Factoring Trinomials (a > 1)

Q: Why do I factor out the GCF before anything else?

Because it usually drops the leading coefficient to $1$, turning a hard problem into an easy one. It also guarantees your final answer is factored completely — a GCF left inside means the factoring is not finished.

A trinomial like $2x^{2} + 10x + 12$ has a leading coefficient bigger than $1$ , so the basic two-numbers trick for $x^{2} + bx + c$ does not apply directly. But very often there is a shortcut hiding in plain sight: all three coefficients share a common factor.

The strategy: pull out the GCF first. Once the greatest common factor is out front, the trinomial left inside frequently has a leading coefficient of $1$ , and you can factor it the easy way — two numbers that multiply to the constant and add to the middle coefficient. The GCF simply stays out front in the final answer.

Step 1: factor out the GCF

Look at all three coefficients. In $2x^{2} + 10x + 12$ , everything is divisible by $2$ , so factor it out: $2(x^{2} + 5x + 6)$ . The leading coefficient inside is now $1$ .

This is why GCF factoring is always the first check. Skipping it either makes the problem much harder or leaves your answer incompletely factored.

Step 2: factor the trinomial inside

Now factor $x^{2} + 5x + 6$ the standard way: find two numbers that multiply to $6$ and add to $5$ . That is $2$ and $3$ , so $x^{2} + 5x + 6 = (x + 2)(x + 3)$ .

Put it together, keeping the GCF out front: $2x^{2} + 10x + 12 = 2(x + 2)(x + 3)$ . The $2$ is part of the answer — never drop it.

Signs work exactly as they do for simple trinomials. A negative constant inside means the two numbers have opposite signs; a positive constant with a negative middle term means both numbers are negative.

Check by multiplying back

Multiply the binomials first, then distribute the GCF. For $2(x + 2)(x + 3)$ : the binomials give $x^{2} + 5x + 6$ , and the $2$ turns that into $2x^{2} + 10x + 12$ — the original trinomial. If you land anywhere else, recheck your two numbers.

If no GCF exists and the leading coefficient still is not $1$ — say $2x^{2} + 7x + 3$ — you need the grouping (AC) method instead. That is the next lesson in this unit.

Worked examples

Example 1: GCF of 2

Factor $2x^{2} + 14x + 24$ .

Factor out the GCF of

2

2(x^{2} + 7x + 12)

Find two numbers with product

12

and sum

7

3 \text{ and } 4

Factor the trinomial

(x + 3)(x + 4)

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

Example 2: negative terms

Factor $3x^{2} - 12x - 36$ .

Factor out the GCF of

3

3(x^{2} - 4x - 12)

Find two numbers with product

-12

and sum

-4

-6 \text{ and } 2

Factor the trinomial

(x - 6)(x + 2)

Answer: $3(x + 2)(x - 6)$

Example 3: GCF of 4

Factor $4x^{2} + 8x - 12$ .

Factor out the GCF of

4

4(x^{2} + 2x - 3)

Find two numbers with product

-3

and sum

2

3 \text{ and } -1

Factor the trinomial

(x + 3)(x - 1)

Check

4(x^{2} + 2x - 3) = 4x^{2} + 8x - 12

Answer: $4(x + 3)(x - 1)$

Try one yourself

Common questions

Why do I factor out the GCF before anything else?

Because it usually drops the leading coefficient to $1$ , turning a hard problem into an easy one. It also guarantees your final answer is factored completely — a GCF left inside means the factoring is not finished.

What if the three terms have no common factor?

Then this shortcut does not apply and you factor with the grouping method: multiply $a \cdot c$ , find two numbers with that product that add to $b$ , split the middle term, and group. See the factoring-by-grouping lesson.

Does the GCF stay in my final answer?

Yes. $2x^{2} + 10x + 12 = 2(x + 2)(x + 3)$ — the $2$ is one of the factors. Writing just $(x + 2)(x + 3)$ multiplies back to $x^{2} + 5x + 6$ , which is not the polynomial you started with.

How do I pick the signs of the two numbers?

Look at the constant term inside first. Positive constant: both numbers share the sign of the middle term. Negative constant: the numbers have opposite signs, and the one with the larger size takes the middle term's sign.

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Factoring Trinomials (a > 1)

Step 1: factor out the GCF

Step 2: factor the trinomial inside

Check by multiplying back

Worked examples

Example 1: GCF of 2

Example 2: negative terms

Example 3: GCF of 4

Try one yourself

Common questions

Keep learning

Factoring by GCF

Factoring Trinomials (a = 1)

Factoring Trinomials by Grouping

Multiplying Polynomials