Factoring Polynomials

Advanced factoring techniques

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: 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)

3Problem 3easy

❓ 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: $(x + 3)(x^2 + 2)$

4Problem 4medium

❓ 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: $(x + 3)(x^2 - 3x + 9)$

5Problem 5medium

❓ 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)

6Problem 6medium

❓ 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)

7Problem 7hard

❓ 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: $(x + 3)(x - 3)(x + 2)(x - 2)$

8Problem 8hard

❓ 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)

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