ALG1-7.3

Factoring: GCF & Grouping

Learn to factor out the GCF (including a negative GCF) and factor four-term polynomials by grouping, then check every answer by re-multiplying.

What you'll do in this lesson

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

What this lesson covers

Multiplying polynomials takes two factors and produces one long expression. Factoring runs that machine backward: you start with the long expression and hunt for the factors that built it. In this lesson you will pull out the greatest common factor of a polynomial, handle the tricky case where the GCF is negative, and break four-term polynomials apart using grouping.

These two moves are the foundation of the rest of the unit. Before you factor a trinomial or recognize a difference of squares, you always check for a common factor first — skipping that step is the single most common reason a factoring answer comes out half-finished. And because factoring is just the distributive property in reverse, you can always check yourself: multiply your factors back out and see whether the original polynomial reappears.

Factoring Is the Distributive Property Run Backward

When you distribute, you take 3x(x+4)3x(x + 4) and spread the 3x3x across both terms to get 3x2+12x3x^2 + 12x. Factoring asks the opposite question: given 3x2+12x3x^2 + 12x, what was multiplied to produce it?

The answer comes from finding what both terms share. Write each term as a product of its pieces:

3x2=3xx3x^2 = 3 \cdot x \cdot x and 12x=223x12x = 2 \cdot 2 \cdot 3 \cdot x.

Both contain a factor of 33 and a factor of xx, so 3x3x can be pulled out front: 3x2+12x=3x(x+4)3x^2 + 12x = 3x(x + 4).

The part left inside the parentheses is what remains after each term is divided by the common factor. That is the mechanical rule: divide every term by the GCF, and the quotients become the terms inside.

Because factoring and distributing are inverse operations, checking is built in. Re-multiply 3x(x+4)3x(x+4) and you get 3x2+12x3x^2 + 12x — the original. Make this check automatic. It takes ten seconds and catches nearly every sign slip and dropped term.

One caution about vocabulary: a factored expression is a product. If your answer still has a plus or minus sign at the top level, connecting two separate chunks, you have not finished factoring. For example, x(x+2)+3(x+2)x(x+2) + 3(x+2) is not factored — it is a sum of two products. Factored means one multiplication statement.

Finding the Greatest Common Factor of a Polynomial

The GCF of a polynomial has a number part and a variable part, and you find them separately.

For the number part, take the greatest common factor of the coefficients (ignore signs for now). For the variable part, take each variable that appears in every term, raised to the smallest exponent it has anywhere in the polynomial. A variable missing from even one term cannot be part of the GCF.

Consider 18x4y224x3y3+30x3y18x^4y^2 - 24x^3y^3 + 30x^3y.
TermCoefficientPower of xxPower of yy
18x4y218x^4y^21842
24x3y3-24x^3y^32433
30x3y30x^3y3031
The GCF of 1818, 2424, and 3030 is 66. The smallest power of xx is 33; the smallest power of yy is 11. So the GCF is 6x3y6x^3y, and18x4y224x3y3+30x3y=6x3y(3xy4y2+5).18x^4y^2 - 24x^3y^3 + 30x^3y = 6x^3y(3xy - 4y^2 + 5).Two places students go wrong. First, when a term is entirely used up by the GCF, they leave a blank instead of writing 11. In 5x2+5x5x^2 + 5x, dividing the second term by 5x5x gives 11, so the answer is 5x(x+1)5x(x + 1), not 5x(x)5x(x). Second, they pull out a common factor that is not the greatest one: 4x2+8x=2(2x2+4x)4x^2 + 8x = 2(2x^2 + 4x) is a true statement but is not fully factored, because 2x2x still divides both inside terms. Always ask whether the leftover terms still share anything.

Pulling Out a Negative GCF

When the leading term of a polynomial is negative, it is standard practice to factor out a negative GCF so the polynomial inside the parentheses starts with a positive term. This matters later: trinomials are far easier to factor when the leading coefficient is positive.

Take 6x3+15x29x-6x^3 + 15x^2 - 9x. The positive GCF is 3x3x, so the negative GCF is 3x-3x. Divide every term by 3x-3x:

6x3÷(3x)=2x2-6x^3 \div (-3x) = 2x^2, 15x2÷(3x)=5x15x^2 \div (-3x) = -5x, and 9x÷(3x)=3-9x \div (-3x) = 3.

So 6x3+15x29x=3x(2x25x+3)-6x^3 + 15x^2 - 9x = -3x(2x^2 - 5x + 3).

Notice that every sign inside flipped compared with factoring out positive 3x3x. That is the whole trap. Students commonly write 3x(2x2+5x3)-3x(2x^2 + 5x - 3) by changing only the first sign, or 3x(2x25x3)-3x(2x^2 - 5x - 3) by forgetting the last one. Dividing each term one at a time, out loud, prevents this — and re-multiplying catches it if it slips through.

A related use of the negative GCF is reversing a binomial. Since 1(x7)=7x-1(x - 7) = 7 - x, the expressions x7x - 7 and 7x7 - x are opposites. Factoring 1-1 out of one turns it into the other, which is exactly the move that rescues a grouping problem when the two binomials come out backward from each other.

One more note: factoring out a negative is a choice, not a law. Both 3x(2x25x+3)-3x(2x^2 - 5x + 3) and 3x(2x2+5x3)3x(-2x^2 + 5x - 3) are correct products. The first form is preferred because it sets up the next factoring step cleanly.

Factoring Four-Term Polynomials by Grouping

When a polynomial has four terms and no single GCF for all four, try grouping. The strategy is to split the polynomial into two pairs, factor the GCF out of each pair, and hope the two leftover binomials match.

Factor x3+5x2+4x+20x^3 + 5x^2 + 4x + 20.

Group the first two and the last two: (x3+5x2)+(4x+20)(x^3 + 5x^2) + (4x + 20). The GCF of the first pair is x2x^2, giving x2(x+5)x^2(x + 5). The GCF of the second pair is 44, giving 4(x+5)4(x + 5). Now the expression reads x2(x+5)+4(x+5)x^2(x+5) + 4(x+5). Both terms share the common binomial factor (x+5)(x + 5), so pull it out front: (x+5)(x2+4)(x + 5)(x^2 + 4).

Subtraction in the third position is where errors cluster. In 2x36x25x+152x^3 - 6x^2 - 5x + 15, rewrite the second group with its sign attached: (2x36x2)+(5x+15)(2x^3 - 6x^2) + (-5x + 15). Factoring 5-5 (not +5+5) out of the second group gives 5(x3)-5(x - 3), matching 2x2(x3)2x^2(x - 3) from the first group. The result is (x3)(2x25)(x - 3)(2x^2 - 5). If you had factored out +5+5 you would get 5(x+3)5(-x + 3), and the binomials would not match.

If the two binomials come out as opposites, such as (x3)(x - 3) and (3x)(3 - x), factor 1-1 out of one group to flip it. If they simply do not match at all, try pairing the terms in a different order — grouping the first with the third and the second with the fourth often works.

Always look for an overall GCF before grouping. Removing it first keeps the numbers small and prevents an incompletely factored answer.

Checking by Re-Multiplying and Knowing When to Stop

Every factoring answer can be verified, so there is no reason to hand in an unchecked one. To check, distribute your factors back out and compare with the original polynomial term by term.

Check (x+5)(x2+4)(x + 5)(x^2 + 4): multiply to get x3+4x+5x2+20x^3 + 4x + 5x^2 + 20, then reorder as x3+5x2+4x+20x^3 + 5x^2 + 4x + 20. It matches, so the factoring is correct.

Check a GCF answer the same way. Does 3x(2x25x+3)-3x(2x^2 - 5x + 3) equal 6x3+15x29x-6x^3 + 15x^2 - 9x? Distributing gives 6x3+15x29x-6x^3 + 15x^2 - 9x. It matches.

Knowing when you are done is the other half. A polynomial is completely factored when none of the remaining factors can be broken down further. After pulling out a GCF, look at what is left inside the parentheses: if it is a trinomial or a difference of squares, upcoming lessons in this unit will let you keep going. After grouping, check each of the two binomial factors the same way.

Some polynomials are prime over the integers — they have no factors besides 11 and themselves. If the coefficients share no common factor and grouping fails in every arrangement, the polynomial may simply be prime, and saying so is a complete answer.

A quick habit that saves grief: count terms first. Two terms suggests GCF or a special form; three suggests GCF then trinomial factoring; four suggests GCF then grouping.

Key terms

Factor (verb).
To rewrite a sum or difference as a product of two or more expressions, undoing the distributive property.
Greatest common factor (GCF).
The largest monomial that divides evenly into every term of a polynomial: the GCF of the coefficients times each shared variable raised to its smallest exponent.
Negative GCF.
A common factor taken out with a negative sign, used when the leading term is negative; every term inside the parentheses changes sign compared with factoring out the positive version.
Factoring by grouping.
A method for four-term polynomials: split into two pairs, factor a GCF from each pair, then factor out the shared binomial.
Common binomial factor.
A two-term expression such as (x+5)(x + 5) that appears as a factor in more than one part of an expression and can be pulled out front.
Completely factored.
Written as a product in which no remaining factor can be factored further over the integers.
Prime polynomial.
A polynomial that cannot be written as a product of two lower-degree polynomials with integer coefficients.
Opposite binomials.
Two binomials like x3x - 3 and 3x3 - x that differ by a factor of 1-1; factoring 1-1 out of one converts it into the other.

Worked example

Factor completely: 2x3+6x210x302x^3 + 6x^2 - 10x - 30. Then check your answer by re-multiplying.
Step 1: Look for an overall GCF of all four terms. The coefficients are 22, 66, 10-10, and 30-30, which share a factor of 22. The last term 30-30 has no xx, so no variable is common. Factor out 22:2x3+6x210x30=2(x3+3x25x15).2x^3 + 6x^2 - 10x - 30 = 2(x^3 + 3x^2 - 5x - 15).Step 2: The expression inside has four terms, so group in pairs, keeping the signs attached:2[(x3+3x2)+(5x15)].2\big[(x^3 + 3x^2) + (-5x - 15)\big].Step 3: Factor a GCF from each pair. From x3+3x2x^3 + 3x^2 pull out x2x^2 to get x2(x+3)x^2(x + 3). From 5x15-5x - 15 pull out 5-5 (negative, so the leftover binomial matches) to get 5(x+3)-5(x + 3). Now:2[x2(x+3)5(x+3)].2\big[x^2(x+3) - 5(x+3)\big].Step 4: Both inner terms share the binomial (x+3)(x + 3). Pull it out:2(x+3)(x25).2(x + 3)(x^2 - 5).Step 5: Is it completely factored? The factor x+3x + 3 is linear, and x25x^2 - 5 is not a difference of squares over the integers because 55 is not a perfect square. Done.

Step 6: Check. Multiply (x+3)(x25)=x35x+3x215(x+3)(x^2-5) = x^3 - 5x + 3x^2 - 15. Multiply by 22: 2x3+6x210x302x^3 + 6x^2 - 10x - 30. This matches the original, so the factoring is correct.

Practice questions

Which expression shows 4x3+12x28x-4x^3 + 12x^2 - 8x with its negative GCF factored out?
  1. 4x(x23x+2)-4x(x^2 - 3x + 2)
  2. 4x(x2+3x2)-4x(x^2 + 3x - 2)
  3. 2x(2x26x+4)-2x(2x^2 - 6x + 4)
  4. 4x(x2+3x2)4x(-x^2 + 3x - 2)

Answer: 4x(x23x+2)-4x(x^2 - 3x + 2)

The coefficients 44, 1212, and 88 share a factor of 44, and every term has at least one xx, so the GCF is 4x4x; since the leading term is negative, use 4x-4x. Divide each term: 4x3÷(4x)=x2-4x^3 \div (-4x) = x^2, 12x2÷(4x)=3x12x^2 \div (-4x) = -3x, and 8x÷(4x)=2-8x \div (-4x) = 2. Note that every sign flipped. The choice 4x(x2+3x2)-4x(x^2 + 3x - 2) changes only some of the signs. The choice 2x(2x26x+4)-2x(2x^2 - 6x + 4) is a true product but 2x-2x is not the greatest common factor. The last choice is a correct product too, but it leaves a negative leading term inside, which is not the requested form.
Factor 15x35x2+6x215x^3 - 5x^2 + 6x - 2 completely, and show the check.

Answer: (3x1)(5x2+2)(3x - 1)(5x^2 + 2)

All four terms share no common factor other than 11 (1515, 55, 66, 22 have no common divisor), so go straight to grouping. Pair them: (15x35x2)+(6x2)(15x^3 - 5x^2) + (6x - 2). From the first pair factor 5x25x^2: 5x2(3x1)5x^2(3x - 1). From the second pair factor 22: 2(3x1)2(3x - 1). The binomials match, so factor out (3x1)(3x-1) to get (3x1)(5x2+2)(3x - 1)(5x^2 + 2). Check by multiplying: 3x5x2=15x33x \cdot 5x^2 = 15x^3, 3x2=6x3x \cdot 2 = 6x, 15x2=5x2-1 \cdot 5x^2 = -5x^2, 12=2-1 \cdot 2 = -2. Combining gives 15x35x2+6x215x^3 - 5x^2 + 6x - 2, the original polynomial.
A student factors 8x312x210x+158x^3 - 12x^2 - 10x + 15 as 4x2(2x3)+5(2x+3)4x^2(2x - 3) + 5(-2x + 3) and then says the polynomial cannot be factored by grouping because the binomials are different. What went wrong, and what is the correct factorization?

Answer: The second group should be factored with a negative GCF, giving 5(2x3)-5(2x - 3); the polynomial factors as (2x3)(4x25)(2x - 3)(4x^2 - 5).

The student's line is algebraically true, but choosing +5+5 as the GCF of 10x+15-10x + 15 produced (2x+3)(-2x + 3), which is the opposite of (2x3)(2x - 3) rather than a mismatch. Factoring 5-5 out instead gives 5(2x3)-5(2x - 3), so the expression becomes 4x2(2x3)5(2x3)4x^2(2x-3) - 5(2x-3) and the shared binomial appears. Pulling it out gives (2x3)(4x25)(2x - 3)(4x^2 - 5). Checking: 2x(4x2)=8x32x(4x^2) = 8x^3, 2x(5)=10x2x(-5) = -10x, 3(4x2)=12x2-3(4x^2) = -12x^2, 3(5)=15-3(-5) = 15, which reassembles the original. Whenever two binomials are opposites, factoring 1-1 out of one group fixes the mismatch.

FAQ

What if the terms have no common factor other than 1?
Then the GCF is 11 and there is nothing useful to pull out, so move on to another method. With four terms, try grouping. With three terms, use trinomial factoring. With two terms, check for a difference of squares. If nothing works in any arrangement, the polynomial may be prime over the integers, and stating that is a complete answer.
Do I always have to factor out a negative when the first term is negative?
It is not required by the rules of algebra — both versions are valid products — but it is the standard convention and it makes life easier. Leaving a positive leading coefficient inside the parentheses lets you apply trinomial and special-form methods without extra sign juggling. Just remember that switching to a negative GCF flips the sign of every single term inside.
What do I do if grouping gives me two binomials that do not match?
First check whether they are opposites, like x4x - 4 and 4x4 - x; if so, factor 1-1 out of one group to flip it. If they are genuinely different, rearrange the four terms and pair them differently — grouping the first with the third and the second with the fourth is the usual second attempt. If no arrangement works, the polynomial does not factor by grouping.
How do I know when a polynomial is completely factored?
Examine each factor you have written and ask whether it can be broken down further. A monomial GCF like 6x2y6x^2y is finished. A linear binomial like x+3x + 3 is finished. But a trinomial, a difference of squares, or a binomial whose terms still share a factor can go further. Re-multiplying confirms your factoring is correct; scanning each factor confirms it is complete.

Learn this with a teacher, not a page

The Crimsora tutor teaches Factoring: GCF & Grouping live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.