General Factoring Strategy

You have now learned several factoring techniques: pulling out the GCF, factoring trinomials, the difference of squares, and factoring by grouping. The challenge in practice is knowing which technique to use and when. This page gives you a systematic strategy — a decision tree — so you never stare at a factoring problem without a plan.

The core principle is simple: follow the same steps every time, in the same order. With enough practice, this decision process becomes automatic.

The Factoring Decision Tree

Follow these steps for every factoring problem:

Step 1 — Factor out the GCF. Always do this first, regardless of the polynomial. If the leading coefficient is negative, factor out the negative along with the GCF.

Step 2 — Count the terms remaining inside the parentheses (or in the original expression if the GCF was 1):

• 2 terms → Check for difference of squares ($a^2 - b^2$). Remember: sum of squares does not factor.
• 3 terms → Use trinomial factoring (find two numbers for $x^2 + bx + c$, or the AC method for $ax^2 + bx + c$).
• 4 terms → Try factoring by grouping (pair the terms, factor each pair, look for a common binomial).

Step 3 — Factor completely. After applying the technique from Step 2, examine each factor to see if it can be factored further. Repeat until every factor is prime (unfactorable over the integers).

Step 4 — Check your work. Multiply all factors back together to verify you get the original polynomial.

The Decision Tree in Action

Let us work through several examples, following the decision tree from start to finish.

Example 1: Factor $3x^2 - 27$

Step 1 — GCF: The GCF of 3 and 27 is 3.

$3x^2 - 27 = 3(x^2 - 9)$

Step 2 — Count terms: Two terms inside the parentheses.

Is $x^2 - 9$ a difference of squares? Yes: $x^2 = (x)^2$ and $9 = (3)^2$.

$3(x + 3)(x - 3)$

Step 3 — Factor completely: Neither $(x + 3)$ nor $(x - 3)$ can be factored further.

Answer: $3(x + 3)(x - 3)$

Example 2: Factor $2x^2 + 14x + 24$

Step 1 — GCF: The GCF of 2, 14, and 24 is 2.

$2x^2 + 14x + 24 = 2(x^2 + 7x + 12)$

Step 2 — Count terms: Three terms. Leading coefficient is 1.

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

$2(x + 3)(x + 4)$

Step 3 — Factor completely: Both binomials are prime.

Answer: $2(x + 3)(x + 4)$

Example 3: Factor $6x^3 + 9x^2 - 6x$

Step 1 — GCF: The GCF of 6, 9, and 6 is 3, and every term has at least $x$. So the GCF is $3x$.

$6x^3 + 9x^2 - 6x = 3x(2x^2 + 3x - 2)$

Step 2 — Count terms: Three terms inside the parentheses. Leading coefficient is 2 (not 1), so use the AC method.

AC product: $2 \times (-2) = -4$. Find two numbers that add to 3 and multiply to $-4$: 4 and $-1$.

Split the middle term:

$3x(2x^2 + 4x - x - 2)$

Group:

$3x\bigl[2x(x + 2) - 1(x + 2)\bigr]$

$3x(x + 2)(2x - 1)$

Step 3 — Factor completely: All three factors are prime.

Answer: $3x(x + 2)(2x - 1)$

Example 4: Factor $x^4 - 16$

Step 1 — GCF: The GCF is 1. Nothing to factor out.

Step 2 — Count terms: Two terms. Difference of squares? Yes: $x^4 = (x^2)^2$ and $16 = (4)^2$.

$x^4 - 16 = (x^2 + 4)(x^2 - 4)$

Step 3 — Factor completely: Is $x^2 + 4$ factorable? No — sum of squares. Is $x^2 - 4$ factorable? Yes — difference of squares:

$x^2 - 4 = (x + 2)(x - 2)$

Answer: $(x^2 + 4)(x + 2)(x - 2)$

Example 5: Factor $2x^3 + 6x^2 - 8x - 24$

Step 1 — GCF: The GCF of 2, 6, 8, and 24 is 2.

$2(x^3 + 3x^2 - 4x - 12)$

Step 2 — Count terms: Four terms inside. Try grouping.

$(x^3 + 3x^2) + (-4x - 12)$

$x^2(x + 3) - 4(x + 3)$

$(x + 3)(x^2 - 4)$

So far: $2(x + 3)(x^2 - 4)$

Step 3 — Factor completely: $x^2 - 4$ is a difference of squares:

$2(x + 3)(x + 2)(x - 2)$

Answer: $2(x + 3)(x + 2)(x - 2)$

Example 6: Factor $-5x^2 + 20$

Step 1 — GCF: The GCF is 5. The leading term is negative, so factor out $-5$:

$-5x^2 + 20 = -5(x^2 - 4)$

Step 2 — Count terms: Two terms. Difference of squares:

$-5(x + 2)(x - 2)$

Answer: $-5(x + 2)(x - 2)$

Example 7: Factor $4x^2 + 12x + 9$

Step 1 — GCF: The GCF is 1.

Step 2 — Count terms: Three terms. The leading coefficient is 4, so try the AC method.

AC product: $4 \times 9 = 36$. Find two numbers that add to 12 and multiply to 36: 6 and 6.

Split the middle term:

$4x^2 + 6x + 6x + 9$

Group:

$2x(2x + 3) + 3(2x + 3)$

$(2x + 3)(2x + 3) = (2x + 3)^2$

This is a perfect square trinomial. The result is a squared binomial.

Answer: $(2x + 3)^2$

How to Recognize Special Patterns Quickly

Over time, you will start to recognize certain patterns before you even reach the decision tree:

Difference of squares: Two terms, both perfect squares, connected by subtraction. Instantly factor as $(a + b)(a - b)$.

Perfect square trinomials: Three terms where the first and last are perfect squares and the middle term is twice their product. Factor as $(a + b)^2$ or $(a - b)^2$:

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

Sum/difference of cubes (a preview for Algebra 2):

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

These are less common in Algebra 1, but knowing they exist helps you avoid wasting time trying other methods.

Knowing When a Polynomial Is Prime

Not every polynomial can be factored over the integers. A polynomial that cannot be factored (other than trivial factors like 1) is called prime or irreducible.

For two terms:

• Sum of squares ($a^2 + b^2$) is prime over the reals
• If the terms are not both perfect squares, check whether a GCF reveals a pattern

For three terms:

• Compute the discriminant $b^2 - 4ac$. If it is not a perfect square (and not zero), the trinomial is prime over the integers
• Example: $x^2 + 3x + 5$ has discriminant $9 - 20 = -11$, which is negative — this trinomial is prime

For four terms:

• If no rearrangement produces a common binomial through grouping, the polynomial may be prime

Real-World Application: Electrician — Simplifying a Power Formula

An electrician working with AC circuits encounters a power expression that needs simplification:

$P = 4I^2R - 16R$

Step 1 — Factor out the GCF: The GCF of $4I^2R$ and $16R$ is $4R$.

$P = 4R(I^2 - 4)$

Step 2 — Count terms: Two terms inside. Difference of squares: $I^2 = (I)^2$ and $4 = (2)^2$.

$P = 4R(I + 2)(I - 2)$

If $R = 10$ ohms and $I = 5$ amps:

$P = 4(10)(5 + 2)(5 - 2) = 40 \times 7 \times 3 = 840 \text{ watts}$

The factored form reveals that the power is zero when $I = 2$ amps — a useful insight for the electrician analyzing the circuit’s behavior at different current levels.

Real-World Application: HVAC — Heat Load Calculation

An HVAC technician calculates the additional heat load (in BTU) for a room expansion:

$Q = 3Lw + 9L + 2w + 6$

The technician wants to factor this to understand the formula’s structure.

Step 1 — GCF: No common factor across all four terms.

Step 2 — Count terms: Four terms. Try grouping.

$(3Lw + 9L) + (2w + 6)$

$3L(w + 3) + 2(w + 3)$

$(w + 3)(3L + 2)$

If the room width $w = 12$ feet and the extension length $L = 8$ feet:

$Q = (12 + 3)(3 \times 8 + 2) = 15 \times 26 = 390 \text{ BTU}$

The factored form shows the heat load depends on $(w + 3)$ and $(3L + 2)$, making it easy to estimate how changing one dimension affects the result.

Common Mistakes to Avoid

1. Skipping the GCF. The most common error is jumping straight to trinomial factoring or the difference of squares without first pulling out the GCF. Always start with Step 1.
2. Stopping too early. After one round of factoring, check every factor to see if it can be factored again. Polynomials like $x^4 - 16$ require two rounds.
3. Applying the wrong technique. A two-term polynomial cannot be factored as a trinomial, and a three-term polynomial is not a grouping problem. Let the number of terms guide your choice.
4. Forgetting that sum of squares is prime. If you end up with $(x^2 + 4)$, leave it alone — do not try to factor it further.
5. Not checking your answer. Always multiply the factored form back out. This catches sign errors, arithmetic mistakes, and incomplete factoring.
6. Mixing up the decision tree order. GCF first, then count terms, then choose the technique. Follow this order every single time.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Always factor out the GCF first — this is Step 1 every single time
• After removing the GCF, count the remaining terms to choose the right technique: 2 terms (difference of squares), 3 terms (trinomial factoring), 4 terms (grouping)
• Factor completely — check every factor to see if it can be factored again
• The sum of squares is prime over the reals — do not try to factor it
• Use the discriminant $b^2 - 4ac$ to determine if a trinomial factors over the integers (it must be a non-negative perfect square)
• Always check by multiplying back out — this is the single best way to catch errors
• Following the same decision tree every time turns factoring from a puzzle into a reliable process

Return to Algebra for more topics in this section.