Factoring quadratic expressions

Why factor quadratics?

Factoring turns ax² + bx + c into a product, which reveals the zeros of the function and helps solve equations.

Step 1: Common factor

Always take out the common factor first.

3x² + 12x = 3x(x + 4)

2x² − 8x − 10 = 2(x² − 4x − 5)

Step 2a: Simple trinomials (a = 1)

Simple trinomial: x² + bx + c. Find two numbers that multiply to c and add to b.

Step 2b: Complex trinomials (a ≠ 1)

Complex trinomial: ax² + bx + c. Use decomposition:

1. Find two numbers that multiply to ac and add to b.
2. Split the middle term using those numbers.
3. Group and factor.

Example: 6x² + 11x + 4

• ac = 24; numbers: 3 and 8
• 6x² + 3x + 8x + 4
• 3x(2x + 1) + 4(2x + 1)
• (3x + 4)(2x + 1)

Example: 4x² − 4x − 15

• ac = −60; numbers: −10 and 6
• 4x² − 10x + 6x − 15
• 2x(2x − 5) + 3(2x − 5)
• (2x + 3)(2x − 5)

Special patterns

• Difference of squares: x² − 25 = (x + 5)(x − 5); 9x² − 4 = (3x + 2)(3x − 2)
• Perfect square: x² + 10x + 25 = (x + 5)²; 4x² − 12x + 9 = (2x − 3)²

Factors give zeros

If y = (x − 3)(x + 2), the zeros are x = 3 and x = −2: the parabola crosses the x-axis there.

Not everything factors

x² + x + 1 has no integer factors. Use the quadratic formula instead. (If b² − 4ac isn’t a perfect square, it won’t factor over the integers.)

Worked examples

Example 1: x² + 9x + 14 → (x + 2)(x + 7).

Example 2: 2x² + 7x − 15 → ac = −30: 10, −3 → 2x² + 10x − 3x − 15 → (2x − 3)(x + 5).

Example 3: 5x² − 45 → 5(x² − 9) → 5(x + 3)(x − 3).

Common mistakes

• Skipping the common factor.
• Sign errors in the two numbers.
• Not checking by expanding.

Try it yourself

1. x² − 7x + 12
2. x² + 3x − 28
3. 3x² + 10x + 8
4. 16x² − 49
5. 2x² − 12x + 18

Answers: 1) (x − 3)(x − 4) 2) (x + 7)(x − 4) 3) (3x + 4)(x + 2) 4) (4x + 7)(4x − 7) 5) 2(x − 3)²

Quick recap

• Take out the common factor first.
• Simple trinomial: multiply to c, add to b. Complex trinomial: decomposition using ac.
• Factors reveal the zeros.