📄 Save or print this lesson:

← Previous Lesson Lesson 2 of 4 Next Lesson →

Lesson 2: Factoring Polynomials

Estimated time: 35-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Factor polynomials using GCF, grouping, and special patterns
Apply sum and difference of cubes formulas
Use the Rational Root Theorem to find potential rational zeros
Perform synthetic division to test potential zeros
Factor polynomials completely using multiple techniques
Find all zeros of polynomial functions algebraically

Review: Basic Factoring Techniques

Before we tackle advanced factoring methods, let's review the fundamental techniques you should already know.

        Basic Factoring Methods
        Greatest Common Factor (GCF): Factor out the largest common factor
Difference of Squares: a2 - b2 = (a + b)(a - b)
Perfect Square Trinomials: a2 ± 2ab + b2 = (a ± b)2
Trinomials: ax2 + bx + c (factor by trial-and-error or AC method)

      

Example 1: Factoring with GCF

Factor completely: 6x⁴ + 9x³ - 15x²

Step 1: Find the GCF

GCF of 6, 9, and 15 is 3

GCF of x⁴, x³, and x² is x²

Overall GCF: 3x²

Step 2: Factor out the GCF

6x⁴ + 9x³ - 15x² = 3x²(2x² + 3x - 5)

Step 3: Factor the remaining trinomial

2x² + 3x - 5 = (2x + 5)(x - 1)

Final Answer: 3x²(2x + 5)(x - 1)

Example 2: Difference of Squares

Factor completely: x⁴ - 16

Step 1: Recognize as difference of squares

x⁴ - 16 = (x²)² - 4²

Step 2: Apply the formula a² - b² = (a + b)(a - b)

x⁴ - 16 = (x² + 4)(x² - 4)

Step 3: Factor further (x² - 4 is also a difference of squares!)

x² - 4 = (x + 2)(x - 2)

Final Answer: (x² + 4)(x + 2)(x - 2)

Note: x² + 4 cannot be factored over the real numbers.

Example 3: Perfect Square Trinomial

Factor: 4x² + 12x + 9

Step 1: Check if it's a perfect square trinomial

First term: 4x² = (2x)²

Last term: 9 = 3²

Middle term: 12x = 2(2x)(3) Checks!

Step 2: Apply the formula a² + 2ab + b² = (a + b)²

Answer: (2x + 3)²

Factoring by Grouping

Factoring by grouping is a powerful technique for polynomials with four or more terms. The idea is to group terms strategically and factor out common factors from each group.

        Steps for Factoring by Grouping
        Group terms in pairs (usually the first two and last two)
Factor out the GCF from each group
If a common binomial factor appears, factor it out
If no common binomial appears, try regrouping differently

      

Example 4: Factoring by Grouping

Factor: x³ + 3x² + 2x + 6

Step 1: Group the first two and last two terms

(x³ + 3x²) + (2x + 6)

Step 2: Factor out the GCF from each group

x²(x + 3) + 2(x + 3)

Step 3: Factor out the common binomial (x + 3)

Answer: (x + 3)(x² + 2)

Example 5: Grouping with Rearrangement

Factor: 6x³ - 9x² + 4x - 6

Step 1: Group and factor

(6x³ - 9x²) + (4x - 6)

3x²(2x - 3) + 2(2x - 3)

Step 2: Factor out (2x - 3)

Answer: (2x - 3)(3x² + 2)

Example 6: Grouping Four-Term Polynomial

Factor: 2x³ - x² - 6x + 3

Step 1: Group terms

(2x³ - x²) + (-6x + 3)

Step 2: Factor each group

x²(2x - 1) - 3(2x - 1)

Step 3: Factor out (2x - 1)

Answer: (2x - 1)(x² - 3)

Sum and Difference of Cubes

Two special formulas allow us to factor the sum or difference of perfect cubes.

Cube Factoring Formulas

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

Memory Tip: "Same, Opposite, Always Positive"

First binomial: SAME sign as the original
Second trinomial: OPPOSITE sign in the middle
Last term in trinomial: ALWAYS POSITIVE

Example 7: Sum of Cubes

Factor: x³ + 27

Step 1: Recognize as sum of cubes

x³ + 27 = x³ + 3³

Here, a = x and b = 3

Step 2: Apply the sum of cubes formula

a³ + b³ = (a + b)(a² - ab + b²)

Step 3: Substitute

(x + 3)(x² - 3x + 9)

Answer: (x + 3)(x² - 3x + 9)

Example 8: Difference of Cubes

Factor: 8x³ - 125

Step 1: Recognize as difference of cubes

8x³ - 125 = (2x)³ - 5³

Here, a = 2x and b = 5

Step 2: Apply the difference of cubes formula

a³ - b³ = (a - b)(a² + ab + b²)

Step 3: Substitute

a = 2x, so a² = 4x²

ab = (2x)(5) = 10x

b² = 25

Answer: (2x - 5)(4x² + 10x + 25)

Example 9: Sum of Cubes with Variables

Factor: 27a³ + 64b³

Step 1: Identify the cubes

27a³ = (3a)³ and 64b³ = (4b)³

Step 2: Apply formula with a = 3a, b = 4b

(3a + 4b)[(3a)² - (3a)(4b) + (4b)²]

Step 3: Simplify

Answer: (3a + 4b)(9a² - 12ab + 16b²)

The Rational Root Theorem

The Rational Root Theorem is a powerful tool for finding potential rational zeros of polynomial functions. It doesn't guarantee which values ARE zeros, but it tells us which values MIGHT be zeros.

Rational Root Theorem: If a polynomial f(x) = a_nxⁿ + ... + a₁x + a₀ has integer coefficients and p/q is a rational zero (in lowest terms), then:

p is a factor of the constant term a₀
q is a factor of the leading coefficient a_n

        Steps for Using the Rational Root Theorem
        List all factors of the constant term (p values)
List all factors of the leading coefficient (q values)
Form all possible fractions p/q (these are potential zeros)
Test each potential zero using substitution or synthetic division
Once you find a zero, factor and continue

      

Example 10: Finding Potential Rational Zeros

List all potential rational zeros of f(x) = 2x³ - 5x² - 4x + 3

Step 1: Identify the constant term and leading coefficient

Constant term: a₀ = 3

Leading coefficient: a_n = 2

Step 2: List factors of the constant term (p)

Factors of 3: ±1, ±3

Step 3: List factors of the leading coefficient (q)

Factors of 2: ±1, ±2

Step 4: Form all possible p/q ratios

Potential rational zeros: ±1, ±3, ±1/2, ±3/2

That's 8 values to test: {-3, -3/2, -1, -1/2, 1/2, 1, 3/2, 3}

Synthetic Division

Synthetic division is a shortcut method for dividing a polynomial by a linear factor (x - c). It's especially useful for testing potential zeros found using the Rational Root Theorem.

        Key Points about Synthetic Division
        Only works for divisors of the form (x - c)
If the remainder is 0, then c is a zero of the polynomial
The quotient gives you the reduced polynomial
Much faster than long division!

      

Example 11: Synthetic Division

Use synthetic division to divide f(x) = 2x³ - 5x² - 4x + 3 by (x - 3)

Step 1: Set up synthetic division

Use c = 3 (from x - 3 = 0)

Write coefficients: 2, -5, -4, 3

    3 |  2   -5   -4    3
      |      6    3   -3
      |___________________
         2    1   -1    0  ← Remainder

Step 2: Interpret the result

Remainder = 0, so x = 3 IS a zero!

Quotient: 2x² + x - 1

Step 3: Write the factorization

f(x) = (x - 3)(2x² + x - 1)

Step 4: Factor the quadratic

2x² + x - 1 = (2x - 1)(x + 1)

Complete factorization: f(x) = (x - 3)(2x - 1)(x + 1)

All zeros: x = 3, x = 1/2, x = -1

Example 12: Complete Factorization Using Rational Root Theorem

Factor completely: f(x) = x³ + 2x² - 5x - 6

Step 1: Find potential rational zeros

Factors of -6: ±1, ±2, ±3, ±6

Factors of 1: ±1

Potential zeros: ±1, ±2, ±3, ±6

Step 2: Test x = 1 using synthetic division

    1 |  1    2   -5   -6
      |       1    3   -2
      |___________________
         1    3   -2   -8  ← Not zero

x = 1 is NOT a zero (remainder ≠ 0)

Step 3: Test x = -1

   -1 |  1    2   -5   -6
      |      -1   -1    6
      |___________________
         1    1   -6    0  ← Zero!

x = -1 IS a zero!

Quotient: x² + x - 6

Step 4: Factor the quotient

x² + x - 6 = (x + 3)(x - 2)

Complete factorization: f(x) = (x + 1)(x + 3)(x - 2)

All zeros: x = -1, x = -3, x = 2

Comprehensive Factoring Strategy

        General Factoring Strategy
        Always factor out GCF first
Check for special patterns (difference of squares, perfect square trinomials, sum/difference of cubes)
For trinomials, try factoring as (ax + b)(cx + d)
For four terms, try factoring by grouping
For higher degree polynomials, use Rational Root Theorem and synthetic division
Check your answer by multiplying back

      

Example 13: Multi-Step Factoring

Factor completely: 2x⁴ - 32

Step 1: Factor out GCF

2x⁴ - 32 = 2(x⁴ - 16)

Step 2: Recognize difference of squares

x⁴ - 16 = (x²)² - 4² = (x² + 4)(x² - 4)

Step 3: Factor x² - 4 further (difference of squares again)

x² - 4 = (x + 2)(x - 2)

Final answer: 2(x² + 4)(x + 2)(x - 2)

Example 14: Combining Techniques

Factor completely: 3x³ - 24

Step 1: Factor out GCF

3x³ - 24 = 3(x³ - 8)

Step 2: Recognize difference of cubes

x³ - 8 = x³ - 2³

Step 3: Apply difference of cubes formula

x³ - 2³ = (x - 2)(x² + 2x + 4)

Final answer: 3(x - 2)(x² + 2x + 4)

Example 15: Higher Degree Polynomial

Find all zeros of f(x) = x⁴ - 5x³ + 5x² + 5x - 6

Step 1: Use Rational Root Theorem

Potential zeros: ±1, ±2, ±3, ±6

Step 2: Test x = 1

    1 |  1   -5    5    5   -6
      |       1   -4    1    6
      |_________________________
         1   -4    1    6    0  ← Zero!

x = 1 is a zero! Quotient: x³ - 4x² + x + 6

Step 3: Test x = 2 on the quotient

    2 |  1   -4    1    6
      |       2   -4   -6
      |___________________
         1   -2   -3    0  ← Zero!

x = 2 is a zero! Quotient: x² - 2x - 3

Step 4: Factor the quadratic

x² - 2x - 3 = (x - 3)(x + 1)

Complete factorization: f(x) = (x - 1)(x - 2)(x - 3)(x + 1)

All zeros: x = -1, 1, 2, 3

Check Your Understanding

1. Factor completely: x³ + 64

Solution:

This is a sum of cubes: x³ + 4³

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

Answer: (x + 4)(x² - 4x + 16)

2. Factor by grouping: 2x³ + 6x² - 5x - 15

Solution:

Group: (2x³ + 6x²) + (-5x - 15)

Factor each group: 2x²(x + 3) - 5(x + 3)

Factor out (x + 3):

Answer: (x + 3)(2x² - 5)

3. List all potential rational zeros of f(x) = 3x⁴ - 2x² + 8

Solution:

Constant term: 8, factors: ±1, ±2, ±4, ±8

Leading coefficient: 3, factors: ±1, ±3

Potential rational zeros:

±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3

4. Factor: 27x³ - 8

Solution:

This is a difference of cubes: (3x)³ - 2³

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

With a = 3x, b = 2:

Answer: (3x - 2)(9x² + 6x + 4)

5. Use synthetic division to determine if x = 2 is a zero of f(x) = x³ - 7x + 6

Solution:

Note: x³ - 7x + 6 = x³ + 0x² - 7x + 6

    2 |  1    0   -7    6
      |       2    4   -6
      |___________________
         1    2   -3    0  ← Remainder = 0

Yes! x = 2 is a zero because the remainder is 0.

Quotient: x² + 2x - 3 = (x + 3)(x - 1)

Complete factorization: (x - 2)(x + 3)(x - 1)

6. Factor completely: 4x³ - 108

Solution:

Step 1: Factor out GCF of 4

4(x³ - 27)

Step 2: Recognize difference of cubes x³ - 3³

4(x - 3)(x² + 3x + 9)

Answer: 4(x - 3)(x² + 3x + 9)

7. Find all zeros: f(x) = x³ - 6x² + 11x - 6

Solution:

Potential zeros: ±1, ±2, ±3, ±6

Testing x = 1: Remainder = 0 (it's a zero!)

Quotient after dividing by (x - 1): x² - 5x + 6

Factor: (x - 2)(x - 3)

All zeros: x = 1, 2, 3

Key Takeaways

Always look for GCF first before attempting other factoring methods
Recognize special patterns: difference of squares, perfect square trinomials, sum/difference of cubes
Factoring by grouping works well for polynomials with four or more terms
Sum of cubes: a³ + b³ = (a + b)(a² - ab + b²)
Difference of cubes: a³ - b³ = (a - b)(a² + ab + b²)
Rational Root Theorem gives potential rational zeros: p/q where p divides constant, q divides leading coefficient
Synthetic division quickly tests potential zeros and gives the quotient polynomial
A remainder of 0 in synthetic division confirms that the tested value is a zero

← Previous: Polynomial Functions Next: Rational Functions →