Factoring Polynomials

Advanced factoring techniques

Written and reviewed by the Study Mondo Education TeamLast updated February 16, 2026

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Factoring Polynomials

Factor by Grouping

Group terms and factor out common factors.

Example: $x^3 + 2x^2 + 3x + 6$ $= x^2(x + 2) + 3(x + 2)$ $= (x + 2)(x^2 + 3)$

Sum and Difference of Cubes

Sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Example: $x^3 - 8$ $= x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)$

Factoring by Substitution

Sometimes substituting can simplify factoring.

Example: $x^4 - 5x^2 + 4$

Let $u = x^2$ : $u^2 - 5u + 4 = (u - 4)(u - 1)$

Substitute back: $= (x^2 - 4)(x^2 - 1)$ $= (x + 2)(x - 2)(x + 1)(x - 1)$

📚 Practice Problems

1Problem 1easy

❓ Question:

Factor completely: 3x² + 12x

💡 Show Solution

Step 1: Find the greatest common factor (GCF): GCF of 3x² and 12x is 3x

Step 2: Factor out the GCF: 3x² + 12x = 3x(x + 4)

Step 3: Verify by expanding: 3x(x + 4) = 3x² + 12x ✓

Answer: 3x(x + 4)

2Problem 2easy

❓ Question:

Factor by grouping: $x^3 + 3x^2 + 2x + 6$

💡 Show Solution

Group the first two and last two terms: $(x^3 + 3x^2) + (2x + 6)$

Factor out common factors: $x^2(x + 3) + 2(x + 3)$

Factor out $(x + 3)$ : $(x + 3)(x^2 + 2)$

Answer: $(x + 3)(x^2 + 2)$

3Problem 3easy

❓ Question:

Factor: x² - 9

💡 Show Solution

Step 1: Recognize as difference of squares: x² - 9 = x² - 3²

Step 2: Apply the formula a² - b² = (a + b)(a - b): x² - 3² = (x + 3)(x - 3)

Step 3: Verify: (x + 3)(x - 3) = x² - 3x + 3x - 9 = x² - 9 ✓

Answer: (x + 3)(x - 3)

4Problem 4easy

❓ Question:

Factor by grouping: $x^3 + 3x^2 + 2x + 6$

💡 Show Solution

Group the first two and last two terms: $(x^3 + 3x^2) + (2x + 6)$

Factor out common factors: $x^2(x + 3) + 2(x + 3)$

Factor out $(x + 3)$ : $(x + 3)(x^2 + 2)$

Answer: $(x + 3)(x^2 + 2)$

5Problem 5medium

❓ Question:

Factor: $x^3 + 27$

💡 Show Solution

This is a sum of cubes: $x^3 + 3^3$

Use the formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Here $a = x$ and $b = 3$ : $x^3 + 27 = (x + 3)(x^2 - 3x + 9)$

Answer: $(x + 3)(x^2 - 3x + 9)$

6Problem 6medium

❓ Question:

Factor: $x^3 + 27$

💡 Show Solution

This is a sum of cubes: $x^3 + 3^3$

Use the formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Here $a = x$ and $b = 3$ : $x^3 + 27 = (x + 3)(x^2 - 3x + 9)$

Answer: $(x + 3)(x^2 - 3x + 9)$

7Problem 7medium

❓ Question:

Factor: x² + 7x + 12

💡 Show Solution

Step 1: Find two numbers that multiply to 12 and add to 7: Factors of 12: 1×12, 2×6, 3×4 Sum check: 1+12=13, 2+6=8, 3+4=7 ✓

Step 2: Write the factored form: x² + 7x + 12 = (x + 3)(x + 4)

Step 3: Verify by expanding: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓

Answer: (x + 3)(x + 4)

8Problem 8hard

❓ Question:

Factor completely: $x^4 - 13x^2 + 36$

💡 Show Solution

This is a quadratic in form. Let $u = x^2$ : $u^2 - 13u + 36$

Factor: $(u - 9)(u - 4)$

Substitute back $u = x^2$ : $(x^2 - 9)(x^2 - 4)$

Both are difference of squares: $= (x + 3)(x - 3)(x + 2)(x - 2)$

Answer: $(x + 3)(x - 3)(x + 2)(x - 2)$

9Problem 9hard

❓ Question:

Factor completely: $x^4 - 13x^2 + 36$

💡 Show Solution

This is a quadratic in form. Let $u = x^2$ : $u^2 - 13u + 36$

Factor: $(u - 9)(u - 4)$

Substitute back $u = x^2$ : $(x^2 - 9)(x^2 - 4)$

Both are difference of squares: $= (x + 3)(x - 3)(x + 2)(x - 2)$

Answer: $(x + 3)(x - 3)(x + 2)(x - 2)$

10Problem 10medium

❓ Question:

Factor completely: 2x² + 11x + 12

💡 Show Solution

Step 1: Use AC method (a=2, c=12): Multiply: 2 × 12 = 24 Find two numbers that multiply to 24 and add to 11: 3 and 8 (3 × 8 = 24, 3 + 8 = 11)

Step 2: Rewrite the middle term: 2x² + 3x + 8x + 12

Step 3: Factor by grouping: (2x² + 3x) + (8x + 12) x(2x + 3) + 4(2x + 3)

Step 4: Factor out common binomial: (2x + 3)(x + 4)

Step 5: Verify: (2x + 3)(x + 4) = 2x² + 8x + 3x + 12 = 2x² + 11x + 12 ✓

Answer: (2x + 3)(x + 4)

11Problem 11hard

❓ Question:

Factor completely: x⁴ - 16

💡 Show Solution

Step 1: Recognize as difference of squares: x⁴ - 16 = (x²)² - 4²

Step 2: Apply difference of squares formula: (x²)² - 4² = (x² + 4)(x² - 4)

Step 3: Check if x² - 4 can be factored further: x² - 4 = x² - 2² = (x + 2)(x - 2)

Step 4: Check if x² + 4 can be factored: x² + 4 cannot be factored over real numbers (sum of squares)

Step 5: Write complete factorization: x⁴ - 16 = (x² + 4)(x + 2)(x - 2)

Step 6: Verify by multiplying: (x² + 4)(x² - 4) = x⁴ - 4x² + 4x² - 16 = x⁴ - 16 ✓

Answer: (x² + 4)(x + 2)(x - 2)

Explain using:

⚠️ Common Mistakes: Factoring Polynomials

Avoid these 3 frequent errors

🌍 Real-World Applications: Factoring Polynomials

See how this math is used in the real world

📝 Worked Example: Solving a Quadratic by Factoring

Problem:

Solve $x^2 - 5x + 6 = 0$ .

2Factor the quadratic

3Set each factor equal to zero

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Polynomial Theorems and Roots

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❓ Frequently Asked Questions

What is Factoring Polynomials?▾

Advanced factoring techniques

How can I study Factoring Polynomials effectively?▾

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 11 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Is this Factoring Polynomials study guide free?▾

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What course covers Factoring Polynomials?▾

Factoring Polynomials is part of the Algebra 2 course on Study Mondo, specifically in the Polynomial Functions section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Factoring Polynomials?▾

Yes, this page includes 11 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Factoring Polynomials

Try the Interactive Version!

Factoring Polynomials

Factor by Grouping

Sum and Difference of Cubes

Factoring by Substitution

📚 Practice Problems

1Problem 1easy

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2Problem 2easy

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3Problem 3easy

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4Problem 4easy

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5Problem 5medium

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6Problem 6medium

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7Problem 7medium

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8Problem 8hard

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9Problem 9hard

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10Problem 10medium

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11Problem 11hard

❓ Question:

⚠️ Common Mistakes: Factoring Polynomials

🌍 Real-World Applications: Factoring Polynomials

📝 Worked Example: Solving a Quadratic by Factoring

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Polynomial Functions

❓ Frequently Asked Questions