Session 01-04 - Advanced Factorization & Radicals

Section 01: Mathematical Foundations & Algebra

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz

Quick Review from Last Session

Complete individually, then we review as group

Simplify: \(\frac{(2x^3)^2 \cdot x^{-5}}{4x^2}\)
Factor: \(9x^2 - 25\)
Solve: \(|2x - 4| > 6\)
Express in scientific notation: \(0.0000234 \times 10^3\)
Factor: \(3x^2 - 12\)

. . .

Ok, lets talk about your solutions together!

Student Presentations

Homework Showcase

20 minutes for discussing your solutions

Present and discuss your solutions from Tasks 01-03
Focus on the most challenging aspects or tasks
Share any challenging aspects or alternative approaches

. . .

Today we build on basic factorization with advanced techniques and introduce radicals!

Advanced Factorization

Repetition: What is Factorization?

Breaking expressions into products of simpler factors

Factorization means writing an expression as a product of its factors.

Example: \(12 = 3 \times 4\) (factoring numbers)
Algebra: \(x^2 + 5x + 6 = (x + 2)(x + 3)\) (factoring polynomials)
Reverse of expansion: \((x + 2)(x + 3) \rightarrow x^2 + 5x + 6\)

. . .

Why Factor?

Because the exam requires you to know this!
Can help us to solve equations later
Cancel common factors in fractions and help to simplify expressions

Factoring Quadratics: \(ax^2 + bx + c\)

When \(a = 1\): Find factors of c that sum to b

Factor \(x^2 + 7x + 12\)

Need factors of 12 that add to 7
Pairs: (1,12), (2,6), (3,4)
Check: \(3 + 4 = 7\) ✓
Result: \((x + 3)(x + 4)\)

Factor \(x^2 - 5x - 14\)

Need factors of -14 that add to -5
Pairs: (-1,14), (1,-14), (-2,7), (2,-7)
Check: \(2 + (-7) = -5\) ✓
Result: \((x + 2)(x - 7)\)

The AC Method for \(ax^2 + bx + c\)

When the leading coefficient \(a \neq 1\)

Factor: \(6x^2 + 13x + 5\)

Step 1: Find \(ac = 6 \times 5 = 30\)
Step 2: Find factors of 30 that sum to 13, e.g. (3,10)
Step 3: Rewrite: \(6x^2 + 3x + 10x + 5\)
Step 4: Group: \(3x(2x + 1) + 5(2x + 1)\)
Step 5: Factor: \((3x + 5)(2x + 1)\)

. . .

Check your solution by expanding: \((3x + 5)(2x + 1) = 6x^2 + 13x + 5\) ✓

The AC Method When \(ac < 0\)

When \(ac\) is negative, factors have opposite signs

Factor: \(6x^2 + 7x - 5\)

Step 1: Find \(ac = 6 \times (-5) = -30\) (negative!)
Step 2: Find factors of -30 that sum to 7, e.g. (10, -3)
- Need one positive, one negative factor!
Step 3: Rewrite: \(6x^2 + 10x - 3x - 5\)
Step 4: Group: \(2x(3x + 5) - 1(3x + 5)\)
Step 5: Factor: \((2x - 1)(3x + 5)\)

. . .

When \(ac < 0\): Look for factor pairs with opposite signs that sum to \(b\)

The Discriminant: Can We Factor?

A mathematical detective tool that saves us time!

The Problem: You’re given \(3x^2 + 7x + 11\) and asked to factor it. Do you: - Spend 10 minutes trying every possible combination? - Or spend 30 seconds checking if it’s even possible?

The Solution: The discriminant! It’s like a “factorability test” that tells us instantly whether we’re wasting our time.

What is a Discriminant? The discriminant \(\Delta = b^2 - 4ac\) is the expression under the square root in the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

How to Calculate \(b^2 - 4ac\):

Step 1: Identify \(a\), \(b\), \(c\) from \(ax^2 + bx + c\)
Step 2: Calculate \(b^2 - 4ac\)
Step 3: Check if result is a perfect square

Perfect Squares and Factoring

Perfect squares are the “clean” numbers of mathematics

Historical Note: Ancient Greeks called these “square numbers” because you can arrange them into perfect square arrays of dots!

Think of perfect squares like this: They’re numbers whose square roots are “nice” whole numbers - no messy decimals!

Perfect squares: \(1 = 1^2\), \(4 = 2^2\), \(9 = 3^2\), \(16 = 4^2\), \(25 = 5^2\), \(36 = 6^2\), etc.

Non-perfect squares: 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21…

Quick Check: To see if a number is a perfect square, find its square root. If the square root is a whole number, it’s a perfect square!

Example: \(\sqrt{25} = 5\) (whole number) → perfect square! But \(\sqrt{21} \approx 4.58...\) (messy decimal) → not perfect!

Fun Fact: The largest perfect square less than 100 is 81. Can you name the next one?

The AC Method Rule

The ultimate time-saver: Know before you try!

The Golden Rule: A quadratic \(ax^2 + bx + c\) can be factored over the integers if and only if the discriminant \(\Delta = b^2 - 4ac\) is a perfect square.

Real-World Application: Imagine you’re an engineer designing a bridge. You need to solve \(2x^2 + 13x + 15 = 0\). Should you factor or use the quadratic formula?

Check: \(\Delta = 13^2 - 4(2)(15) = 169 - 120 = 49 = 7^2\) ✓
Since it’s a perfect square, factoring will work: \((2x + 3)(x + 5)\)
Time saved: 5 minutes of trial and error!

Counter-example: For \(x^2 + 3x - 3\): \(\Delta = 9 + 12 = 21\) (not perfect) → Don’t waste time trying to factor!

Pro Tip: Always check the discriminant first! It’s like checking if a door is locked before trying to open it.

Factoring by Grouping

Group terms with common factors

Factor \(x^3 + 2x^2 - 3x - 6\)

Group: \((x^3 + 2x^2) + (-3x - 6)\)
Factor each: \(x^2(x + 2) - 3(x + 2)\)
Common factor: \((x + 2)(x^2 - 3)\)

Factor \(2x^3 - x^2 - 8x + 4\)

Group: \((2x^3 - x^2) + (-8x + 4)\)
Factor: \(x^2(2x - 1) - 4(2x - 1)\)
Result: \((2x - 1)(x^2 - 4)\)
Even further: \((2x - 1)(x + 2)(x - 2)\)

Sum and Difference of Cubes

These patterns are worth memorizing!

Pattern	Formula
Sum of Cubes	\(a^3 + b^3 = (a+b)(a^2-ab+b^2)\)
Difference of Cubes	\(a^3 - b^3 = (a-b)(a^2+ab+b^2)\)

Examples:

\(x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)\)
\(x^3 - 27 = x^3 - 3^3 = (x - 3)(x^2 + 3x + 9)\)
\(8x^3 + 125 = (2x)^3 + 5^3 = (2x + 5)(4x^2 - 10x + 25)\)

Individual Exercise 01

Go ahead and factor these completely

\(3x^2 + 10x + 8\)
\(x^3 - 64\)
\(2x^3 + 3x^2 - 8x - 12\)
\(4x^2 - 11x - 3\)
\(27x^3 + 8\)

Break - 10 Minutes

Roots and Radicals

Understanding Roots

Roots ask: “What number gives me this when raised to a power?”

\(\sqrt{25} = 5\) because \(5^2 = 25\)
\(\sqrt[3]{8} = 2\) because \(2^3 = 8\)
\(\sqrt[4]{81} = 3\) because \(3^4 = 81\)

. . .

The Sign Rules

Square roots (and even roots): Always positive by convention
- \(\sqrt{9} = 3\) (not \(-3\), even though \((-3)^2 = 9\))
Cube roots (and odd roots): Keep the original sign
- \(\sqrt[3]{-8} = -2\) because \((-2)^3 = -8\)

Properties of Radicals

These properties allow us to simplify

Property	Formula	Example
Product	\(\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}\)	\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
Quotient	\(\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}\)	\(\sqrt{\frac{16}{4}} = \frac{\sqrt{16}}{\sqrt{4}} = 2\)
Power	\(\sqrt[n]{a^m} = a^{m/n}\)	\(\sqrt[3]{x^6} = x^2\)

. . .

Key idea: Look for perfect squares, cubes, etc. that you can “pull out” of the radical!

Simplifying Radicals

Strategy: Extract perfect powers from under the radical

Simplify \(\sqrt{72}\)

Factor: \(72 = 36 \times 2 = 6^2 \times 2\)
Extract: \(\sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2}\)
Result: \(6\sqrt{2}\)

Simplify \(\sqrt{50x^5y^3}\)

Factor: \(50 = 25 \times 2\), \(x^5 = x^4 \cdot x\), \(y^3 = y^2 \cdot y\)
Extract: \(\sqrt{25x^4y^2 \cdot 2xy}\)
Result: \(5x^2y\sqrt{2xy}\)

Operations with Radicals

Can only combine like radicals!

Example: Simplify \(3\sqrt{12} + 2\sqrt{27} - \sqrt{48}\)

Simplify each term:
- \(3\sqrt{12} = 3 \cdot 2\sqrt{3} = 6\sqrt{3}\)
- \(2\sqrt{27} = 2 \cdot 3\sqrt{3} = 6\sqrt{3}\)
- \(\sqrt{48} = 4\sqrt{3}\)
Combine: \(6\sqrt{3} + 6\sqrt{3} - 4\sqrt{3} = 8\sqrt{3}\)

. . .

Always simplify radicals first before combining!

Rationalizing Denominators

What is Rationalizing?

Removing radicals from denominators

Rationalize means to rewrite a fraction
We want no square roots (or other radicals) in the denominator.

Before: \(\frac{1}{\sqrt{2}}\) (radical in denominator)
After: \(\frac{\sqrt{2}}{2}\) (no radical in denominator)

. . .

Why Rationalize?

Easier calculation before calculators, standard form for mathematical expressions, and often simplifies further operations.

Simple Radical Denominators

Basic principle: Multiply by a form of 1 that eliminates the radical

Rationalize \(\frac{3}{\sqrt{5}}\)

Multiply by \(\frac{\sqrt{5}}{\sqrt{5}}\)
\(\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5}\)

Rationalize \(\frac{3}{\sqrt[3]{2}}\)

We could make the denominator a perfect cube
Multiply by \(\frac{\sqrt[3]{4}}{\sqrt[3]{4}}\) (since \(2 \times 4 = 8 = 2^3\))
Result: \(\frac{3\sqrt[3]{4}}{2}\)

Using Conjugates

A conjugate flips the sign between terms

Definition: The conjugate of \(a + b\sqrt{c}\) is \(a - b\sqrt{c}\)

Example: Conjugate of \(\sqrt{3} + 1\) is \(\sqrt{3} - 1\)
Key property: \((a + b)(a - b) = a^2 - b^2\) (difference of squares)
Why it works: The radical terms cancel out when multiplied!

Conjugates

Use the conjugate to eliminate radicals

Example: Rationalize \(\frac{2}{\sqrt{3} + 1}\)

Multiply by conjugate: \(\frac{\sqrt{3} - 1}{\sqrt{3} - 1}\)
Numerator: \(2(\sqrt{3} - 1) = 2\sqrt{3} - 2\)
Denominator: \((\sqrt{3})^2 - 1^2 = 3 - 1 = 2\)
Result: \(\frac{2\sqrt{3} - 2}{2} = \sqrt{3} - 1\)

Pair Exercise 01

Work together on these problems

Simplify: \(\sqrt{72} + \sqrt{50} - \sqrt{98}\)
Rationalize: \(\frac{4}{\sqrt{6} - \sqrt{2}}\)
Simplify: \(\sqrt[3]{54x^7y^5}\)
Simplify: \(\frac{\sqrt{45x^3}}{\sqrt{5x}}\)

Coffee Break - 15 Minutes

Complex Algebraic Manipulation

Combining All Techniques

Use factorization, exponents, and radicals together

Example: Simplify \(\frac{x^2 - 4}{x^2 - x - 6} \cdot \frac{x^2 - 9}{x + 2}\)

Factor everything:
- \(x^2 - 4 = (x + 2)(x - 2)\)
- \(x^2 - x - 6 = (x - 3)(x + 2)\)
- \(x^2 - 9 = (x + 3)(x - 3)\)
Rewrite: \(\frac{(x + 2)(x - 2)}{(x - 3)(x + 2)} \cdot \frac{(x + 3)(x - 3)}{x + 2}\)
Result: \(\frac{(x - 2)(x + 3)}{x + 2}\)

Complex Fractions

Simplify: \(\frac{\frac{x}{3}}{\frac{2}{x}}\)

Remember: Dividing by a fraction means multiply by its reciprocal
\(\frac{\frac{x}{3}}{\frac{2}{x}} = \frac{x}{3} \times \frac{x}{2}\)
Multiply: \(\frac{x \times x}{3 \times 2} = \frac{x^2}{6}\)

Practice

Individual Exercise 02

Apply all techniques together

Factor completely: \(8x^3 - 125\)
Simplify: \(\sqrt{75x^3} - x\sqrt{12x} + 2\sqrt{27x^3}\)
Rationalize: \(\frac{3}{2 - \sqrt{3}}\)
Simplify: \(\frac{x^3 - 8}{x^2 - 4} \div \frac{x^2 + 2x + 4}{x + 2}\)

Wrap-up

Key Takeaways

AC method handles quadratics with \(a \neq 1\)
Grouping wcan work for four-term polynomials
Cube formulas follow specific patterns
Radicals simplify by extracting perfect powers
Rationalization uses conjugates for binomials

For Next Time

Homework: Complete Tasks 01-04

Preview of Session 01-05:

Logarithms and their properties
Binomial theorem and Pascal’s triangle
Advanced algebraic applications

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Practice factorization and radicals - they’re essential for all advanced math!

Questions & Discussion

Open Floor

Your questions and insights are welcome!

Which factorization method is most challenging?
Any confusion about radicals or rationalization?
Real-world applications you’re curious about?

See you next session!

Keep practicing - mastery comes through repetition!