What do I do if no pair of numbers works?

First make sure you listed every factor pair of c (or of ac), including the negative pairs when the product is negative. If you have checked them all and none gives the required sum, the trinomial is prime over the integers, and "prime" is the complete answer. Do not force a factorization that fails the FOIL check.

Do I have to use grouping when the leading coefficient is not 1, or can I just guess and check?

Guess-and-check with binomials works and some students prefer it, especially when a and c have few factors. The ac-method is more reliable because it never depends on luck: the product and sum conditions either have an integer solution or they do not. Whichever method you use, verify with FOIL.

Does the order of my two binomial factors matter?

No. Multiplication is commutative, so (x+2)(x+5) and (x+5)(x+2) are the same answer. Likewise, when you split the middle term in the ac-method, either order of the two numbers leads to the same final factorization, though the intermediate grouping steps look different.

Why does the ac-method work at all?

If ax^2+bx+c factors as (px+q)(rx+s), then a=pr, c=qs, and b=ps+qr. The two pieces of the middle term, ps and qr, multiply to ps· qr=(pr)(qs)=ac and add to b. So finding two numbers with product ac and sum b is exactly recovering those two middle pieces, and grouping reassembles the binomials.

ALG1-7.4

Factoring Trinomials

Learn to factor x² + bx + c with sum-product and ax² + bx + c with the ac-method and grouping — GCF first, signs decoded, every answer checked by FOIL.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Factoring Trinomials, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Factoring Trinomials free

What this lesson covers

Multiplying two binomials is the easy direction:

(x+3)(x+5)

becomes

x^2+8x+15

in a few seconds. Factoring runs the movie backwards. You are handed

x^2+8x+15

and asked which two binomials were multiplied to build it. That reverse question is the heart of this lesson, and it is the skill that later lets you solve quadratic equations, simplify rational expressions, and find where a parabola crosses the x-axis.

You already know how to pull out a greatest common factor and how to factor a four-term expression by grouping. Here those two tools get combined into a reliable routine: pull the GCF out first, use the sum-product idea when the leading coefficient is 1, use the ac-method with grouping when it is not, and multiply back with FOIL every single time to confirm the answer.

The Sum-Product Method When the Lead Coefficient Is 1

Multiply

(x+m)(x+n)

and you get

x^2+(m+n)x+mn

. Read that identity carefully: the middle coefficient is the sum of the two numbers, and the constant is their product. So factoring

x^2+bx+c

is nothing more than hunting for two numbers whose product is

c

and whose sum is

b

.

Take

x^2+11x+24

. Product

24

, sum

11

. List the factor pairs of

24

1

and

24

2

and

12

3

and

8

4

and

6

. Only

3

and

8

add to

11

, so

x^2+11x+24=(x+3)(x+8)

.

The organized way to do this is to list every factor pair of

c

first and only then check sums. Students who guess randomly tend to stop at the first pair that multiplies correctly and forget to check the sum, which is the single most common error in this section.

When the constant is negative, the two numbers have opposite signs, so their "sum" is really a difference. For

x^2-2x-35

, you need a product of

-35

and a sum of

-2

: the pair is

5

and

-7

, giving

(x+5)(x-7)

.

One more caution: the order of the two binomials never matters, because multiplication is commutative.

(x+3)(x+8)

and

(x+8)(x+3)

are the same answer. What does matter is that each binomial contains

x

and the correct signed number.

Using the Signs of b and c to Narrow the Search

Before listing anything, look at the signs of

b

and

c

. They tell you the signs of the two numbers you are hunting for, which cuts the work roughly in half.

Sign of $c$	Sign of $b$	Signs of the two numbers	Example
positive	positive	both positive	$x^2+9x+20=(x+4)(x+5)$
positive	negative	both negative	$x^2-9x+20=(x-4)(x-5)$
negative	positive	one of each, larger absolute value positive	$x^2+3x-10=(x+5)(x-2)$
negative	negative	one of each, larger absolute value negative	$x^2-3x-10=(x-5)(x+2)$

Why does this work? A positive product means the two numbers match in sign, and then their sum carries that shared sign. A negative product means they disagree in sign, and the sign of the sum is inherited from whichever number is farther from zero.

A frequent slip is writing

x^2-9x+20=(x+4)(x-5)

. Check the constant:

4\cdot(-5)=-20

, not

+20

, so the signs are wrong. The middle term misses too: it comes out to

-x

, not

-9x

. Testing the product as well as the sum catches this instantly.

Also be alert to trinomials that simply do not factor over the integers. For

x^2+4x+7

, the only integer pair with product

7

1

and

7

, whose sum is

8

. No pair works, so the trinomial is prime over the integers. Saying "prime" is a complete answer; inventing a factorization that does not multiply back is not.

The ac-Method with Grouping When a Is Not 1

When the leading coefficient is something other than

1

, as in

3x^2+10x+8

, the sum-product shortcut no longer reads straight off the expression, because the leading coefficient gets tangled into the middle term. The fix is the ac-method, which converts the problem into a grouping problem you already know how to finish.

Multiply

a\cdot c

. Here

a=3

and

c=8

, so

ac=24

. Now find two numbers with product

24

and sum

b=10

: those are

4

and

6

. Split the middle term using them:

3x^2+10x+8=3x^2+4x+6x+8

Group in pairs and factor each pair:

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

. The matching binomial

(3x+4)

is the signal that the split was done correctly, and factoring it out gives

(3x+4)(x+2)

.

A few practical points. The two numbers can be written in either order when you split;

3x^2+6x+4x+8

groups to

3x(x+2)+4(x+2)=(x+2)(3x+4)

, the same answer. If the second group starts with a minus sign, factor out a negative so the parentheses match: for

2x^2-7x+3

you get

2x^2-6x-x+3=2x(x-3)-1(x-3)=(x-3)(2x-1)

. Forgetting that

-1

and writing

2x(x-3)-(x-3)

without factoring cleanly is where grouping usually breaks down.

If no integer pair multiplies to

ac

and adds to

b

, the trinomial is prime over the integers.

GCF First, FOIL Check Last

Every factoring problem starts the same way: look for a greatest common factor of all three terms and pull it out. Skipping this step is the most common reason an answer comes out incomplete, because the leftover trinomial is harder to factor and the final form is not fully factored.

Compare the two routes on

2x^2+14x+24

. Straight to the ac-method:

ac=48

, pair

6

and

8

, split, group, and you arrive at

(2x+6)(x+4)

— which is not finished, since

2x+6=2(x+3)

. Pulling the

2

out first gives

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

in one clean pass. Same answer, far less work, and no risk of stopping early.

The GCF can include variables:

5x^3-20x^2-105x=5x(x^2-4x-21)=5x(x-7)(x+3)

. Keep that GCF written in front through every later step; dropping it partway through is a silent error that only the check will catch.

Speaking of the check: multiply your factors back out with FOIL and confirm you land on the original trinomial. For

2(x+3)(x+4)

, first do

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

, then distribute the

2

to get

2x^2+14x+24

. Match. This takes about fifteen seconds and turns factoring from guesswork into something you can be certain about. If the check fails, the middle term is almost always the culprit — recheck the sum of your two numbers.

Key terms

Trinomial.: A polynomial with exactly three terms, such as $x^2+7x+12$ or $6x^2-x-15$ .
Sum-product method.: For $x^2+bx+c$ , finding two numbers whose product is $c$ and whose sum is $b$ , then writing the factors as $(x+m)(x+n)$ .
ac-method.: For $ax^2+bx+c$ , finding two numbers whose product is $a\cdot c$ and whose sum is $b$ , splitting the middle term with them, and factoring by grouping.
Greatest common factor (GCF).: The largest monomial that divides every term of a polynomial; always factored out before anything else.
Factoring by grouping.: Splitting a four-term polynomial into two pairs, factoring each pair, and factoring out the shared binomial.
Prime (irreducible) trinomial.: A trinomial that cannot be written as a product of polynomials with integer coefficients, such as $x^2+4x+7$ .
FOIL.: First, Outer, Inner, Last — the order used to multiply two binomials, used here to verify a factorization.
Completely factored.: Written as a product in which no factor can be broken down further, including having no remaining common factor.

Worked example

Factor completely:

6x^3-3x^2-45x

Step 1 — GCF first. Each term is divisible by

3

and by

x

, so the GCF is

3x

. Factoring it out gives

3x(2x^2-x-15)

. Write the

3x

in front and carry it through the rest of the problem.

Step 2 — Identify the pieces. Inside the parentheses,

a=2

b=-1

c=-15

. Since

a\neq 1

, use the ac-method.

Step 3 — Find the pair. Compute

ac=2\cdot(-15)=-30

. You need two numbers with product

-30

and sum

-1

. Because the product is negative, the numbers have opposite signs. Pairs of

30

1

and

30

2

and

15

3

and

10

5

and

6

. The pair

5

and

6

differs by

1

, and since the sum must be negative the larger number takes the minus sign:

-6

and

5

. Check:

(-6)(5)=-30

and

-6+5=-1

. Correct.

Step 4 — Split the middle term.

2x^2-x-15=2x^2-6x+5x-15

.

Step 5 — Group and factor each pair.

(2x^2-6x)+(5x-15)=2x(x-3)+5(x-3)

. The parentheses match, which confirms the split.

Step 6 — Factor out the common binomial.

2x(x-3)+5(x-3)=(x-3)(2x+5)

.

Step 7 — Restore the GCF. The complete factorization is

3x(x-3)(2x+5)

.

Step 8 — Check by FOIL.

(x-3)(2x+5)=2x^2+5x-6x-15=2x^2-x-15

. Multiply by

3x

6x^3-3x^2-45x

. This matches the original, so the factorization is confirmed.

Practice questions

Which of the following is the complete factorization of

x^2-7x+12

$(x-3)(x-4)$
$(x+3)(x+4)$
$(x-2)(x-6)$
$(x-1)(x-12)$

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

You need two numbers with product

12

and sum

-7

. Because

c

is positive and

b

is negative, both numbers must be negative. The negative factor pairs of

12

are

-1

and

-12

(sum

-13

-2

and

-6

(sum

-8

), and

-3

and

-4

(sum

-7

). Only the last pair works, so the factorization is

(x-3)(x-4)

. The pair

(x+3)(x+4)

gives the right constant but a middle term of

+7x

, and

(x-2)(x-6)

gives

-8x

. FOIL confirms:

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

Factor

4x^2+4x-15

completely, showing the ac-method steps, then verify your answer.

Answer: $(2x+5)(2x-3)$

There is no common factor of

4

4

, and

-15

other than

1

, so go straight to the ac-method. Here

ac=4\cdot(-15)=-60

and

b=4

. Two numbers with product

-60

and sum

4

are

10

and

-6

. Split the middle term:

4x^2+10x-6x-15

. Group:

2x(2x+5)-3(2x+5)

. Notice the

-3

must be factored from

-6x-15

so that the parentheses match. Factor out the common binomial:

(2x+5)(2x-3)

. Check by FOIL:

4x^2-6x+10x-15=4x^2+4x-15

, which matches.

A student factors

3x^2+18x+24

(3x+12)(x+2)

and stops. Is the answer correct? Is it complete? Explain what should have been done differently.

Answer: The multiplication is correct, but the answer is not completely factored; the finished form is $3(x+2)(x+4)$ .

FOIL shows

(3x+12)(x+2)=3x^2+6x+12x+24=3x^2+18x+24

, so the product is right. However,

3x+12

still has a common factor of

3

, so the expression is not completely factored:

(3x+12)=3(x+4)

, making the full factorization

3(x+2)(x+4)

. The efficient route is to pull the GCF of

3

out at the very start:

3(x^2+6x+8)

, then find two numbers with product

8

and sum

6

, namely

2

and

4

, giving

3(x+2)(x+4)

directly. This is exactly why the GCF step comes before the sum-product or ac-method.

FAQ

What do I do if no pair of numbers works?: First make sure you listed every factor pair of $c$ (or of $ac$ ), including the negative pairs when the product is negative. If you have checked them all and none gives the required sum, the trinomial is prime over the integers, and "prime" is the complete answer. Do not force a factorization that fails the FOIL check.
Do I have to use grouping when the leading coefficient is not 1, or can I just guess and check?: Guess-and-check with binomials works and some students prefer it, especially when $a$ and $c$ have few factors. The ac-method is more reliable because it never depends on luck: the product and sum conditions either have an integer solution or they do not. Whichever method you use, verify with FOIL.
Does the order of my two binomial factors matter?: No. Multiplication is commutative, so $(x+2)(x+5)$ and $(x+5)(x+2)$ are the same answer. Likewise, when you split the middle term in the ac-method, either order of the two numbers leads to the same final factorization, though the intermediate grouping steps look different.
Why does the ac-method work at all?: If $ax^2+bx+c$ factors as $(px+q)(rx+s)$ , then $a=pr$ , $c=qs$ , and $b=ps+qr$ . The two pieces of the middle term, $ps$ and $qr$ , multiply to $ps\cdot qr=(pr)(qs)=ac$ and add to $b$ . So finding two numbers with product $ac$ and sum $b$ is exactly recovering those two middle pieces, and grouping reassembles the binomials.

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