Factoring Trinomials (a = 1)

Q: What if the trinomial starts with something other than x², like 2x² + 7x + 3?

That's a leading coefficient greater than $1$, and it needs an extra step — most classes teach the AC method or factoring by grouping. Master the $x^2 + bx + c$ case first, because the same two-number thinking sits inside the harder version.

Factoring a trinomial means rewriting something like $x^2 + 7x + 12$ as a product of two binomials: $(x + 3)(x + 4)$ . It's multiplication run in reverse — you're asking what two factors were multiplied to build the trinomial in the first place.

When the trinomial starts with plain $x^2$ (a leading coefficient of $1$ ), there's one move that handles every problem: find two numbers that multiply to the constant $c$ and add to the middle coefficient $b$ . Once you can spot that pair quickly, factoring becomes a number puzzle, and it's the doorway to solving quadratic equations.

The two-number method

A trinomial $x^2 + bx + c$ factors as $(x + p)(x + q)$ where $p \cdot q = c$ and $p + q = b$ . That's the whole method: hunt for two numbers whose product is the constant term and whose sum is the middle coefficient.

Start by listing factor pairs of $c$ , then check which pair adds to $b$ . For $x^2 + 7x + 12$ , the pairs for $12$ are $1$ and $12$ , $2$ and $6$ , $3$ and $4$ . Only $3 + 4 = 7$ , so the answer is $(x + 3)(x + 4)$ .

Always check your work by multiplying the binomials back out. If you don't land exactly on the original trinomial, one of your numbers is wrong — usually a sign.

Let the signs do the thinking

The signs of $b$ and $c$ tell you what kind of pair to look for before you test a single number. If $c$ is positive, both numbers have the same sign — both positive when $b$ is positive, both negative when $b$ is negative.

If $c$ is negative, the two numbers have opposite signs, and the number with the larger absolute value takes the sign of $b$ . So in $x^2 + 3x - 18$ , you need one positive and one negative number, and the bigger one is positive: $6$ and $-3$ .

Reading the signs first cuts the search in half. In $x^2 - 9x + 20$ , positive $c$ and negative $b$ mean both numbers are negative — you only test $-1$ and $-20$ , $-2$ and $-10$ , $-4$ and $-5$ .

When nothing works

Some trinomials don't factor with integers. If you've listed every factor pair of $c$ and none adds to $b$ — try $x^2 + 3x + 5$ , where no pair of integers multiplies to $5$ and adds to $3$ — the trinomial is prime. That's a legitimate answer, and it's also the cue that a quadratic equation built from it needs the quadratic formula instead.

Worked examples

Example 1: both numbers positive

Factor $x^2 + 8x + 15$ .

Identify the targets: multiply to

15

, add to

8

Factor pairs of

15

1

and

15

3

and

5

The pair

3

and

5

adds to

8

3 + 5 = 8

Write the factors

(x + 3)(x + 5)

Check:

(x + 3)(x + 5) = x^2 + 5x + 3x + 15 = x^2 + 8x + 15

✓

Answer: $(x + 3)(x + 5)$

Example 2: both numbers negative

Factor $x^2 - 9x + 20$ .

Targets: multiply to

+20

, add to

-9

Positive product, negative sum — both numbers are negative

Test negative pairs of

20

-4

and

-5

works

(-4) + (-5) = -9

Write the factors

(x - 4)(x - 5)

Answer: $(x - 4)(x - 5)$

Example 3: opposite signs, positive middle term

Factor $x^2 + 3x - 18$ .

Targets: multiply to

-18

, add to

+3

Negative product — one number positive, one negative, and the larger one is positive

Test pairs:

6

and

-3

works

6 \cdot (-3) = -18, \quad 6 + (-3) = 3

Write the factors

(x + 6)(x - 3)

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

Example 4: opposite signs, negative middle term

Factor $x^2 - 2x - 24$ .

Targets: multiply to

-24

, add to

-2

Negative product — opposite signs, and the larger number is negative

Test pairs:

4

and

-6

works

4 \cdot (-6) = -24, \quad 4 + (-6) = -2

Write the factors

(x + 4)(x - 6)

Check:

(x + 4)(x - 6) = x^2 - 6x + 4x - 24 = x^2 - 2x - 24

✓

Answer: $(x + 4)(x - 6)$

Try one yourself

Common questions

Does the order of the factors matter?

No. $(x + 3)(x + 4)$ and $(x + 4)(x + 3)$ are the same answer, because multiplication works in either order. Write whichever comes to you first.

What if the trinomial starts with something other than x², like 2x² + 7x + 3?

That's a leading coefficient greater than $1$ , and it needs an extra step — most classes teach the AC method or factoring by grouping. Master the $x^2 + bx + c$ case first, because the same two-number thinking sits inside the harder version.

Why do we factor trinomials at all?

Mostly to solve quadratic equations. Once $x^2 + 7x + 12 = 0$ becomes $(x + 3)(x + 4) = 0$ , the zero product property says one of the factors must equal zero, so $x = -3$ or $x = -4$ . Factoring turns one hard equation into two easy ones.

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Factoring Trinomials (a = 1)

The two-number method

Let the signs do the thinking

When nothing works

Worked examples

Example 1: both numbers positive

Example 2: both numbers negative

Example 3: opposite signs, positive middle term

Example 4: opposite signs, negative middle term

Try one yourself

Common questions

Keep learning

Quadratic Formula

The Distributive Property

Combining Like Terms