:

AC Method for ax² + bx + c:

1. Multiply a × c (the first and last coefficients)
2. Find two numbers that multiply to (a × c) and add to b
3. Rewrite the middle term using those two numbers
4. Factor by grouping

Example -- AC Method Step by Step

Factor: 2x² + 7x + 3

Step 1: a × c = 2 × 3 = 6 Step 2: Find two numbers that multiply to 6 and add to 7 1 × 6 = 6 and 1 + 6 = 7 ✓ Found them: 1 and 6 Step 3: Rewrite the middle term (7x) as 1x + 6x: 2x² + 1x + 6x + 3 Step 4: Factor by grouping: (2x² + 1x) + (6x + 3) x(2x + 1) + 3(2x + 1) <-- both groups have (2x + 1)! (2x + 1)(x + 3) Answer: (2x + 1)(x + 3) Verify: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓

Example -- AC Method with Negatives

Factor: 3x² - 10x + 8

a × c = 3 × 8 = 24. Need two numbers that multiply to 24 and add to -10.

Since product is positive and sum is negative, both numbers are negative: (-4)(-6) = 24 and -4 + (-6) = -10.

Rewrite: 3x² - 4x - 6x + 8

Group: x(3x - 4) - 2(3x - 4) = (3x - 4)(x - 2)

Technique 4: Difference of Squares

This is one of the cleanest patterns in all of algebra. Whenever you see something squared minus something else squared, it factors instantly:

a² - b² = (a + b)(a - b)

Why does this work? Think about it backwards. If you expand (a + b)(a - b) using FOIL:

(a + b)(a - b) F: a × a = a² O: a × -b = -ab I: b × a = +ab <-- the middle terms cancel out! L: b × -b = -b² a² - ab + ab - b² = a² - b² The "outer" and "inner" terms always cancel. That is why the result has no middle term.

Example -- Basic Difference of Squares

Factor: x² - 49

Is this "something² minus something²"? Yes: x² = (x)² and 49 = (7)².

Answer: (x + 7)(x - 7)

Verify: (x + 7)(x - 7) = x² - 7x + 7x - 49 = x² - 49. Correct.

Example -- Harder Difference of Squares

Factor: 4x² - 25

4x² = (2x)² and 25 = (5)². So a = 2x, b = 5.

Answer: (2x + 5)(2x - 5)

Example -- Sneaky Difference of Squares

Factor: x&sup4; - 16

x&sup4; = (x²)² and 16 = (4)². So: (x² + 4)(x² - 4).

But wait -- x² - 4 is also a difference of squares! Factor it again:

Answer: (x² + 4)(x + 2)(x - 2)

(x² + 4 cannot be factored further because it is a sum of squares.)

Technique 5: Perfect Square Trinomials

Sometimes a trinomial is a perfect square in disguise. If the first and last terms are perfect squares, check if the middle term is twice the product of their square roots:

a² + 2ab + b² = (a + b)²
a² - 2ab + b² = (a - b)²

Is x² + 6x + 9 a perfect square trinomial? First term: x² = (x)² ✓ perfect square Last term: 9 = (3)² ✓ perfect square Middle term: 6x = 2(x)(3)? 2 × x × 3 = 6x ✓ yes! Answer: (x + 3)² Visual (area model): x 3 +----------+------+ | | | x | x² | 3x | | | | +----------+------+ | | | 3 | 3x | 9 | | | | +----------+------+ It is a perfect SQUARE -- both sides are the same (x + 3).

Example -- Perfect Square with Subtraction

Factor: 4x² - 20x + 25

4x² = (2x)², and 25 = (5)². Is the middle 2(2x)(5) = 20x? Yes.

Answer: (2x - 5)²

Technique 6: Factoring by Grouping (4 Terms)

When you have 4 terms, try splitting them into two groups of 2, factor each group separately, and see if a common factor emerges.

Example -- Step by Step

Factor: x³ + 3x² + 2x + 6

Step 1: Group into pairs: (x³ + 3x²) + (2x + 6) Step 2: Factor each group: x²(x + 3) + 2(x + 3) ↑ ↑ both groups contain (x + 3) -- this is the key! Step 3: Factor out the common (x + 3): (x + 3)(x² + 2) Answer: (x + 3)(x² + 2) Verify: (x + 3)(x² + 2) = x³ + 2x + 3x² + 6 = x³ + 3x² + 2x + 6 ✓

Factoring Mistakes to Avoid

a² + b² does NOT factor. x² + 9 cannot be factored. Only a² minus b² factors. This is the most common mistake.
Forgetting the GCF. Always check for a common factor first. 2x² + 4x + 2 looks hard until you pull out the 2: 2(x² + 2x + 1) = 2(x + 1)².
Sign errors in the AC method. When you split the middle term, make sure the signs are right. Always multiply back out to verify.
Stopping too early. After factoring once, check if you can factor further. x&sup4; - 16 = (x² + 4)(x² - 4) is not fully factored -- the second piece factors again into (x+2)(x-2).

Factoring Summary -- The Whole System

Every factoring problem in algebra uses one or more of these techniques:

GCF: Pull out the biggest common factor. Always first.
Trinomial (a=1): Find two numbers that multiply to c and add to b.
Trinomial (a≠1): AC method -- multiply a×c, find the pair, rewrite middle term, group.
Difference of squares: a² - b² = (a+b)(a-b).
Perfect square: a² ± 2ab + b² = (a ± b)².
Grouping: Split 4 terms into 2 pairs, factor each, pull out common binomial.

If you can do these 6 techniques, you can factor anything that shows up in algebra, calculus, or a technical interview. Practice each one until it is automatic.

8. Quadratic Equations

Standard Form

ax² + bx + c = 0 (where a ≠ 0)

A quadratic equation is any equation where the highest power of the variable is 2. The graph of a quadratic function is a parabola -- a U-shaped curve that opens up (if a > 0) or down (if a < 0).

Method 1: Solving by Factoring

If you can factor the quadratic, set each factor equal to zero and solve.

Example -- Solving by Factoring

Solve: x² + 5x + 6 = 0

Step 1: Factor: (x + 2)(x + 3) = 0

Step 2: Set each factor to zero:

x + 2 = 0 → x = -2

x + 3 = 0 → x = -3

Solutions: x = -2 or x = -3

Example -- Factoring with Leading Coefficient

Solve: 2x² + 7x + 3 = 0

Step 1: Find two numbers that multiply to 2 * 3 = 6 and add to 7. That's 1 and 6.

Step 2: Rewrite the middle term: 2x² + x + 6x + 3 = 0

Step 3: Factor by grouping: x(2x + 1) + 3(2x + 1) = 0

Step 4: Factor out (2x + 1): (2x + 1)(x + 3) = 0

Solutions: x = -1/2 or x = -3

Method 2: The Quadratic Formula

Factoring is great when it works, but sometimes the numbers are ugly and you cannot find the pair. The quadratic formula is the guaranteed method -- it works for every quadratic equation, always. It is the universal key.

x = (-b ± √(b² - 4ac)) / 2a

For the equation ax² + bx + c = 0, just plug in the values of a, b, and c. That is literally all you do.

What Each Part of the Formula Means

This formula looks scary but every piece has a job. Let's break it down:

-b ± √(b² - 4ac) x = ───────────────────── 2a Let's label each piece: -b = the "center" of the two answers. The two solutions are always equally spaced on either side of -b/2a. ± = "plus or minus" -- this is why you get TWO answers. One uses +, the other uses -. b² - 4ac = the "discriminant" -- this decides how many solutions exist (see below). √(b² - 4ac) = how far apart the two answers are from the center. 2a = scales everything based on the leading coefficient.

Why Does This Formula Exist?

The quadratic formula is what you get when you "complete the square" on the general equation ax² + bx + c = 0 (instead of a specific one). Someone did the algebra once, for the general case, and got this formula. Now you never have to complete the square again -- just plug in a, b, c.

Think of it like a function in programming. Instead of writing the same logic every time, someone wrote a reusable function: solve(a, b, c) that returns the answer for any quadratic.

When to Use the Quadratic Formula

Always works -- even when factoring is impossible
Use it when the numbers are ugly (fractions, large numbers, no obvious factor pairs)
Use it when you need exact answers with square roots (not just integer solutions)
Real-world uses: projectile motion (when does the ball hit the ground?), break-even analysis (when does profit = cost?), optimization problems, physics simulations

Example -- Quadratic Formula

Solve: x² - 4x - 5 = 0

Here a = 1, b = -4, c = -5.

x = (-(-4) ± √((-4)² - 4(1)(-5))) / 2(1)

x = (4 ± √(16 + 20)) / 2

x = (4 ± √36) / 2

x = (4 ± 6) / 2

Two solutions:

x = (4 + 6) / 2 = 10 / 2 = 5

x = (4 - 6) / 2 = -2 / 2 = -1

Solutions: x = 5 or x = -1

Example -- Quadratic Formula (No Nice Roots)

Solve: 2x² + 3x - 4 = 0

Here a = 2, b = 3, c = -4.

x = (-3 ± √(9 + 32)) / 4

x = (-3 ± √41) / 4

Since √41 ≈ 6.403:

x ≈ (-3 + 6.403) / 4 ≈ 0.851

x ≈ (-3 - 6.403) / 4 ≈ -2.351

Solutions: x = (-3 + √41) / 4 or x = (-3 - √41) / 4

The Discriminant -- Peek at the Answer Before Solving

The expression under the square root, b² - 4ac, is called the discriminant. It is like a preview -- before you even solve the equation, the discriminant tells you how many answers you will get.

Why? Because the ± in the formula gives you two answers: one with + and one with -. But if the thing inside the square root is zero, both answers are the same. And if it is negative, you cannot take the square root at all (no real answer).

Discriminant	Number of Solutions	What It Means
b² - 4ac > 0	Two real solutions	Parabola crosses x-axis twice
b² - 4ac = 0	One real solution	Parabola just touches x-axis (the vertex lands exactly on it)
b² - 4ac < 0	No real solutions	Parabola floats above (or below) the x-axis and never touches it

What the discriminant looks like visually (the parabola y = ax² + bx + c): b²-4ac > 0 b²-4ac = 0 b²-4ac < 0 Two solutions One solution No solutions \ / | * \ / * * * ──*───*──x── ───────*────x── ──────*─────*──x── (curve never Crosses x-axis Touches x-axis reaches x-axis) at TWO points at ONE point

Method 3: Completing the Square

This technique rewrites the quadratic by creating a perfect square trinomial. It's the method used to derive the quadratic formula itself.

Example -- Completing the Square

Solve: x² + 6x + 2 = 0

Step 1: Move the constant: x² + 6x = -2

Step 2: Take half of 6, which is 3, and square it to get 9. Add 9 to both sides:

x² + 6x + 9 = -2 + 9

(x + 3)² = 7

Step 3: Take the square root of both sides:

x + 3 = ±√7

Step 4: Subtract 3:

x = -3 ± √7

Solutions: x = -3 + √7 ≈ -0.354 and x = -3 - √7 ≈ -5.646

Common Mistake

In the quadratic formula, the entire numerator (-b ± √(b² - 4ac)) is divided by 2a, not just the square root part. Many students write x = -b ± √(b² - 4ac) / 2a, which would incorrectly divide only the √ part by 2a. Use parentheses to keep it clear.

Why This Matters for CS

Quadratic equations show up in physics simulations (projectile motion follows a parabola), game development (collision detection), and algorithm analysis. Some recurrence relations that describe algorithm runtimes lead to quadratic equations. The quadratic formula itself is a great example of how a general formula can solve a whole class of problems -- the same mindset behind writing reusable functions in code.

9. Exponent Rules (Advanced)

Exponents are shorthand for repeated multiplication. x³ means x * x * x. The base is x and the exponent (or power) is 3.

The Core Rules

Rule	Formula	Example
Product Rule	x^a * x^b = x^a+b	x³ * x² = x⁵
Quotient Rule	x^a / x^b = x^a-b	x⁵ / x² = x³
Power Rule	(x^a)^b = x^ab	(x³)² = x⁶
Product to Power	(xy)^a = x^a * y^a	(2x)³ = 8x³
Quotient to Power	(x/y)^a = x^a / y^a	(x/3)² = x²/9

Zero and Negative Exponents

x⁰ = 1 (for any x ≠ 0)

x^-n = 1 / xⁿ

Why does x⁰ = 1? Think of it this way: x³ / x³ = 1 (anything divided by itself is 1). But by the quotient rule, x³ / x³ = x^3-3 = x⁰. So x⁰ must equal 1.

Why does x^-n = 1/xⁿ? Same logic: x² / x⁵ = 1/x³ (dividing out). But by the quotient rule, x² / x⁵ = x^2-5 = x^-3. So x^-3 = 1/x³.

Example -- Negative Exponents

Simplify: 2^-3

2^-3 = 1 / 2³ = 1/8

Fractional Exponents (Roots)

x^1/n = ⁿ√x (the nth root of x)

x^m/n = (ⁿ√x)^m = ⁿ√(x^m)

Example -- Fractional Exponents

Simplify: 8^2/3

Step 1: 8^1/3 = ³√8 = 2

Step 2: 2² = 4

Answer: 8^2/3 = 4

Scientific Notation

Scientific notation expresses very large or very small numbers using powers of 10:

a × 10ⁿ (where 1 ≤ a < 10)

Example -- Scientific Notation

4,500,000 = 4.5 × 10⁶ (moved the decimal 6 places left)

0.00032 = 3.2 × 10^-4 (moved the decimal 4 places right)

Common Mistake

The product rule (x^a * x^b = x^a+b) only works when the bases are the same. You can simplify x³ * x² = x⁵, but you CANNOT simplify x³ * y² this way. Also, don't confuse the product rule (add exponents) with the power rule (multiply exponents).

Complete Laws of Exponents Summary

Here are all the exponent laws in one place. Understanding why each works will help you remember them.

Law 1: x¹ = x — Any number to the first power is itself

Law 2: x⁰ = 1 — Any non-zero number to the zero power is 1

Law 3: x^-n = 1/xⁿ — Negative exponent means "reciprocal"

Law 4: x^m × xⁿ = x^m+n — Multiplying same bases: ADD exponents

Law 5: x^m ÷ xⁿ = x^m-n — Dividing same bases: SUBTRACT exponents

Law 6: (x^m)ⁿ = x^m×n — Power of a power: MULTIPLY exponents

Law 7: (xy)ⁿ = xⁿ × yⁿ — Power of a product

Law 8: (x/y)ⁿ = xⁿ / yⁿ — Power of a quotient

Law 9: x^m/n = ⁿ√(x^m) = (ⁿ√x)^m — Fractional exponents = roots

Deep Dive: Why These Laws Work

Why x⁰ = 1?

The Logic

Consider: x³ ÷ x³ = 1 (anything divided by itself is 1)

But using Law 5: x³ ÷ x³ = x^3-3 = x⁰

Therefore: x⁰ = 1

Why x^-n = 1/xⁿ?

The Logic

Consider: x² ÷ x⁵ = x × x / (x × x × x × x × x) = 1/(x × x × x) = 1/x³

But using Law 5: x² ÷ x⁵ = x^2-5 = x^-3

Therefore: x^-3 = 1/x³

Practice Problems: Simplifying Exponents

Example 1: Product Rule

Simplify: 2⁴ × 2³

= 2⁴⁺³ = 2⁷ = 128

Example 2: Quotient Rule

Simplify: 5⁸ ÷ 5⁵

= 5^8-5 = 5³ = 125

Example 3: Power Rule

Simplify: (3²)⁴

= 3^2×4 = 3⁸ = 6561

Example 4: Negative Exponents

Simplify: 4^-2

= 1/4² = 1/16 = 0.0625

Example 5: Fractional Exponents

Simplify: 27^2/3

= (³√27)² = 3² = 9

Or: ³√(27²) = ³√729 = 9

Example 6: Combined Rules

Simplify: (x³y²)⁴

= (x³)⁴ × (y²)⁴

= x¹² × y⁸ = x¹²y⁸

Example 7: Complex Expression

Simplify: (2x³)² × (3x^-2)

= 2² × x⁶ × 3 × x^-2

= 4 × 3 × x^6-2

= 12x⁴

Moving Between Numerator and Denominator

A negative exponent in the numerator becomes positive when moved to the denominator, and vice versa:

x^-n = 1/xⁿ
1/x^-n = xⁿ

Moving rule: When you move a term across the fraction bar, flip the sign of its exponent.

Example: Simplify to Positive Exponents Only

Write without negative exponents: x^-3y² / z^-1

= y²z / x³

(x^-3 moves to denominator as x³, z^-1 moves to numerator as z)

Common Exponent Patterns

Expression	Result	Pattern
2⁰	1	Anything to the 0 is 1
(-3)²	9	Negative base, even exponent = positive
(-3)³	-27	Negative base, odd exponent = negative
-3²	-9	This is -(3²), NOT (-3)²
0⁵	0	0 to any positive power is 0
0⁰	undefined	Mathematically ambiguous

Watch the Negative Sign!

-3² ≠ (-3)²

-3² means -(3²) = -9 (the negative is not part of the base)

(-3)² means (-3) × (-3) = 9 (the negative IS part of the base)

This distinction is crucial in programming too: -3**2 vs (-3)**2

Why This Matters for CS

Exponents are fundamental to understanding big O notation. When we say an algorithm is O(n²) or O(2ⁿ), we're using exponents to describe how fast the runtime grows. Scientific notation is essentially how floating-point numbers are stored in computer memory -- as a mantissa times a power of 2. Understanding exponent rules helps you simplify algorithm complexity expressions like O(n² * n³) = O(n⁵).

Big-O Simplification Mindset

When you see a complex expression and need to find the Big-O, your brain might want to simplify it perfectly. Stop. Big-O is approximation math, not exact math. You only need to find the biggest term.

The Big-O Rules (memorize these three):

1. Drop constants: O(5n²) = O(n²). Constants don't affect growth rate.

2. Drop lower-order terms: O(n² + n) = O(n²). Smaller terms become irrelevant as n grows.

3. The highest exponent wins: In n⁴ + n² + n, the n⁴ dominates. O(n⁴).

Full Walkthrough: f(n) = 3n × (40 + n³/2)

Step 1: Expand using distribution

= 3n × 40 + 3n × n³/2

= 120n + 3n⁴/2

Step 2: Find the Big-O

We have two terms: 120n and 3n⁴/2

Which has the higher exponent?

120n = 120n¹ → exponent is 1
3n⁴/2 = (3/2)n⁴ → exponent is 4

Step 3: Apply the rules

Drop the lower term (120n): gone.

Drop the constant (3/2): gone.

What remains: O(n⁴)

More Big-O Simplification Practice

Expression	Expand	Dominant Term	Big-O
5n + 2	(already expanded)	5n	O(n)
n² + 100n + 500	(already expanded)	n²	O(n²)
n(n+1)/2	n²/2 + n/2	n²/2	O(n²)
3n⁴ + n² + 7	(already expanded)	3n⁴	O(n⁴)
2ⁿ + n¹⁰⁰	(already expanded)	2ⁿ	O(2ⁿ)

Note: Exponentials (2ⁿ) always beat polynomials (n¹⁰⁰), no matter how big the exponent. 2ⁿ grows faster than n^1000000 for large enough n.

The Mental Shift

Your brain wants to simplify 3n⁴/2 into a clean number. For Big-O, you don't need to. The moment you see n⁴, you're done. You don't care about the 3 or the /2. Engineers ask "what dominates when n is huge?" not "is this algebraically pristine?"

Think of it like this: if n = 1,000,000 then n⁴ = 1,000,000,000,000,000,000,000,000. The 120n term adds 120,000,000 to that. That's 0.000000000000012% of the total. It literally doesn't matter.

When You DO Need Exact Algebra

Big-O lets you skip simplification. But when solving equations, computing actual values, or proving formulas, you need exact algebra. The skill is knowing which mode you're in:

Big-O mode: Find the dominant term, drop everything else. Approximation is the goal.
Exact mode: Every step must be precise. Fractions, signs, and constants all matter.

Most algorithm analysis lives in Big-O mode. Most debugging and implementation lives in exact mode.

10. Logarithms

The One Sentence You Need

Forget the formal definition for a second. Here's the intuition that matters:

log₂(n) = "How many times do I halve n to reach 1?"

That's it. That's the whole idea. Everything else follows from this.

The Halving Game -- Play It Right Now

Start with 32. Keep halving until you hit 1. Count the steps:

32 → 16 → 8 → 4 → 2 → 1 5 steps

So log₂(32) = 5. Done.

Try 1,000,000:

1,000,000 → 500,000 → 250,000 → 125,000 → 62,500 → 31,250 → 15,625 → ~7,813 → ~3,906 → ~1,953 → ~977 → ~488 → ~244 → ~122 → ~61 → ~31 → ~15 → ~8 → ~4 → ~2 → 1

About 20 steps. log₂(1,000,000) ≈ 20. A million items reduced to 20 steps by halving!

The 20 Questions Analogy

The game "20 Questions" is literally binary search. With each yes/no question, you eliminate half the possibilities.

After 1 question: 500,000 possibilities remain
After 2 questions: 250,000
After 10 questions: ~1,000
After 20 questions: ~1

With 20 yes/no questions, you can find any item among a million. That's the power of log₂(n). It measures how many times you split the problem in half.

The Formal Definition

Now that you have the intuition, here's the math: a logarithm answers "What power do I raise this base to in order to get this number?" It is the inverse of exponentiation.

log_b(x) = y   means   b^y = x

"log base b of x equals y" means "b raised to the y equals x"

Exponentiation: 2⁵ = 32   (I know the power, what's the result?)
Logarithm: log₂(32) = 5   (I know the result, what's the power?)

Example -- Reading Logarithms

log₂(8) = ?

Ask: "2 to what power gives 8?" 2³ = 8, so log₂(8) = 3

Or think: "How many halvings? 8 → 4 → 2 → 1 = 3 halvings"

log₁₀(1000) = ?

Ask: "10 to what power gives 1000?" 10³ = 1000, so log₁₀(1000) = 3

Or think: "How many digits minus 1?" 1000 has 4 digits, so log₁₀(1000) = 3

log₅(25) = ?

Ask: "5 to what power gives 25?" 5² = 25, so log₅(25) = 2

The log₂ Table You Must Memorize

These values come up constantly in CS. Don't memorize them by rote -- see the pattern:

n	log₂(n)	Why (halving)	Where You See This
1	0	Already at 1, zero halvings	Base case
2	1	2 → 1 (1 halving)	One comparison
4	2	4 → 2 → 1	2-bit number
8	3	8 → 4 → 2 → 1	3-bit number, byte = 2³
16	4	4 halvings	Hex digit (0-F)
32	5	5 halvings	32-bit int
64	6	6 halvings	64-bit systems
128	7	7 halvings	ASCII character set
256	8	8 halvings	1 byte = 2&sup8; = 256 values
1,024	10	10 halvings	1 KB (kibibyte)
1,048,576	20	20 halvings	1 MB
~1 billion	30	30 halvings	1 GB
~4.3 billion	32	32 halvings	IPv4 addresses, 32-bit max

The Mind-Blowing Takeaway

log₂ grows absurdly slowly. You need to DOUBLE n just to add 1 to log₂(n). Going from 1 million to 1 billion (1000x more data) only adds 10 more steps. This is why O(log n) algorithms are so fast -- they barely notice when you throw more data at them.

log₂ in Actual Algorithms

Binary Search -- The Classic O(log n)

You have a sorted array of 1,000 elements. How many comparisons does binary search need?

# Array of 1000 sorted elements
# Step 1: Check middle (index 500). Too high? Search left half (0-499).
# Step 2: Check middle of left half (index 250). Too low? Search right (251-499).
# Step 3: Check middle (index 375). Keep halving...
# ...
# After ~10 steps: found it!
# 
# Why 10? Because log₂(1000) ≈ 10.
# Each step halves the search space: 1000 → 500 → 250 → 125 → 63 → 32 → 16 → 8 → 4 → 2 → 1

Linear search would check up to 1,000 elements. Binary search: 10. That's the power of log n.

Height of a Balanced Binary Tree

A balanced binary tree with n nodes has height log₂(n).

Why? Each level doubles the number of nodes: 1, 2, 4, 8, 16, ...

So a tree with 1 million nodes is only about 20 levels deep.

That's why tree lookups (BST, AVL, Red-Black) are O(log n) -- you descend at most log₂(n) levels.

How Many Bits to Represent n?

The number of bits needed to represent n in binary is ⌊log₂(n)⌋ + 1.

Why? Each bit doubles the range: 1 bit → 2 values, 2 bits → 4 values, k bits → 2^k values.

To represent 1000: log₂(1000) ≈ 9.97, so you need 10 bits. Check: 2¹⁰ = 1024 ≥ 1000.

Common and Natural Logs

log₂(x) = "CS log" -- the one you'll use 90% of the time. How many halvings / doublings.
log₁₀(x) = "common log" -- roughly counts the number of digits. log₁₀(1000) = 3 and 1000 has 4 digits.
ln(x) = log_e(x) = "natural log" -- where e ≈ 2.71828. Shows up in calculus, probability, and continuous growth/decay.

Why Base Doesn't Matter in Big-O

In Big-O notation, we just write O(log n) without specifying the base. Why?

The change of base formula: log₂(n) = log₁₀(n) / log₁₀(2) ≈ 3.32 × log₁₀(n)

So log₂(n) and log₁₀(n) differ only by a constant factor (3.32), and Big-O ignores constants. All log bases are equivalent up to a constant, so O(log₂ n) = O(log₁₀ n) = O(ln n) = O(log n).

Logarithm Rules

Rule	Formula	In Words
Product Rule	log(ab) = log(a) + log(b)	Log of a product = sum of logs
Quotient Rule	log(a/b) = log(a) - log(b)	Log of a quotient = difference of logs
Power Rule	log(aⁿ) = n × log(a)	Exponent comes down as multiplier
Change of Base	log_b(x) = log(x) / log(b)	Convert between bases
Identity	log_b(b) = 1	Any base logged to itself is 1
Identity	log_b(1) = 0	Log of 1 is always 0
Inverse	b^log_b(x) = x	Exponent and log cancel out

Example -- Using Log Rules

Expand: log(x²y/z)

Step 1: Apply quotient rule: log(x²y) - log(z)

Step 2: Apply product rule: log(x²) + log(y) - log(z)

Step 3: Apply power rule: 2log(x) + log(y) - log(z)

Answer: 2log(x) + log(y) - log(z)

Example -- Why the Product Rule Makes Sense

Think about it in binary (base 2):

log₂(8 × 4) = log₂(32) = 5

log₂(8) + log₂(4) = 3 + 2 = 5 ✓

Why does this work? Multiplying by 8 means "shift left 3 bits" and multiplying by 4 means "shift left 2 bits." Total shift: 3 + 2 = 5. Logs count the shifts!

Solving Exponential Equations with Logs

When the variable is in the exponent, use logarithms to bring it down.

Example -- Solving an Exponential Equation

Solve: 2^x = 32

Method 1 (recognition): 2⁵ = 32, so x = 5.

Method 2 (halving intuition): 32 → 16 → 8 → 4 → 2 → 1. That's 5 halvings, so x = 5.

Method 3 (logs): Take log of both sides:

log(2^x) = log(32)

x × log(2) = log(32)

x = log(32) / log(2) = 5

Example -- When You Can't Guess

Solve: 3^x = 20

Take log of both sides: x × log(3) = log(20)

x = log(20) / log(3) = 1.3010 / 0.4771 ≈ 2.727

CS Example -- How Long Until My Data Doubles?

Your database grows 5% per month. When will it double?

You need: 1.05^x = 2

x = log(2) / log(1.05) = 0.3010 / 0.02119 ≈ 14.2 months

Rule of 72 shortcut: 72 / growth rate ≈ doubling time. So 72 / 5 ≈ 14.4 months. Close enough!

Thinking in Logs: The Cheat Sheet

When you see these patterns, think "logarithm":

You See...	Think...	Example
"Halving repeatedly"	log₂(n) steps	Binary search, divide & conquer
"How many digits?"	log₁₀(n) + 1 digits	Number of digits in n
"How many bits?"	log₂(n) + 1 bits	Binary representation
"Tree height"	log₂(n) levels	Balanced BST, heap
"Splitting into k parts"	log_k(n) levels	B-tree (k-way split)
"When does it double?"	Solve with logs	Growth rate analysis
"Undoing an exponent"	Apply log to both sides	Solving 2^x = 1024

Common Mistakes

log(a + b) is NOT log(a) + log(b). There is no log rule for sums. The product rule says log(a × b) = log(a) + log(b), but this only works for multiplication, not addition. This is the #1 algebra error students make with logs.

log(0) is undefined. You can't raise any positive number to a power and get 0. And log of negatives doesn't exist in real numbers either.

Quick Self-Test: Can You Answer These Instantly?

If you can answer these without a calculator, you understand log₂:

log₂(64) = ? (6 -- because 2&sup6; = 64)
log₂(1) = ? (0 -- because 2&sup0; = 1)
log₂(1024) = ? (10 -- because 2¹⁰ = 1024)
How many binary search steps for 1 billion items? (~30)
Height of a balanced BST with 1 million nodes? (~20)

11. Sequences and Series

A sequence is an ordered list of numbers following a pattern. A series is the sum of a sequence. Understanding these is essential for algorithm analysis -- every time you analyze a loop or recursion, you're working with sequences and sums.

What is a Sequence? (The Basics)

Think of a sequence as a numbered list where each position has exactly one value:

Simple Examples

Natural numbers: 1, 2, 3, 4, 5, 6, ... (position 1 = 1, position 2 = 2, etc.)

Even numbers: 2, 4, 6, 8, 10, ... (position n = 2n)

Squares: 1, 4, 9, 16, 25, ... (position n = n²)

Powers of 2: 1, 2, 4, 8, 16, 32, ... (position n = 2^(n-1))

We write a_n to mean "the nth term of the sequence." So if our sequence is 2, 4, 6, 8, 10, ... then:

a₁ = 2 (first term)
a₂ = 4 (second term)
a₃ = 6 (third term)
a_n = 2n (formula for any term)

What is a Series? (Sum of a Sequence)

A series is what you get when you add up the terms of a sequence:

Sequence vs Series

Sequence: 1, 2, 3, 4, 5 (a list)

Series: 1 + 2 + 3 + 4 + 5 = 15 (a sum)

Sequence: 1, 2, 4, 8 (a list)

Series: 1 + 2 + 4 + 8 = 15 (a sum)

Why Does This Matter for Programming?

When you write a loop, you're essentially generating a sequence. When you count how many times that loop runs, you're computing a series. For example:

# This loop generates the sequence 0, 1, 2, 3, 4
for i in range(5):
    print(i)

# How many iterations? That's 5 (the sum of "1" five times)
# If you nested loops, you'd need series formulas to count total iterations

Arithmetic Sequences

In an arithmetic sequence, each term differs from the previous by a constant amount called the common difference (d).

How to Recognize Arithmetic Sequences

Subtract any term from the next. If you always get the same number, it's arithmetic!

3, 7, 11, 15, 19, ...

7 - 3 = 4, 11 - 7 = 4, 15 - 11 = 4, 19 - 15 = 4 → d = 4 ✓

10, 7, 4, 1, -2, ...

7 - 10 = -3, 4 - 7 = -3, 1 - 4 = -3 → d = -3 ✓ (can be negative!)

General term: aₙ = a₁ + (n - 1)d

Where:
• a₁ = first term
• d = common difference
• n = term number
• aₙ = the nth term

Step-by-Step: Finding the nth Term

Sequence: 3, 7, 11, 15, 19, ...

Step 1: Identify a₁ (first term) = 3

Step 2: Find d (common difference) = 7 - 3 = 4

Step 3: Write the formula: aₙ = 3 + (n - 1)(4)

Step 4: Simplify: aₙ = 3 + 4n - 4 = 4n - 1

Find the 10th term:

a₁₀ = 4(10) - 1 = 40 - 1 = 39

Find the 100th term:

a₁₀₀ = 4(100) - 1 = 400 - 1 = 399

Notice: with the formula, jumping to the 100th term is instant -- no need to calculate all 99 previous terms!

Common Mistake: Off-By-One Error

The formula uses (n - 1), NOT n. Why? Because the first term doesn't "add" any d yet:

a₁ = a₁ + (1-1)d = a₁ + 0d = a₁ ✓
a₂ = a₁ + (2-1)d = a₁ + 1d ✓
a₃ = a₁ + (3-1)d = a₁ + 2d ✓

To get from term 1 to term n, you add d exactly (n-1) times.

Arithmetic Series (Sum of Arithmetic Sequence)

The sum of the first n terms of an arithmetic sequence:

Sum formula: Sₙ = n/2 × (a₁ + aₙ) = n/2 × (first + last)

Or equivalently: Sₙ = n/2 × (2a₁ + (n-1)d)

Understanding the Formula Intuitively

Why does Sₙ = n/2 × (first + last) work?

Take 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10:

Pair the first with the last: 1 + 10 = 11

Pair the second with second-to-last: 2 + 9 = 11

Pair 3 + 8 = 11, 4 + 7 = 11, 5 + 6 = 11

We have 5 pairs (that's n/2 = 10/2), each summing to 11 (that's first + last).

Total: 5 × 11 = 55

Example -- Sum of 1 to 100

Find: 1 + 2 + 3 + ... + 100

This is an arithmetic sequence with a₁ = 1, aₙ = 100, n = 100

S₁₀₀ = 100/2 × (1 + 100) = 50 × 101 = 5,050

Legend has it that Gauss solved this as a child by noticing 1+100 = 2+99 = 3+98 = ... = 101, and there are 50 such pairs.

Example -- Sum of First n Odd Numbers

Find: 1 + 3 + 5 + 7 + 9 (first 5 odd numbers)

a₁ = 1, d = 2, n = 5, a₅ = 1 + (5-1)(2) = 9

S₅ = 5/2 × (1 + 9) = 2.5 × 10 = 25

Fun fact: The sum of the first n odd numbers is always n². Try it: 1+3=4=2², 1+3+5=9=3², 1+3+5+7=16=4²!

CS Application: Loop Analysis

The sum 1 + 2 + 3 + ... + n = n(n+1)/2 appears constantly in algorithm analysis:

for i in range(n):
    for j in range(i):  # Inner loop runs 0, 1, 2, ..., n-1 times
        print(i, j)
# Total iterations: 0 + 1 + 2 + ... + (n-1) = n(n-1)/2 = O(n²)

This is why nested triangular loops are O(n²), not O(n)!

Geometric Sequences

In a geometric sequence, each term is multiplied by a constant called the common ratio (r).

How to Recognize Geometric Sequences

Divide any term by the previous term. If you always get the same number, it's geometric!

2, 6, 18, 54, 162, ...

6/2 = 3, 18/6 = 3, 54/18 = 3, 162/54 = 3 → r = 3 ✓

100, 50, 25, 12.5, ...

50/100 = 0.5, 25/50 = 0.5, 12.5/25 = 0.5 → r = 0.5 ✓ (can be a fraction!)

Arithmetic vs Geometric:

Arithmetic: add the same amount (+d) → linear growth

Geometric: multiply by the same factor (×r) → exponential growth

General term: aₙ = a₁ × r^(n-1)

Where:
• a₁ = first term
• r = common ratio
• n = term number

Step-by-Step: Finding the nth Term

Sequence: 2, 6, 18, 54, 162, ...

Step 1: Identify a₁ = 2

Step 2: Find r = 6/2 = 3

Step 3: Write the formula: aₙ = 2 × 3^(n-1)

Find the 8th term:

a₈ = 2 × 3^(8-1) = 2 × 3⁷ = 2 × 2187 = 4,374

Notice how fast geometric sequences grow! The 8th term is over 4,000 even though we started at 2.

Example -- Exponential Growth (Compound Interest)

You invest $1,000 at 10% annual interest. How much after 5 years?

Each year, you multiply by 1.10 (100% + 10% = 110% = 1.10)

a₁ = 1000, r = 1.10, n = 6 (year 0 through year 5)

a₆ = 1000 × 1.10⁵ = 1000 × 1.61051 = $1,610.51

Example -- Exponential Decay (Half-Life)

A radioactive substance has a half-life of 1 hour. Starting with 800g, how much after 4 hours?

Each hour, you multiply by 0.5 (half remains)

a₁ = 800, r = 0.5, n = 5 (hour 0 through hour 4)

a₅ = 800 × 0.5⁴ = 800 × 0.0625 = 50g

Exponential Growth: Why It Destroys Everything

Humans are terrible at intuiting exponential growth. Our brains think linearly, so exponentials always surprise us. Let's fix that.

The Rice and Chessboard Problem

A king agrees to pay 1 grain of rice on square 1, 2 grains on square 2, 4 on square 3, doubling each square on a 64-square chessboard.

Square 1: 1 grain

Square 10: 512 grains (barely a pinch)

Square 20: 524,288 grains (a small bag)

Square 30: ~537 million grains (~13 tons of rice)

Square 40: ~550 billion grains (~13,000 tons)

Square 64: 2⁶³ = 9,223,372,036,854,775,808 grains

That's ~230 billion tons of rice -- more than all rice ever produced in human history.

This is why 2ⁿ algorithms are death sentences. The numbers look fine until they suddenly don't.

The Growth Rate Comparison Table (Burn This Into Your Brain)

n	log₂(n)	n	n log n	n²	2ⁿ	n!
1	0	1	0	1	2	1
5	2.3	5	12	25	32	120
10	3.3	10	33	100	1,024	3,628,800
20	4.3	20	86	400	1,048,576	~2.4 × 10¹⁸
30	4.9	30	147	900	~1 billion	~2.7 × 10³²
50	5.6	50	282	2,500	~10¹⁵	~3 × 10⁶⁴
100	6.6	100	664	10,000	~10³⁰	~9 × 10¹⁵⁷

At n=100: log n is 7, n² is 10,000, but 2ⁿ has 30 digits. Your computer running 10 billion operations/second would need 10²⁰ seconds for 2¹⁰⁰ operations. The universe is only 4 × 10¹⁷ seconds old.

Exponential Growth in CS: Recognizing the Pattern

You're dealing with exponential growth when:

Each step doubles the work: Recursive Fibonacci, brute-force subset enumeration
You're trying "all combinations": Password cracking, the traveling salesman problem
A function calls itself multiple times: f(n) = f(n-1) + f(n-2) -- each call spawns two more

And exponential decay (good!) when:

Each step halves the problem: Binary search, balanced tree operations
Probability drops exponentially: Hash collision chains, retry backoff
Cache hit rates: Working set usually follows an exponentially decaying access pattern

The Doubling Time Rule

If something grows by x% each period, it doubles in roughly 72/x periods.

Examples:

Data growing at 10% per month → doubles in ~7.2 months
Users growing at 5% per week → doubles in ~14.4 weeks
CPU speed doubling every 18 months (Moore's Law) → ~100% growth every 18 months

This is critical for capacity planning. If your server handles today's load fine but traffic doubles every 3 months, you have a 3-month runway, not years.

Geometric Series (Sum of Geometric Sequence)

Finite sum (r ≠ 1): Sₙ = a₁ × (1 - rⁿ)/(1 - r)

Alternative form: Sₙ = a₁ × (rⁿ - 1)/(r - 1) (more intuitive when r > 1)

Infinite sum (|r| < 1): S∞ = a₁/(1 - r)

Understanding Finite Geometric Sums

Find: 1 + 2 + 4 + 8 + 16 + 32 (sum of first 6 terms)

a₁ = 1, r = 2, n = 6

S₆ = 1 × (2⁶ - 1)/(2 - 1) = (64 - 1)/1 = 63

Shortcut for powers of 2: 1 + 2 + 4 + ... + 2^(n-1) = 2ⁿ - 1

So 1 + 2 + 4 + 8 + 16 + 32 = 2⁶ - 1 = 64 - 1 = 63 ✓

Why the 2ⁿ - 1 Pattern Matters

This pattern appears EVERYWHERE in computer science:

Binary numbers: An 8-bit number ranges from 0 to 2⁸ - 1 = 255
Max unsigned int (32-bit): 2³² - 1 = 4,294,967,295
Perfect binary tree: A complete binary tree of height h has 2^(h+1) - 1 nodes

Here's why: a binary tree of height 3 has 1 + 2 + 4 + 8 = 15 = 2⁴ - 1 nodes.

CS Application: Binary Numbers & Recursive Calls

The sum 1 + 2 + 4 + 8 + ... + 2^(n-1) = 2ⁿ - 1 explains:

Binary numbers: An n-bit unsigned integer can represent values from 0 to 2ⁿ - 1
Recursive branching: A binary tree of depth n has at most 2ⁿ - 1 nodes
Naive Fibonacci: Each call spawns 2 more calls → exponential growth

# Total calls in a binary recursive tree of depth n:
# Level 0: 1 call
# Level 1: 2 calls
# Level 2: 4 calls
# ...
# Level n-1: 2^(n-1) calls
# Total: 1 + 2 + 4 + ... + 2^(n-1) = 2ⁿ - 1 = O(2ⁿ)

Example -- Infinite Geometric Sum

Find: 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

a₁ = 1, r = 1/2 (|r| < 1, so it converges)

S∞ = 1/(1 - 1/2) = 1/(1/2) = 2

No matter how many terms you add, you'll never exceed 2!

Summation Notation (Sigma Notation)

Sigma notation (Σ) is a compact way to write sums. Think of it as a for loop in math form:

Σ notation: Σᵢ₌₁ⁿ f(i) = f(1) + f(2) + f(3) + ... + f(n)

Read it as: "Sum of f(i) as i goes from 1 to n"

The parts:
• i = the loop variable (called the "index")
• 1 = starting value (bottom of Σ)
• n = ending value (top of Σ)
• f(i) = what you're adding each iteration (the body)

Summations ARE For Loops

Every summation is literally a for loop. Every for loop that accumulates a total is a summation.

# This Python code...
total = 0
for i in range(1, n+1):
    total += f(i)

# ...is EXACTLY this math:
# Σᵢ₌₁ⁿ f(i)

Once you see this connection, summations stop being scary and start being familiar.

Examples -- Expanding Summations

Σᵢ₌₁⁵ i = 1 + 2 + 3 + 4 + 5 = 15

Σᵢ₌₁⁴ i² = 1² + 2² + 3² + 4² = 1 + 4 + 9 + 16 = 30

Σᵢ₌₀³ 2ⁱ = 2⁰ + 2¹ + 2² + 2³ = 1 + 2 + 4 + 8 = 15

Σᵢ₌₁³ (2i + 1) = 3 + 5 + 7 = 15

Translating Loops to Summations (The Skill That Matters)

This is the real reason you learn summations: to analyze code complexity. Here's the systematic method:

Pattern 1: Simple Loop → Simple Sum

for i in range(n):       # i goes from 0 to n-1
    do_something()         # O(1) work each time

# Total work: Σᵢ₌₀ⁿ⁻¹ 1 = n
# Big-O: O(n)

Pattern 2: Triangular Nested Loop → Arithmetic Sum

for i in range(n):            # i = 0, 1, 2, ..., n-1
    for j in range(i):         # inner runs 0, 1, 2, ..., n-1 times
        do_something()

# Inner loop runs: 0 + 1 + 2 + 3 + ... + (n-1)
# As a summation: Σᵢ₌₀ⁿ⁻¹ i = n(n-1)/2
# Big-O: O(n²)

Pattern 3: Halving Loop → Geometric/Log

i = n
while i >= 1:
    do_something()             # O(1) work
    i = i // 2                 # halve each iteration

# Iterations: n, n/2, n/4, ..., 2, 1
# Number of iterations: log₂(n)
# Big-O: O(log n)

Pattern 4: Nested with Halving → n log n

for i in range(n):            # O(n) outer iterations
    j = n
    while j >= 1:              # O(log n) inner iterations
        do_something()
        j = j // 2

# Total: Σᵢ₌₀ⁿ⁻¹ log(n) = n × log(n)
# Big-O: O(n log n)

Pattern 5: Doubling Inner Loop → Geometric Sum

for i in range(n):            # outer: n iterations
    for j in range(2**i):      # inner: 1, 2, 4, 8, ... 2^(n-1) iterations
        do_something()

# Total: Σᵢ₌₀ⁿ⁻¹ 2ⁱ = 2ⁿ - 1
# Big-O: O(2ⁿ)

The Cheat Sheet: Loop Pattern → Summation → Big-O

Loop Pattern	Summation	Result	Big-O
Simple loop (0 to n)	Σ 1	n	O(n)
Nested, both 0 to n	ΣΣ 1	n²	O(n²)
Triangular (j < i)	Σᵢ₌₀ⁿ i	n(n-1)/2	O(n²)
Halving (i = i/2)	Σ 1 (log n times)	log n	O(log n)
Outer n × inner log n	Σ log n	n log n	O(n log n)
Doubling inner (2ⁱ)	Σ 2ⁱ	2ⁿ - 1	O(2ⁿ)

Important Sum Formulas

These formulas appear constantly in algorithm analysis. Memorize them!

Sum	Formula	CS Application
1 + 2 + 3 + ... + n	n(n+1)/2	Triangular nested loops → O(n²)
1² + 2² + 3² + ... + n²	n(n+1)(2n+1)/6	Some triple nested patterns
1 + 2 + 4 + ... + 2^(n-1)	2ⁿ - 1	Binary trees, recursive branching → O(2ⁿ)
1 + r + r² + ... + rⁿ	(rⁿ⁺¹ - 1)/(r - 1)	Geometric growth patterns
n + n/2 + n/4 + ... + 1	≈ 2n	Work in merge sort levels → O(n)
1 + 1/2 + 1/3 + ... + 1/n	≈ ln(n)	Harmonic series (QuickSort avg case)

Harmonic Series

The harmonic series is: 1 + 1/2 + 1/3 + 1/4 + ... + 1/n

Hₙ = 1 + 1/2 + 1/3 + ... + 1/n ≈ ln(n) + 0.577...

CS Application: Expected Comparisons

The harmonic series appears in:

QuickSort expected comparisons: Average case involves harmonic sums
Coupon collector problem: Expected tries to collect all n coupons ≈ n × Hₙ ≈ n ln(n)
Hash table analysis: Expected probe count with linear probing

Connecting Sequences to Recursion

Many sequences are defined recursively -- each term depends on previous terms.

Famous Recursive Sequences

Fibonacci: F(n) = F(n-1) + F(n-2), with F(0) = 0, F(1) = 1

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

Factorial as a sequence: n! = n × (n-1)!

1, 1, 2, 6, 24, 120, 720, ...

Powers of 2: a(n) = 2 × a(n-1), with a(0) = 1

1, 2, 4, 8, 16, 32, ...

Why Naive Fibonacci is Slow

The Fibonacci recurrence T(n) = T(n-1) + T(n-2) creates exponential calls because each call branches into two more. The number of calls follows the Fibonacci sequence itself, which grows like φⁿ (where φ ≈ 1.618). This is why fib(50) with naive recursion is impossibly slow, but dynamic programming solves it in O(n).

The Big Picture

Sequences and series are the mathematical language of algorithm analysis:

Counting loop iterations = summing an arithmetic series
Recursive branching = geometric series
Master Theorem = solving recurrence relations (recursive sequences)
Amortized analysis = understanding how costs average over a sequence of operations

Every time you analyze code complexity, you're working with these formulas.

12. Practice Quiz

Test your understanding of the key algebra concepts covered on this page. Click an answer to check it.

1. Simplify: 3(2x - 4) + 5x

Answer: B) 11x - 12
First distribute: 3(2x - 4) = 6x - 12. Then combine with +5x: 6x - 12 + 5x = 11x - 12. Remember to distribute the 3 to both terms inside the parentheses.

2. Solve for x: 5x - 3 = 2x + 12

Answer: A) x = 5
Subtract 2x from both sides: 3x - 3 = 12. Add 3 to both sides: 3x = 15. Divide by 3: x = 5. Check: 5(5) - 3 = 22 and 2(5) + 12 = 22. Correct!

3. Factor completely: x² - 9

Answer: B) (x + 3)(x - 3)
This is a difference of squares. x² - 9 = x² - 3² = (x + 3)(x - 3). Remember: a² - b² = (a + b)(a - b). Note that (x - 3)² would expand to x² - 6x + 9, which is different.

4. What is the value of the discriminant for 2x² + 3x + 5 = 0, and what does it tell you?

Answer: C) -31; no real solutions
The discriminant is b² - 4ac = 3² - 4(2)(5) = 9 - 40 = -31. Since the discriminant is negative, the square root of a negative number is not a real number, so this equation has no real solutions. The parabola doesn't cross the x-axis.

5. Simplify using log rules: log₂(8) + log₂(4)

Answer: B) 5
Method 1 (evaluate each): log₂(8) = 3 (since 2³ = 8) and log₂(4) = 2 (since 2² = 4). So 3 + 2 = 5.
Method 2 (product rule): log₂(8) + log₂(4) = log₂(8 * 4) = log₂(32) = 5 (since 2⁵ = 32).
Note: log₂(12) is wrong -- the product rule says log(a) + log(b) = log(a*b), NOT log(a+b).

← Pre-Algebra Geometry →

Table of Contents

1. Variables and Expressions

What Are Variables?

Algebraic Expressions

Simplifying Expressions

The Distributive Property

Distributing with Fractions

Splitting Fractions (When You Can and Can't)

Common Denominator: Making Fractions Combine

2. Solving Linear Equations

One-Step Equations

Two-Step Equations

Multi-Step Equations (Variables on Both Sides)

Equations with Fractions

3. Inequalities

Solving Inequalities

Graphing Inequalities on a Number Line

4. Functions

What Is a Function?

Domain and Range

Linear Functions: y = mx + b

Slope Formula

Understanding Slope Intuitively

Finding the Equation of a Line

Graphing Lines Step-by-Step

Special Forms of Linear Equations

5. Systems of Equations

Method 1: Substitution

Method 2: Elimination

Understanding Solutions Graphically

More Practice Examples

6. Polynomials

What Are Polynomials?

Adding and Subtracting Polynomials

Multiplying Polynomials (FOIL)

Special Products (Memorize These!)

Perfect Square Trinomials

Difference of Squares

Sum and Difference of Cubes

Multiplying Polynomials with More Than Two Terms

Polynomial Vocabulary

Polynomial Division (Optional Advanced Topic)

7. Factoring

What is Factoring, Really?

First, Understand FOIL (So You Can Reverse It)

The Area Model -- A Visual Way to See It

Factoring Strategy -- Follow This Every Time

Technique 1: Greatest Common Factor (GCF)

Technique 2: Factoring Trinomials (x² + bx + c)

Technique 3: Factoring Harder Trinomials (ax² + bx + c, where a ≠ 1)

Technique 4: Difference of Squares

Technique 5: Perfect Square Trinomials

Technique 6: Factoring by Grouping (4 Terms)

8. Quadratic Equations

Standard Form

Method 1: Solving by Factoring

Method 2: The Quadratic Formula

What Each Part of the Formula Means

The Discriminant -- Peek at the Answer Before Solving

Method 3: Completing the Square

9. Exponent Rules (Advanced)

The Core Rules

Zero and Negative Exponents

Fractional Exponents (Roots)

Scientific Notation

Complete Laws of Exponents Summary

Deep Dive: Why These Laws Work

Why x0 = 1?

Why x-n = 1/xn?

Practice Problems: Simplifying Exponents

Moving Between Numerator and Denominator

Common Exponent Patterns

Big-O Simplification Mindset

10. Logarithms

The One Sentence You Need

The Formal Definition

The log2 Table You Must Memorize

log2 in Actual Algorithms

Common and Natural Logs

Why x⁰ = 1?

Why x^-n = 1/xⁿ?

The log₂ Table You Must Memorize

log₂ in Actual Algorithms

5. Simplify using log rules: log₂(8) + log₂(4)