Worked Examples
Learn by example! Below you'll find detailed solutions for each of Factor Friend's mathematical tools. Each example shows the complete work with step-by-step explanations to help you understand the mathematical reasoning.
1. Factor & Prime Factorization
Example 1.1: Finding All Factors of 180
Problem: Find all factors and factor pairs of 180, and express it as a product of primes.
Prime Factorization: Divide by smallest primes repeatedly
180 ÷ 2 = 90
90 ÷ 2 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5
5 is prime
180 ÷ 2 = 90
90 ÷ 2 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5
5 is prime
Result: 180 = 2² × 3² × 5
Factor Count: Add 1 to each exponent and multiply
(2 + 1) × (2 + 1) × (1 + 1) = 3 × 3 × 2 = 18 factors
(2 + 1) × (2 + 1) × (1 + 1) = 3 × 3 × 2 = 18 factors
Factor Pairs:
1 × 180
2 × 90
3 × 60
4 × 45
5 × 36
6 × 30
9 × 20
10 × 18
12 × 15
1 × 180
2 × 90
3 × 60
4 × 45
5 × 36
6 × 30
9 × 20
10 × 18
12 × 15
Answer: 180 has 18 factors. Prime factorization: 2² × 3² × 5
Example 1.2: Perfect Square Recognition
Problem: Is 144 a perfect square? Find its factors.
Prime Factorization: 144 = 2⁴ × 3²
Perfect Square Test: All exponents are even, so 144 = 12²
Factor Count: (4 + 1) × (2 + 1) = 15 factors
Note: The middle factor pair is (12, 12), confirming it's a perfect square
Answer: Yes, 144 = 12². It has 15 factors including 12 as the square root.
💡 Quick Tip: A number is a perfect square if and only if all exponents in its prime factorization are even. Perfect squares always have an odd number of total factors.
2. Least Common Multiple (LCM)
Example 2.1: LCM of Two Numbers
Problem: Find the LCM of 12 and 18.
Prime Factorizations:
12 = 2² × 3
18 = 2 × 3²
12 = 2² × 3
18 = 2 × 3²
For LCM, take highest power of each prime:
Highest power of 2: 2² (from 12)
Highest power of 3: 3² (from 18)
Highest power of 2: 2² (from 12)
Highest power of 3: 3² (from 18)
Multiply: LCM = 2² × 3² = 4 × 9 = 36
Answer: LCM(12, 18) = 36
Example 2.2: LCM of Three Numbers
Problem: Find the LCM of 8, 12, and 15.
Prime Factorizations:
8 = 2³
12 = 2² × 3
15 = 3 × 5
8 = 2³
12 = 2² × 3
15 = 3 × 5
Identify all primes: 2, 3, and 5
Take highest powers:
2³ (from 8)
3¹ (appears in both 12 and 15)
5¹ (from 15)
2³ (from 8)
3¹ (appears in both 12 and 15)
5¹ (from 15)
Calculate: LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120
Answer: LCM(8, 12, 15) = 120
💡 Real-World Application: LCM is used when finding common denominators for adding fractions. For example, to add 1/12 + 1/18, you need to find LCM(12, 18) = 36, then convert both fractions: 3/36 + 2/36 = 5/36.
3. Greatest Common Factor (GCF)
Example 3.1: GCF for Simplifying Fractions
Problem: Find GCF(48, 60) to simplify the fraction 48/60.
Prime Factorizations:
48 = 2⁴ × 3
60 = 2² × 3 × 5
48 = 2⁴ × 3
60 = 2² × 3 × 5
Find common primes: Both have 2 and 3
Take lowest powers of common primes:
Lowest power of 2: 2² (from 60)
Lowest power of 3: 3¹ (both have this)
Lowest power of 2: 2² (from 60)
Lowest power of 3: 3¹ (both have this)
Calculate: GCF = 2² × 3 = 4 × 3 = 12
Simplify fraction: 48/60 = (48÷12)/(60÷12) = 4/5
Answer: GCF(48, 60) = 12, so 48/60 simplifies to 4/5
Example 3.2: GCF of Three Numbers
Problem: Find the GCF of 24, 36, and 60.
Prime Factorizations:
24 = 2³ × 3
36 = 2² × 3²
60 = 2² × 3 × 5
24 = 2³ × 3
36 = 2² × 3²
60 = 2² × 3 × 5
Common primes: 2 and 3 (all three numbers have both)
Take lowest powers:
2² (lowest among 2³, 2², 2²)
3¹ (lowest among 3¹, 3², 3¹)
2² (lowest among 2³, 2², 2²)
3¹ (lowest among 3¹, 3², 3¹)
Calculate: GCF = 2² × 3 = 4 × 3 = 12
Answer: GCF(24, 36, 60) = 12
💡 Quick Fact: If two numbers have no common prime factors, their GCF is 1, and they're called "relatively prime" or "coprime." For example, GCF(15, 28) = 1.
4. Quadratic Equation Factoring
Example 4.1: Simple Trinomial (a = 1)
Problem: Factor and solve x² + 7x + 12 = 0
Find factors of c = 12 that sum to b = 7:
Factors of 12: 1×12, 2×6, 3×4
3 + 4 = 7 ✓
Factors of 12: 1×12, 2×6, 3×4
3 + 4 = 7 ✓
Write factored form: (x + 3)(x + 4) = 0
Solve each factor:
x + 3 = 0 → x = -3
x + 4 = 0 → x = -4
x + 3 = 0 → x = -3
x + 4 = 0 → x = -4
Answer: x² + 7x + 12 = (x + 3)(x + 4), solutions: x = -3 or x = -4
Example 4.2: Difference of Squares
Problem: Factor x² - 25 = 0
Recognize pattern: A² - B² = (A + B)(A - B)
Identify A and B:
x² = (x)², so A = x
25 = 5², so B = 5
x² = (x)², so A = x
25 = 5², so B = 5
Apply formula: (x + 5)(x - 5) = 0
Solve: x = -5 or x = 5
Answer: x² - 25 = (x + 5)(x - 5), solutions: x = ±5
Example 4.3: Complex Trinomial (a ≠ 1)
Problem: Factor 6x² + 11x + 4 = 0
Calculate a × c: 6 × 4 = 24
Find factors of 24 that sum to 11:
Factors: 1×24, 2×12, 3×8, 4×6
3 + 8 = 11 ✓
Factors: 1×24, 2×12, 3×8, 4×6
3 + 8 = 11 ✓
Rewrite middle term: 6x² + 3x + 8x + 4
Factor by grouping:
3x(2x + 1) + 4(2x + 1)
(3x + 4)(2x + 1) = 0
3x(2x + 1) + 4(2x + 1)
(3x + 4)(2x + 1) = 0
Solve:
3x + 4 = 0 → x = -4/3
2x + 1 = 0 → x = -1/2
3x + 4 = 0 → x = -4/3
2x + 1 = 0 → x = -1/2
Answer: 6x² + 11x + 4 = (3x + 4)(2x + 1), solutions: x = -4/3 or x = -1/2
Example 4.4: Using Quadratic Formula
Problem: Solve x² - 4x + 1 = 0 (no rational factors)
Identify coefficients: a = 1, b = -4, c = 1
Calculate discriminant: D = b² - 4ac = 16 - 4 = 12
Apply formula: x = (4 ± √12) / 2
Simplify: √12 = 2√3
x = (4 ± 2√3) / 2 = 2 ± √3
x = (4 ± 2√3) / 2 = 2 ± √3
Answer: x = 2 + √3 ≈ 3.73 or x = 2 - √3 ≈ 0.27
💡 Remember: The discriminant (D = b² - 4ac) tells you about the roots:
• D > 0 and perfect square → two rational roots
• D > 0 but not perfect square → two irrational roots
• D = 0 → one repeated root
• D < 0 → two complex roots
• D > 0 and perfect square → two rational roots
• D > 0 but not perfect square → two irrational roots
• D = 0 → one repeated root
• D < 0 → two complex roots
5. Quadratic Formula Solver
The quadratic formula tool shows the complete derivation for solving any quadratic equation, with detailed steps for calculating the discriminant and simplifying the solutions.
Example 5.1: Rational Solutions
Problem: Solve 2x² + 7x + 3 = 0 using the quadratic formula
Identify coefficients: a = 2, b = 7, c = 3
Calculate discriminant: D = b² - 4ac = 49 - 24 = 25
Since D = 25 is a perfect square: We'll get rational solutions
Apply formula: x = (-7 ± √25) / 4 = (-7 ± 5) / 4
Two solutions:
x = (-7 + 5) / 4 = -2/4 = -1/2
x = (-7 - 5) / 4 = -12/4 = -3
x = (-7 + 5) / 4 = -2/4 = -1/2
x = (-7 - 5) / 4 = -12/4 = -3
Answer: x = -1/2 or x = -3
Example 5.2: Complex Solutions
Problem: Solve x² + 2x + 5 = 0
Coefficients: a = 1, b = 2, c = 5
Discriminant: D = 4 - 20 = -16
Since D < 0: We have complex solutions involving i
Apply formula: x = (-2 ± √(-16)) / 2 = (-2 ± 4i) / 2
Simplify: x = -1 ± 2i
Answer: x = -1 + 2i or x = -1 - 2i (complex conjugate pair)
💡 Quadratic Formula Benefits: Unlike factoring, the quadratic formula works for ALL quadratic equations, including those with irrational or complex solutions. Our tool shows every computational step!
6. Radical Simplification
Example 6.1: Simplifying Square Roots
Problem: Simplify √180
Prime factorization: 180 = 2² × 3² × 5
Group perfect squares: √(2² × 3² × 5)
Extract perfect squares: 2 × 3 × √5 = 6√5
Answer: √180 = 6√5
Example 6.2: Cube Roots
Problem: Simplify ∛216
Prime factorization: 216 = 2³ × 3³
Group by index 3: ∛(2³ × 3³)
Extract perfect cubes: 2 × 3 = 6
Answer: ∛216 = 6 (it's a perfect cube!)
Example 6.3: Rationalizing Denominators
Problem: Simplify √(3/8)
Separate: √3 / √8
Simplify denominator: √8 = √(2² × 2) = 2√2
Rationalize: Multiply by √2/√2
(√3 / 2√2) × (√2/√2) = √6 / 4
(√3 / 2√2) × (√2/√2) = √6 / 4
Answer: √(3/8) = √6 / 4
💡 Rationalization Rule: Never leave radicals in the denominator of a final answer. Multiply by the appropriate form of 1 to eliminate them.
7. Vertex Form Conversion
Example 7.1: Converting to Vertex Form
Problem: Convert y = x² + 6x + 5 to vertex form
Start: y = x² + 6x + 5
Complete the square: Take half of b, then square it
(6/2)² = 9
(6/2)² = 9
Add and subtract 9: y = x² + 6x + 9 - 9 + 5
Factor perfect square: y = (x + 3)² - 4
Vertex: h = -3, k = -4, so vertex is (-3, -4)
Answer: y = (x + 3)² - 4, vertex: (-3, -4)
Example 7.2: When a ≠ 1
Problem: Convert y = 2x² - 8x + 3 to vertex form
Factor out a = 2: y = 2(x² - 4x) + 3
Complete the square inside: (-4/2)² = 4
Add and subtract: y = 2(x² - 4x + 4 - 4) + 3
Distribute the -4: y = 2(x² - 4x + 4) - 8 + 3
Factor and simplify: y = 2(x - 2)² - 5
Vertex: (2, -5)
Answer: y = 2(x - 2)² - 5, vertex: (2, -5)
💡 Graph Reading: In vertex form y = a(x - h)² + k:
• (h, k) is the vertex (minimum if a > 0, maximum if a < 0)
• |a| controls how "wide" or "narrow" the parabola is
• a < 0 means the parabola opens downward
• (h, k) is the vertex (minimum if a > 0, maximum if a < 0)
• |a| controls how "wide" or "narrow" the parabola is
• a < 0 means the parabola opens downward
8. Polynomial Long Division
Example 8.1: Basic Division
Problem: Divide (x³ + 5x² + 8x + 4) by (x + 2)
First term: x³ ÷ x = x²
Multiply back: x²(x + 2) = x³ + 2x²
Subtract: 3x² + 8x + 4
Multiply back: x²(x + 2) = x³ + 2x²
Subtract: 3x² + 8x + 4
Second term: 3x² ÷ x = 3x
Multiply back: 3x(x + 2) = 3x² + 6x
Subtract: 2x + 4
Multiply back: 3x(x + 2) = 3x² + 6x
Subtract: 2x + 4
Third term: 2x ÷ x = 2
Multiply back: 2(x + 2) = 2x + 4
Subtract: 0
Multiply back: 2(x + 2) = 2x + 4
Subtract: 0
Answer: Quotient = x² + 3x + 2, Remainder = 0
Therefore: (x³ + 5x² + 8x + 4) = (x + 2)(x² + 3x + 2)
Therefore: (x³ + 5x² + 8x + 4) = (x + 2)(x² + 3x + 2)
Example 8.2: Division with Remainder
Problem: Divide (2x³ - 3x² + 4x - 5) by (x - 1)
Quotient terms: 2x², -x, 3
(Working through the same process as above)
(Working through the same process as above)
Final remainder: -2
Answer: (2x³ - 3x² + 4x - 5) / (x - 1) = 2x² - x + 3 - 2/(x - 1)
💡 Division Algorithm: For polynomials, Dividend = Divisor × Quotient + Remainder
This is just like regular number division, but with polynomials!
This is just like regular number division, but with polynomials!
9. Arithmetic Series
Example 9.1: Sum of Arithmetic Series (Given First and Last Terms)
Problem: Find the sum of the arithmetic series: 5 + 10 + 15 + ... + 100
Identify the components:
First term (a₁) = 5
Last term (aₙ) = 100
Common difference (d) = 10 - 5 = 5
First term (a₁) = 5
Last term (aₙ) = 100
Common difference (d) = 10 - 5 = 5
Find the number of terms:
Use the formula: aₙ = a₁ + (n - 1)d
100 = 5 + (n - 1)(5)
100 = 5 + 5n - 5
100 = 5n
n = 20 terms
Use the formula: aₙ = a₁ + (n - 1)d
100 = 5 + (n - 1)(5)
100 = 5 + 5n - 5
100 = 5n
n = 20 terms
Calculate the sum:
Use the formula: Sₙ = n(a₁ + aₙ)/2
S₂₀ = 20(5 + 100)/2
S₂₀ = 20(105)/2
S₂₀ = 2,100/2
S₂₀ = 1,050
Use the formula: Sₙ = n(a₁ + aₙ)/2
S₂₀ = 20(5 + 100)/2
S₂₀ = 20(105)/2
S₂₀ = 2,100/2
S₂₀ = 1,050
Answer: The sum is 1,050
Example 9.2: Sum of First n Natural Numbers
Problem: Find the sum of the first 50 natural numbers (1 + 2 + 3 + ... + 50)
Identify the series:
This is an arithmetic series with a₁ = 1, aₙ = 50, d = 1, n = 50
This is an arithmetic series with a₁ = 1, aₙ = 50, d = 1, n = 50
Apply the sum formula:
Sₙ = n(a₁ + aₙ)/2
S₅₀ = 50(1 + 50)/2
S₅₀ = 50(51)/2
S₅₀ = 2,550/2
S₅₀ = 1,275
Sₙ = n(a₁ + aₙ)/2
S₅₀ = 50(1 + 50)/2
S₅₀ = 50(51)/2
S₅₀ = 2,550/2
S₅₀ = 1,275
Verification using Gauss's method:
Pair first and last: 1 + 50 = 51
Pair second and second-to-last: 2 + 49 = 51
There are 25 such pairs: 25 × 51 = 1,275 ✓
Pair first and last: 1 + 50 = 51
Pair second and second-to-last: 2 + 49 = 51
There are 25 such pairs: 25 × 51 = 1,275 ✓
Answer: The sum of the first 50 natural numbers is 1,275
Example 9.3: Finding Sum Using Number of Terms
Problem: Find the sum of the first 15 terms of the series: 3 + 7 + 11 + 15 + ...
Identify components:
First term (a₁) = 3
Common difference (d) = 7 - 3 = 4
Number of terms (n) = 15
First term (a₁) = 3
Common difference (d) = 7 - 3 = 4
Number of terms (n) = 15
Find the last term:
aₙ = a₁ + (n - 1)d
a₁₅ = 3 + (15 - 1)(4)
a₁₅ = 3 + 14(4)
a₁₅ = 3 + 56
a₁₅ = 59
aₙ = a₁ + (n - 1)d
a₁₅ = 3 + (15 - 1)(4)
a₁₅ = 3 + 14(4)
a₁₅ = 3 + 56
a₁₅ = 59
Calculate the sum:
Sₙ = n(a₁ + aₙ)/2
S₁₅ = 15(3 + 59)/2
S₁₅ = 15(62)/2
S₁₅ = 930/2
S₁₅ = 465
Sₙ = n(a₁ + aₙ)/2
S₁₅ = 15(3 + 59)/2
S₁₅ = 15(62)/2
S₁₅ = 930/2
S₁₅ = 465
Answer: The sum of the first 15 terms is 465
Example 9.4: Sum of Even Numbers
Problem: Find the sum of all even numbers from 2 to 100
Identify the series:
Series: 2 + 4 + 6 + 8 + ... + 100
a₁ = 2, aₙ = 100, d = 2
Series: 2 + 4 + 6 + 8 + ... + 100
a₁ = 2, aₙ = 100, d = 2
Find the number of terms:
aₙ = a₁ + (n - 1)d
100 = 2 + (n - 1)(2)
100 = 2 + 2n - 2
100 = 2n
n = 50
aₙ = a₁ + (n - 1)d
100 = 2 + (n - 1)(2)
100 = 2 + 2n - 2
100 = 2n
n = 50
Calculate the sum:
Sₙ = n(a₁ + aₙ)/2
S₅₀ = 50(2 + 100)/2
S₅₀ = 50(102)/2
S₅₀ = 5,100/2
S₅₀ = 2,550
Sₙ = n(a₁ + aₙ)/2
S₅₀ = 50(2 + 100)/2
S₅₀ = 50(102)/2
S₅₀ = 5,100/2
S₅₀ = 2,550
Answer: The sum of all even numbers from 2 to 100 is 2,550
Key Formulas for Arithmetic Series:
• Number of terms: n = [(aₙ - a₁) / d] + 1
• nth term: aₙ = a₁ + (n - 1)d
• Sum formula: Sₙ = n(a₁ + aₙ)/2 or Sₙ = n[2a₁ + (n - 1)d]/2
• Always verify that the last term is reachable by adding the common difference!
• Number of terms: n = [(aₙ - a₁) / d] + 1
• nth term: aₙ = a₁ + (n - 1)d
• Sum formula: Sₙ = n(a₁ + aₙ)/2 or Sₙ = n[2a₁ + (n - 1)d]/2
• Always verify that the last term is reachable by adding the common difference!
Practice Makes Perfect
These examples demonstrate the core concepts behind each of Factor Friend's tools. The best way to master these techniques is through practice. Head over to our tools page and try solving your own problems!