How do you factor a trinomial when the leading coefficient is not 1?

Use the AC method: multiply the leading coefficient $a$ by the constant $c$, then find two numbers that multiply to $ac$ and add to $b$. Rewrite the middle term using those two numbers and factor by grouping. For example, to factor $2x^2 + 7x + 3$, compute $ac = 6$. The numbers 1 and 6 multiply to 6 and add to 7, so rewrite as $2x^2 + x + 6x + 3$, then group: $x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$.

Factoring — Definition, Formula & Examples

Factoring is the process of rewriting a mathematical expression as a product of two or more simpler expressions. For example, you can factor

x^2 + 5x + 6

into

(x + 2)(x + 3)

To factor a polynomial $p(x)$ over the integers is to express it as a product $p(x) = a \cdot f_1(x) \cdot f_2(x) \cdots f_k(x)$ , where $a$ is an integer and each $f_i(x)$ is a polynomial of degree at least 1 with integer coefficients that cannot be factored further (i.e., each $f_i$ is irreducible over $\mathbb{Z}$ ).

Key Formula

a^2 - b^2 = (a + b)(a - b)

Where:

$a$ = Any expression (variable, number, or combination)
$b$ = Any expression (variable, number, or combination)

How It Works

Factoring reverses the multiplication of polynomials. You start with a single expression and break it into pieces whose product equals the original. The most common techniques are: pulling out a greatest common factor (GCF), factoring trinomials of the form

ax^2 + bx + c

, using the difference of squares identity, and grouping terms. Once a polynomial is fully factored, you can set each factor equal to zero to solve equations — this is the zero-product property. Choosing the right method depends on the number of terms, the degree, and any recognizable patterns in the expression.

Worked Example

Problem: Factor the trinomial

x^2 + 7x + 12

completely.

Step 1: Identify two numbers that multiply to 12 (the constant term) and add to 7 (the coefficient of

x

p \cdot q = 12 \quad \text{and} \quad p + q = 7

Step 2: Test factor pairs of 12: (1,12), (2,6), (3,4). The pair 3 and 4 satisfies both conditions because

3 \times 4 = 12

and

3 + 4 = 7

p = 3, \quad q = 4

Step 3: Write the factored form using these values.

x^2 + 7x + 12 = (x + 3)(x + 4)

Answer:

(x + 3)(x + 4)

Another Example

Problem: Factor

6x^2 - 54

completely.

Step 1: Factor out the greatest common factor. Both terms share a factor of 6.

6x^2 - 54 = 6(x^2 - 9)

Step 2: Recognize that

x^2 - 9

is a difference of squares:

x^2 - 3^2

x^2 - 9 = (x + 3)(x - 3)

Step 3: Combine everything for the fully factored result.

6x^2 - 54 = 6(x + 3)(x - 3)

Answer:

6(x + 3)(x - 3)

Why It Matters

Factoring is essential in Algebra 1, Algebra 2, and Precalculus for solving quadratic equations, simplifying rational expressions, and finding polynomial roots. Engineers and physicists use it to solve design equations, while data scientists factor expressions when simplifying statistical models. Mastering factoring also builds the foundation for partial fraction decomposition in calculus.

Common Mistakes

Mistake: Forgetting to factor out the GCF first, which makes the remaining expression harder — or impossible — to factor by other methods.

Correction: Always check for a greatest common factor before trying any other technique. For instance,

3x^2 + 12x + 12

should first become

3(x^2 + 4x + 4)

, which then factors as

3(x + 2)^2

Mistake: Incorrectly applying the difference of squares pattern to a sum of squares, writing

x^2 + 9 = (x + 3)(x - 3)

Correction: The identity

a^2 - b^2 = (a+b)(a-b)

only works for a difference (minus sign). The sum

a^2 + b^2

does not factor over the real numbers.