# Factorization of Quadratic Polynomials - Methods

TL;DR

Factorizing a quadratic polynomial \( ax^2 + bx + c \) means rewriting it as a product of two linear factors, usually by splitting the middle term so the two pieces multiply to \( ac \) and add to \( b \). This article covers four methods, six worked examples, and the sign errors that derail most attempts.

## What Does It Mean to Factorize a Quadratic Polynomial?

**Factorizing a quadratic polynomial means expressing \( ax^2 + bx + c \) as a product of two linear factors of the form \( (px+q)(rx+s) \).** When the quadratic equals zero, those factors provide the roots directly.

If \( ax^2 + bx + c = (x-h)(x-k) \), then \( h \) and \( k \) are the roots of the quadratic. Not every quadratic factors neatly over the rational numbers. When it does not, the [quadratic formula](/content/math/algebra/quadratic-equations/index.html) still finds the roots.

## How Do You Factorize a Quadratic Polynomial?

There are four standard methods:

- **Splitting the middle term**: Find two numbers that multiply to \( ac \) and add to \( b \).
- **Algebraic identities**: When the quadratic is a perfect square trinomial or a difference of squares.
- **Grouping**: Useful when a common factor appears in pairs.
- **The quadratic formula**: When no clean factors exist.

### Splitting the middle term, step by step

For \( ax^2 + bx + c \), find two numbers whose product is \( a \times c \) and whose sum is \( b \).

1. Compute \( ac \).
2. Find two numbers multiplying to \( ac \) and adding to \( b \).
3. Split the middle term.
4. Group and factor each pair.
5. Factor out the common bracket.

### Examples of Factorization of Quadratic Polynomials

**Example 1: Factorize \( x^2 + 8x + 15 \)**

1. \( ac = 1 \times 15 = 15 \)
2. Two numbers are 3 and 5.
3. Split: \( x^2 + 3x + 5x + 15 \)
4. Group: \( x(x+3) + 5(x+3) \)
5. Factor: \( (x+3)(x+5) \)

**Final answer:** \( (x+3)(x+5) \)

**Example 2: Factorize \( x^2 - 2x - 15 \)**

1. Product is \( -15 \) and sum is \( -2 \).
2. Correct numbers are -5 and 3.
3. Split: \( x^2 - 5x + 3x - 15 \)
4. Group: \( x(x - 5) + 3(x - 5) \)
5. Factor: \( (x - 5)(x + 3) \)

**Final answer:** \( (x-5)(x+3) \)

**Example 3: Factorize \( 6x^2 - 5x - 6 \)**

1. \( ac = 6 \times (-6) = -36 \)
2. Numbers are -9 and +4.
3. Split: \( 6x^2 - 9x + 4x - 6 \)
4. Group: \( 3x(2x - 3) + 2(2x - 3) \)
5. Factor: \( (2x - 3)(3x + 2) \)

**Final answer:** \( (2x-3)(3x+2) \)

### Common Mistakes in Factorization

1. **Getting the signs of the two numbers backwards.** Check both conditions: product equals \( ac \) and sum equals \( b \).
2. **Forgetting the leading coefficient when \( a \neq 1\).** Always use \( a \times c \) as the product target.
3. **Assuming every quadratic factors over the rationals.** Check the [discriminant](/content/math/algebra/discriminant/index.html).

## Practice Questions on Factorization of Quadratic Polynomials

1. Factorize \( x^2 + 7x + 12 \).
2. Factorize \( x^2 - x - 12 \).
3. Factorize \( 2x^2 + 7x + 3 \).
4. Factorize \( x^2 - 14x + 49 \).
5. Factorize \( 25x^2 - 9 \).
6. Factorize \( 3x^2 + 5x - 4 \).

**Answers**

1. \( (x+3)(x+4) \)
2. \( (x-4)(x+3) \)
3. \( (2x+1)(x+3) \)
4. \( (x-7)^2 \)
5. \( (5x+3)(5x-3) \)
6. No rational factors; roots are \( x = \frac{-1 \pm \sqrt{13}}{3} \).

## Key Takeaways

- Factorization rewrites \( ax^2 + bx + c \) as a product of two linear factors.
- Splitting the middle term is the main method.
- Perfect square trinomials and differences of squares factor instantly via identities.
- If no rational factors exist, use the quadratic formula while checking the discriminant first.