Session 01-04 - Advanced Factorization & Radicals

Section 01: Mathematical Foundations & Algebra

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz

Quick Review from Last Session

Complete individually, then we review as group

  1. Simplify: \(\frac{(2x^3)^2 \cdot x^{-5}}{4x^2}\)

  2. Factor: \(9x^2 - 25\)

  3. Solve: \(|2x - 4| > 6\)

  4. Express in scientific notation: \(0.0000234 \times 10^3\)

  5. 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?

How do we know if factoring is even possible?

The discriminant \(\Delta = b^2 - 4ac\) tells us instantly:

  • Step 1: Identify \(a\), \(b\), \(c\) from \(ax^2 + bx + c\)
  • Step 2: Calculate \(b^2 - 4ac\)
  • Step 3: If the result is a perfect square, factoring will work!

A perfect square is a number whose square root is a whole number: \(1, 4, 9, 16, 25, 36, 49, 64, 81, ...\)

Applying the Discriminant

The rule: \(ax^2 + bx + c\) factors over integers iff \(\Delta = b^2 - 4ac\) is a perfect square.

Example: Factor \(2x^2 + 13x + 15\)

  • Check: \(\Delta = 13^2 - 4(2)(15) = 169 - 120 = 49 = 7^2\) ✓
  • Since it’s a perfect square, factoring will work: \((2x + 3)(x + 5)\)

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

Always check the discriminant first — it saves you from trial and error!

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

Work individually for 8 minutes

  1. \(3x^2 + 10x + 8\)

  2. \(x^3 - 64\)

  3. \(2x^3 + 3x^2 - 8x - 12\)

  4. \(4x^2 - 11x - 3\)

  5. \(27x^3 + 8\)

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 in pairs for 8 minutes

  1. Simplify: \(\sqrt{72} + \sqrt{50} - \sqrt{98}\)

  2. Rationalize: \(\frac{4}{\sqrt{6} - \sqrt{2}}\)

  3. Simplify: \(\sqrt[3]{54x^7y^5}\)

  4. 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

Work individually for 10 minutes

  1. Factor completely: \(8x^3 - 125\)

  2. Simplify: \(\sqrt{75x^3} - x\sqrt{12x} + 2\sqrt{27x^3}\)

  3. Rationalize: \(\frac{3}{2 - \sqrt{3}}\)

  4. 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 can 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

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!