Factoring Polynomials: Strategies, Examples, and How to Spot the Right Method

Why factor polynomials?

Factoring rewrites a polynomial as a product of simpler polynomials. The process is the reverse of expanding (FOIL or distribution). Factoring matters because once a polynomial is in factored form, you can:

• Solve polynomial equations using the zero-product property: if $P(x) Q(x) = 0$, then $P(x) = 0$ or $Q(x) = 0$.
• Identify the $x$-intercepts of the corresponding graph.
• Simplify rational expressions by canceling common factors.

This article walks through the five techniques you should reach for, in roughly the order you should try them.

Step 0: Always factor out the GCF first

Before anything else, look for a greatest common factor (GCF) shared by every term — a number, a power of $x$, or both. Pulling it out makes the remaining polynomial smaller and often easier to factor further.

Example 1. Factor $6x^3 + 9x^2$.

Both terms share $3x^2$:

Example 2. Factor $4x^4 - 12x^3 + 8x^2$.

The GCF is $4x^2$:

After factoring out $4x^2$, the remaining trinomial $x^2 - 3x + 2$ factors further (Method 1 below).

Method 1: Factoring trinomials $x^2 + bx + c$

For a monic trinomial (leading coefficient $1$), look for two numbers that multiply to $c$ and add to $b$. Those numbers become the constants of the two factors.

Example 3. Factor $x^2 + 7x + 12$.

Find two numbers that multiply to $12$ and add to $7$. Those are $3$ and $4$:

Example 4. Factor $x^2 - 5x + 6$.

We need numbers that multiply to $+6$ and add to $-5$. Both are negative: $-2$ and $-3$.

If $c$ is negative, the two factors have opposite signs.

Method 2: Factoring trinomials $ax^2 + bx + c$ (with $a \ne 1$)

When the leading coefficient is not $1$, the most reliable approach is the AC method:

1. Multiply $a \cdot c$.
2. Find two numbers that multiply to $a \cdot c$ and add to $b$.
3. Split the middle term into those two pieces.
4. Factor by grouping.

Example 5. Factor $2x^2 + 7x + 3$.

1. $a \cdot c = 2 \cdot 3 = 6$.
2. Find numbers that multiply to $6$ and add to $7$: those are $6$ and $1$.
3. Split: $2x^2 + 6x + x + 3$.
4. Group: $(2x^2 + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$.

You can sanity-check by FOILing: $(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$. ✓

Method 3: Difference of squares

Whenever you see $a^2 - b^2$, factor on sight using

This pattern only works for a difference; $a^2 + b^2$ does not factor over the real numbers.

Example 6. Factor $x^2 - 25$.

This is $x^2 - 5^2$, so $x^2 - 25 = (x + 5)(x - 5)$.

Example 7. Factor $9x^2 - 16$.

Recognize this as $(3x)^2 - 4^2$:

Method 4: Perfect-square trinomials

If a trinomial fits the pattern

you can factor it as a single square. The middle term is the giveaway: it must be exactly $2$ times the product of the square roots of the first and last terms.

Example 8. Factor $x^2 + 6x + 9$.

The first and last terms are perfect squares ($x^2$ and $3^2$), and the middle term equals $2 \cdot x \cdot 3 = 6x$. ✓ So $x^2 + 6x + 9 = (x + 3)^2$.

Example 9. Factor $4x^2 - 20x + 25$.

First term is $(2x)^2$, last term is $5^2$, middle is $-2 \cdot (2x) \cdot 5 = -20x$. ✓ So $4x^2 - 20x + 25 = (2x - 5)^2$.

Method 5: Factor by grouping (four-term polynomials)

When you have four terms, try grouping them in pairs and factoring each pair.

Example 10. Factor $x^3 + 2x^2 + 3x + 6$.

Group: $(x^3 + 2x^2) + (3x + 6) = x^2(x + 2) + 3(x + 2)$.

Both pieces share $(x + 2)$, so factor it out:

If the two binomials don't match after the first grouping, try rearranging the terms before pairing.

A quick decision flow

When you see a polynomial, work through this short list:

1. Always factor out the GCF first. Do not skip this — it makes everything simpler.
2. Two terms? Check for a difference of squares.
3. Three terms with leading coefficient $1$? Try Method 1 (find numbers that multiply to $c$ and add to $b$).
4. Three terms with leading coefficient $\ne 1$? Use the AC method (Method 2). Also check if it is a perfect-square trinomial.
5. Four terms? Try grouping.

If none of these works, the polynomial may be prime (not factorable over the integers), or you may need higher-degree techniques like synthetic division or the rational root theorem (Algebra II topics).

Common mistakes

• Forgetting the GCF. $4x^2 - 16 = 4(x^2 - 4) = 4(x + 2)(x - 2)$. Skipping the GCF makes the next steps harder and the factored form incomplete.
• Trying to "factor" a sum of squares. $a^2 + b^2$ is not $(a + b)^2$. Sums of squares do not factor over the real numbers (only over the complex numbers).
• Mixing up signs in difference of squares. $a^2 - b^2 = (a + b)(a - b)$, not $(a - b)^2$. The two factors have opposite signs in front of $b$.
• Not checking by expanding. After factoring, multiply your factors back out. If you don't recover the original polynomial, something went wrong.