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➕ Algebra 1 · Factoring

Factoring by Grouping

Factor four-term polynomials by grouping: pair the terms, pull out each GCF, then the common binomial. Includes sign care, rearranging, and a trinomial link.

Grades 9, 10⏱ 16 min✔ AI-checkedCCSS A-SSE.A.2
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Written by AI, then independently re-solved and checked by separate AI reviewers and automated answer checks. Written by Claude (Sonnet 5.5) from the concept template. Automated gate: schema, notation and every math answer verified by evaluated expressions. Two blind re-solves by separate AI solvers with no access to the answer key: every keyed answer matched. Two prose audits by a separate, stronger AI reviewer (Claude Opus 5.5): every finding fixed, no wrong answer keys found. Final small edits re-checked by the automated gate (no human teacher review yet). Last checked 2026-11-08.

  • A-SSE.A.2: Use the structure of an expression to identify ways to rewrite it.

By the end you can…

  • Factor a four-term polynomial such as x3 + 3x2 + 2x + 6 by pairing the terms, taking out each pair’s GCF and pulling out the common binomial.
  • Handle subtraction and hidden 1s by taking out a negative GCF when needed, then check by expanding and factor completely.
  • Rearrange terms when the first pairing fails, and recognize when grouping will not work.
  • Use grouping to split the middle term of a trinomial ax2 + bx + c.

Lesson (school style)

Factoring by grouping is a method for some four-term polynomials, like x3 + 3x2 + 2x + 6, where no single factor (other than 1) divides all four terms. (If all four terms do share a GCF, take it out first.) You split the terms into two pairs, take the greatest common factor (GCF) out of each pair, and look for the same binomial left over in both. If it matches, that binomial is a common factor, and you pull it out. Factoring is the reverse of multiplying, so you can always check by expanding.

When to try grouping

Try grouping when a polynomial has four terms and no GCF (other than 1) divides all four. If all four terms do share a GCF, take it out first and then look at what is left inside. The terms can use several variables: xy + 3x + 4y + 12 is a four-term polynomial too.

Grouping does not always work. It works only for certain four-term polynomials: those where, after each pair’s GCF is taken out, the two leftover binomials are the same (or become the same after a sign fix, covered below).

The method on one example

Factor x3 + 3x2 + 2x + 6.

  1. Look for a GCF of all four terms. There is none: the coefficients 1, 3, 2 and 6 share no factor except 1, and the constant 6 has no x.
  2. Pair the terms, the first two and the last two, keeping each sign with its term: (x3 + 3x2) + (2x + 6).
  3. Take out each pair’s GCF. The GCF of x3 and 3x2 is x2, so x3 + 3x2 = x2(x + 3). The GCF of 2x and 6 is 2, so 2x + 6 = 2(x + 3).
  4. Compare the binomials. Both pairs left (x + 3), so the expression is now x2(x + 3) + 2(x + 3).
  5. Pull out the common binomial: (x + 3)(x2 + 2).
  6. Check by expanding: (x + 3)(x2 + 2) = x3 + 2x + 3x2 + 6, which is x3 + 3x2 + 2x + 6 in standard form (descending powers).

Why it works

After step 4 you have two terms, x2(x + 3) and 2(x + 3), and both contain the factor (x + 3). The distributive property run backward says x2 · B + 2 · B = (x2 + 2) · B, and here B is the whole binomial x + 3. Treating (x + 3) as one chunk is the key move. That is reading the structure of the expression.

Notice that x2(x + 3) + 2(x + 3) is still a sum of two terms, so it is not factored yet. It is factored only when you write it as a product: (x + 3)(x2 + 2).

Sign care: when the third term is subtracted

Signs belong to the terms they sit in front of. In x3 − 4x2 − 3x + 12, the second pair is −3x + 12: a negative first term and a positive second term.

Pair the terms: (x3 − 4x2) + (−3x + 12). The first pair gives x2(x − 4). For the second pair, take out the negative of the GCF, −3 (not 3): −3x + 12 = −3(x − 4), because (−3)(x) = −3x and (−3)(−4) = +12. Now both binomials are (x − 4):

x2(x − 4) − 3(x − 4) = (x − 4)(x2 − 3)

What goes wrong if you take out +3? You get 3(−x + 4). The binomial (−x + 4) is the opposite of (x − 4), so the pairs do not match. You can still rescue it by taking a −1 out of (−x + 4), since −x + 4 = −(x − 4), but it is quicker to take out the negative at the start.

The principle: choose the sign of each pair’s GCF so that the leftover binomials match. In a polynomial like this one, where the first pair leaves a binomial that starts with a positive x, a negative third term means the second pair needs a negative GCF, and a positive third term, as in x3 + 3x2 + 2x + 6, means a positive GCF.

Hidden 1s and −1s

Sometimes a pair has no visible common factor except 1 or −1. Take that out, and do not let it disappear. In x3 + 5x2 − x − 5 the second pair is −x − 5, and its GCF is −1:

x3 + 5x2 − x − 5 = x2(x + 5) − 1(x + 5) = (x + 5)(x2 − 1)

If you drop the −1, you would write (x + 5)(x2), which is wrong. Expanding shows it: (x + 5)(x2) = x3 + 5x2, which is missing the last two terms.

Then look at each factor. x2 − 1 is a difference of squares (see the Difference of Squares lesson), so (x + 5)(x2 − 1) = (x + 5)(x + 1)(x − 1), which is completely factored.

When the first pairing fails: rearrange

Suppose the terms arrive in a different order: 6xy + 4 + 3x + 8y. Pair the first two and the last two: (6xy + 4) + (3x + 8y). The first pair has the GCF 2, giving 2(3xy + 2), but 3x + 8y has no common factor other than 1. The pairs do not leave matching binomials.

Rearranging is allowed, because the order of addition does not change the value, as long as every sign stays with its term. Put terms that share a variable next to each other: 6xy + 3x + 8y + 4. Now (6xy + 3x) + (8y + 4) = 3x(2y + 1) + 4(2y + 1) = (2y + 1)(3x + 4).

Check by expanding: (2y + 1)(3x + 4) = 6xy + 8y + 3x + 4, the same four terms.

For a polynomial in one variable, write it in standard form first. That usually gives the right pairing already.

When grouping will not work

Try x3 + 2x2 + 3x + 5. The first pairing gives x2(x + 2) + (3x + 5), and (x + 2) does not match (3x + 5). Rearranging to (x3 + 3x) + (2x2 + 5) gives x(x2 + 3) + (2x2 + 5), and (x2 + 3) does not match (2x2 + 5) either. The only other way to pair the terms, (x3 + 5) + (2x2 + 3x), leaves (x3 + 5) and x(2x + 3), which do not match. A sign fix cannot help, because these binomials differ by more than a sign.

When no arrangement gives matching binomials, grouping fails. That does not prove the polynomial is prime. Other tools can still work: a GCF you missed, a pattern such as a difference of squares, or methods from later courses. For example, x3 + x2 + x − 3 fails every pairing, yet it equals (x − 1)(x2 + 2x + 3). The polynomial x3 + 2x2 + 3x + 5 does happen to be prime (it cannot be factored using integers), but showing that needs a tool from a later course. For now, the practical signal is simple: if you cannot make the binomials match, grouping does not apply.

Grouping and trinomials: splitting the middle term

Grouping also factors many trinomials ax2 + bx + c, by turning them into four terms. Find two integers m and n with

  • m × n = a × c, and
  • m + n = b.

Rewrite bx as mx + nx, then group. Why a × c? When m × n = a × c, the numbers a and m are in the same ratio as n and c, so the binomials ax + m and nx + c are multiples of each other and the pairs match. (This assumes none of a, m, n and c is 0.)

For 2x2 + 7x + 6: a × c = 12, and the pair 3 and 4 has product 12 and sum 7. So 2x2 + 3x + 4x + 6 = x(2x + 3) + 2(2x + 3) = (2x + 3)(x + 2).

If you take out any GCF first and then no pair of integers has product a × c and sum b, the trinomial is prime. (For trinomials this search is a complete test. That is different from four-term polynomials, where failing to group does not prove a polynomial is prime.) This is the idea behind the ac method in the Factoring Trinomials lesson, which covers trinomials in full. Here it is only a short link.

Once a quadratic is factored, you can read off where it equals zero, because a product is 0 when one of its factors is 0. For example, (2x + 3)(x + 2) = 0 when x = −3/2 or x = −2. You will solve equations this way in Solving Quadratics by Factoring.

Always finish with a check

Expand your answer. You should get the original polynomial back (the terms may appear in a different order). Then look at every factor and ask whether it can be factored further, such as a difference of squares or a GCF you can still take out. Keep any GCF you took out at the start in the final answer.

Key words

Factoring by grouping
Splitting a four-term polynomial into two pairs, taking the GCF out of each pair, and pulling out the binomial that both pairs leave behind.
Common binomial factor
A binomial that appears as a factor in every term, such as (x + 3) in x2(x + 3) + 2(x + 3). Pulling it out gives (x + 3)(x2 + 2).
Greatest common factor (GCF)
The largest factor that divides every term of a polynomial or pair of terms. The GCF itself is positive: for 2x + 6 it is 2. When the signs call for it, you pull out the negative of the GCF instead, such as −3 from −3x + 12 (called a negative GCF in this lesson).
Binomial
A polynomial with exactly two terms, such as x + 3.
Standard form (descending powers)
A polynomial written with the highest power first, then the next highest, and so on, such as x3 + 3x2 + 2x + 6.
Prime (cannot be factored using integers)
A polynomial that cannot be written as a product of simpler polynomials with integer coefficients, other than by taking out 1 or −1. For example, x2 + 2 is prime.
Completely factored
Written as a product of any GCF and prime factors, so that no factor can be factored further using integers.

Worked examples

Example 1. Factor x3 + 3x2 + 2x + 6.

  1. Look for a GCF of all four terms. The coefficients 1, 3, 2 and 6 have no common factor other than 1, and the constant 6 has no x, so there is none.
  2. Pair the terms, keeping signs: (x3 + 3x2) + (2x + 6).
  3. Factor each pair: x3 + 3x2 = x2(x + 3) and 2x + 6 = 2(x + 3). The expression is now x2(x + 3) + 2(x + 3).
  4. Both terms contain (x + 3). Pull it out: (x + 3)(x2 + 2).
  5. Check by expanding: (x + 3)(x2 + 2) = x3 + 2x + 3x2 + 6. In standard form that is x3 + 3x2 + 2x + 6, which matches.
  6. The factor x2 + 2 cannot be factored using integers, so the answer is complete.

Answer: Answer: (x + 3)(x2 + 2)

Example 2. Factor x3 − 4x2 − 3x + 12.

  1. There is no GCF for all four terms. Pair them, keeping each sign with its term: (x3 − 4x2) + (−3x + 12).
  2. First pair: x3 − 4x2 = x2(x − 4).
  3. Second pair: the first term is −3x, so take out −3: −3x + 12 = −3(x − 4). Check: (−3)(x) = −3x and (−3)(−4) = +12.
  4. The expression is now x2(x − 4) − 3(x − 4), and both binomials are (x − 4).
  5. Pull it out: (x − 4)(x2 − 3).
  6. Check by expanding: (x − 4)(x2 − 3) = x3 − 3x − 4x2 + 12, which is x3 − 4x2 − 3x + 12 in standard form.
  7. What if you took out +3? Then −3x + 12 = 3(−x + 4), and (−x + 4) does not match (x − 4). It is the opposite of it, so you would have to take out a −1 as well to fix it.

Answer: Answer: (x − 4)(x2 − 3)

Example 3. Factor x3 + 5x2 − x − 5 completely.

  1. Pair the terms: (x3 + 5x2) + (−x − 5).
  2. First pair: x2(x + 5). Second pair: its GCF is −1, because −x − 5 = −1(x + 5). Do not drop the 1.
  3. The expression is x2(x + 5) − 1(x + 5). Pull out (x + 5): (x + 5)(x2 − 1).
  4. Inspect each factor. x2 − 1 = x2 − 12 is a difference of squares, so it factors again: (x + 1)(x − 1).
  5. The completely factored form is (x + 5)(x + 1)(x − 1).
  6. Check by expanding: (x + 1)(x − 1) = x2 − 1, and (x + 5)(x2 − 1) = x3 − x + 5x2 − 5, which is x3 + 5x2 − x − 5 in standard form.

Answer: Answer: (x + 5)(x + 1)(x − 1)

Example 4. Factor 3x3 + 6x2 + 15x + 30 completely.

  1. Take out the GCF of all four terms first. Every coefficient is divisible by 3: 3x3 + 6x2 + 15x + 30 = 3(x3 + 2x2 + 5x + 10).
  2. Group what is inside the parentheses: (x3 + 2x2) + (5x + 10) = x2(x + 2) + 5(x + 2) = (x + 2)(x2 + 5).
  3. Keep the 3 out front: 3(x + 2)(x2 + 5).
  4. Check by expanding: (x + 2)(x2 + 5) = x3 + 5x + 2x2 + 10, so 3(x + 2)(x2 + 5) = 3x3 + 15x + 6x2 + 30, which is the original in a different order.
  5. If you skipped the GCF, grouping would still give (x + 2)(3x2 + 15), which equals the original. But it is not completely factored, because 3x2 + 15 = 3(x2 + 5) still has a common factor.

Answer: Answer: 3(x + 2)(x2 + 5)

Example 5. Factor 6xy + 4 + 3x + 8y.

  1. There is no GCF for all four terms. Try the pairs as given: (6xy + 4) + (3x + 8y). The first pair has GCF 2, giving 2(3xy + 2). The second pair, 3x + 8y, has no common factor other than 1. The binomials cannot match, so rearrange.
  2. The order of addition does not change the value, so reorder the terms, keeping each sign with its term. Put terms that share a factor next to each other: 6xy and 3x share 3x, and 8y and 4 share 4. That gives 6xy + 3x + 8y + 4.
  3. Pair again: (6xy + 3x) + (8y + 4). First pair: 3x(2y + 1). Second pair: 4(2y + 1).
  4. Both binomials are (2y + 1): 3x(2y + 1) + 4(2y + 1) = (2y + 1)(3x + 4).
  5. Check by expanding: (2y + 1)(3x + 4) = 6xy + 8y + 3x + 4, which has the same four terms as the original.

Answer: Answer: (2y + 1)(3x + 4)

Example 6. Try to factor x3 + 2x2 + 3x + 5 by grouping.

  1. There is no GCF for all four terms. Pair the terms: (x3 + 2x2) + (3x + 5).
  2. First pair: x2(x + 2). Second pair: 3 and 5 have no common factor other than 1, so it stays 1(3x + 5). The binomials (x + 2) and (3x + 5) do not match.
  3. Rearrange: (x3 + 3x) + (2x2 + 5). First pair: x(x2 + 3). Second pair: no common factor, so it stays (2x2 + 5). The binomials (x2 + 3) and (2x2 + 5) do not match either.
  4. The last way to pair the terms is (x3 + 5) + (2x2 + 3x). The first pair has no common factor, and the second is x(2x + 3), so (x3 + 5) and (2x + 3) do not match either.
  5. A negative GCF would only change signs, and these binomials differ by more than a sign. So grouping does not work.
  6. Grouping failing does not by itself make a polynomial prime. Some polynomials that fail every pairing still factor in other ways. This particular one does happen to be prime, but showing that needs a tool from a later course. In general, “grouping fails” means “try another method,” not automatically “prime.”

Answer: Answer: Grouping does not work for x3 + 2x2 + 3x + 5. (This polynomial happens to be prime, but grouping failing does not prove that.)

Example 7. Factor 2x2 + 7x + 6 by splitting the middle term.

  1. Here a = 2, b = 7 and c = 6, so a × c = 12. Look for two integers with product 12 and sum 7.
  2. The pairs with product 12 are 1 and 12, 2 and 6, and 3 and 4 (and the same pairs with both negative). Only 3 + 4 = 7.
  3. Split 7x as 3x + 4x: 2x2 + 3x + 4x + 6.
  4. Group: (2x2 + 3x) + (4x + 6) = x(2x + 3) + 2(2x + 3) = (2x + 3)(x + 2).
  5. Check by expanding: (2x + 3)(x + 2) = 2x2 + 4x + 3x + 6 = 2x2 + 7x + 6.
  6. Splitting as 4x + 3x instead gives 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3), the same product. The order of the two split terms does not matter.

Answer: Answer: (2x + 3)(x + 2)

Example 8. Factor 6x2 − 5x − 6 by splitting the middle term.

  1. Here a × c = 6 × (−6) = −36 and b = −5. Look for two integers with product −36 and sum −5.
  2. The product is negative, so one number is positive and one is negative. Try pairs: 1 and −36 (sum −35), 2 and −18 (sum −16), 3 and −12 (sum −9), 4 and −9 (sum −5). That works.
  3. Split −5x as 4x − 9x: 6x2 + 4x − 9x − 6.
  4. Group: (6x2 + 4x) + (−9x − 6) = 2x(3x + 2) − 3(3x + 2). The second pair needed −3: −9x − 6 = −3(3x + 2). Then pull out the binomial: (3x + 2)(2x − 3).
  5. Check by expanding: (3x + 2)(2x − 3) = 6x2 − 9x + 4x − 6 = 6x2 − 5x − 6.

Answer: Answer: (3x + 2)(2x − 3)

In plain English

Say you have four terms, like x3 + 3x2 + 2x + 6. No single number or letter divides all four. But you can split them into two pairs.

In the first pair, x3 + 3x2, you can pull out x2. That leaves x2(x + 3). In the second pair, 2x + 6, you can pull out 2. That leaves 2(x + 3).

Now look closely. Both pairs have (x + 3) left over. That matching bracket is a common factor, so you can pull it out too. The answer is (x + 3)(x2 + 2).

If the brackets do not match, check your signs, or put the terms in a different order. Sometimes nothing works, and that is okay. Always multiply your answer back out to check it.

See it

Pair the terms, factor each pair, pull out the match x3 + 3x2 + 2x + 6 pair 1 pair 2 x2 (x + 3) + 2 (x + 3) GCF of pair 1 is x2 GCF of pair 2 is 2 Same binomial in both: (x + 3) pull out the common binomial (x + 3)(x2 + 2) Check: (x + 3)(x2 + 2) = x3 + 2x + 3x2 + 6 = x3 + 3x2 + 2x + 6
Pair the terms, take out each pair’s GCF, and pull out the binomial both pairs share: x3 + 3x2 + 2x + 6 = (x + 3)(x2 + 2).
The sign you take out decides whether the binomials match For x3 − 4x2 − 3x + 12, the first pair gives x2(x − 4). Now the second pair, −3x + 12: Take out +3: no match −3x + 12 = 3(−x + 4) x2(x − 4) + 3(−x + 4) (x − 4) vs (−x + 4) Opposites, so they do not match. You would need to take out a −1 as well. Take out −3: match −3x + 12 = −3(x − 4) x2(x − 4) − 3(x − 4) (x − 4) and (x − 4) The same binomial: pull it out. x2 − 3 is what is left over. Answer: (x − 4)(x2 − 3) A mismatch of opposites can be fixed by taking out −1: −x + 4 = −(x − 4). Check: (x − 4)(x2 − 3) = x3 − 3x − 4x2 + 12, the original polynomial.
Choose the sign of each pair’s GCF so the binomials match. Taking out −3 gives (x − 4) twice, while +3 gives the opposite binomial (−x + 4).

The same idea, the fun way

Pick the one that clicks. Each one explains the same concept from a different angle.

Relationships

Same trip, opposite view

Two friends describe the same walk. Maya says, “The park is 4 blocks east of the school.” Noor says, “The school is 4 blocks west of the park.” It is the same fact, seen from opposite ends. To compare their directions, one of them has to flip the point of view.

How it connects: Think of positions along a street, with the school at 4 and the park at x. How far east the park is from the school is x − 4, and how far east the school is from the park is 4 − x. They look alike but are opposites: 4 − x = −(x − 4), like east versus west. To make them match, take a −1 out of one of them, and that −1 stays in the product as a sign change. The limit: this only fixes binomials that are opposites. (x − 4) and (x + 4), or (x + 2) and (3x + 5), are not mirror images, so no −1 will fix them.

Proverb

If at first you don’t succeed, try, try again

This proverb means that when your first attempt fails, you should try again, often in a new way.

How it connects: If the first pairing does not give matching binomials, rearrange the terms or fix a sign and try again. 6xy + 4 + 3x + 8y fails when you pair the first two terms and the last two as written, but 6xy + 3x + 8y + 4 works. The limit: the number of tries is finite. After a sign check and the other pairings fail to match, as with x3 + 2x2 + 3x + 5, stop and say grouping does not work (that does not prove the polynomial is prime). Trying forever will not help.

Idiom

Common ground

To find “common ground” means to find something both sides agree on, even if they disagree about everything else.

How it connects: The pairs x2(x + 3) and 2(x + 3) differ in front, but they share common ground: (x + 3). That shared binomial is what you pull out. The limit: sometimes there is no common ground. The binomials (x + 2) and (3x + 5) share nothing, and then grouping fails.

Food

The combo meal

A combo meal costs c + 3 dollars: a sandwich for c dollars plus fries for 3 dollars. Your team orders n combos, and the coaches order 4 more of the same combo. The total cost in dollars is n(c + 3) + 4(c + 3). Since every combo is identical, that is (n + 4) combos at (c + 3) dollars each.

How it connects: n(c + 3) + 4(c + 3) = (n + 4)(c + 3). The combo is the common binomial. Check with n = 10 and c = 5: 10(8) + 4(8) = 112, and 14 × 8 = 112. The limit: it works because the coaches’ combos are exactly the same as the team’s. If their combo cost c + 5 dollars, there would be no common factor to pull out.

Gaming

The matching card game

In a memory card game you flip two cards and keep them only if they match. A near-miss does not count, so you flip them back and try other cards.

How it connects: After you take out each pair’s GCF, the two leftover binomials are your two cards. They must match to pull out. (x − 4) and (x − 4) match. (x − 4) and (−x + 4) are a near-miss, and here math is kinder than the game: they are opposites, so taking out a −1 turns one into the other. The limit: in the game you can keep flipping cards, but a polynomial has only a few ways to pair its terms, and some, like x3 + 2x2 + 3x + 5, have no matching pair at all.

Everyday joke

Taking out the negative

Why do math teachers love taking out the negative? It leaves everyone inside the parentheses feeling more positive.

How it connects: −3x + 12 = −3(x − 4): taking out −3 leaves the binomial (x − 4), whose first term is positive. The limit: the negative is not gone. It moved out front as a factor, and −3(x − 4) still equals −3x + 12. Not everything inside turns positive: the −4 is still negative. Only the first term flips to positive. Also, taking out a negative is only the right move when it makes the binomials match.

Memory help

Acronym

Great Pairs Match Exactly

Great Pairs Match Exactly. G: take out the GCF of all four terms first. P: Pair up the terms (first two, last two) and take out each pair’s GCF. M: the leftover binomials must Match. If they are opposites, take out a negative to fix it. E: Extract the common binomial as a factor, then Expand to check.

Rhyme

The grouping rhyme

Pair them two by two, take each GCF through. Same bracket twice? Pull it out and you’re through. Brackets disagree? Check the sign you took out, or reorder the terms and try anew. Expand to see if your answer is true.

Picture it

Underline the leftovers

After you factor each pair, underline the binomial left in each one. If the two underlined binomials look identical, pull them out. If one is the mirror image of the other, like x − 4 and −x + 4, fix the sign. If they are truly different, rearrange the terms, or stop because grouping does not apply.

Sketch it

Two bubbles

Draw a bubble around the first two terms and another around the last two. Write each bubble’s GCF outside it and the leftover binomial inside it. If both bubbles hold the same binomial, you have found a common factor. If not, redraw the bubbles with the terms in a different order.

Trap alert

A sum is not a product

x2(x + 3) + 2(x + 3) is not yet a product of factors. It is a sum of two terms. Pull out the common binomial to get the product (x + 3)(x2 + 2).

Trap alert

The sign belongs to the term

In x3 − 4x2 − 3x + 12 the second pair is −3x + 12, not 3x + 12. Take out −3 to get −3(x − 4), which matches the first pair’s (x − 4).

Study notes

Key ideas

  • Factoring by grouping works for certain four-term polynomials: those whose two pairs leave the same binomial. If all four terms share a GCF, take it out first.
  • Pair the terms, take out each pair’s GCF, and look for the same binomial in both. Then pull it out: ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y).
  • Choose the sign of each GCF so the binomials match. Taking out a −1 turns (−x + 4) into −(x − 4).
  • If the first pairing does not work, check signs and then try rearranging. If no arrangement matches, grouping fails. That does not prove the polynomial is prime.
  • Grouping also splits the middle term of a trinomial: find m and n with m × n = a × c and m + n = b.
  • Always check by expanding, and factor completely.

Steps

  1. G: take out the GCF of all four terms, if there is one.
  2. P: pair the terms (first two, last two), keeping each sign with its term, and take out each pair’s GCF.
  3. M: make sure the leftover binomials match. Fix the sign with a negative GCF, or rearrange the terms.
  4. E: extract the common binomial, then expand to check.
  5. Look at each factor and keep factoring if you can.

Formulas

  • Grouping pattern: ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)
  • Taking out a negative: −px + pq = −p(x − q)
  • Splitting the middle term: ax2 + bx + c = ax2 + mx + nx + c, where m + n = b and m × n = a × c

Watch out

  • x2(x + 3) + 2(x + 3) is still a sum. Pull out the binomial to get a product.
  • The sign belongs to the term: −3x + 12 = −3(x − 4), not 3(x − 4).
  • Do not drop a 1 or a −1: x2(x + 5) − 1(x + 5) = (x + 5)(x2 − 1), not (x + 5)(x2).
  • When all four terms share a GCF, take it out first, or you will have a common factor left over at the end.
  • If the binomials do not match, do not force them. Check signs, try rearranging, or accept that grouping fails (which does not mean the polynomial is prime).
  • To split the middle term, the product is a × c and the sum is b. It is not the other way around.

Why it works

  • Why does grouping work? x2(x + 3) + 2(x + 3) has (x + 3) in both terms, so the distributive property run backward gives (x2 + 2)(x + 3).
  • Why does the sign matter? (x − 4) and (4 − x) are opposites, not equal. Only identical binomials can be pulled out as one factor.
  • Why may you rearrange the terms? Addition can be done in any order, as long as each sign stays with its term.

Common mistakes

✗ Mistakex2(x + 3) + 2(x + 3) is the final answer.
✓ InsteadThat is still a sum of two terms. Pull out the common binomial to get the product (x + 3)(x2 + 2).
✗ MistakeIn x3 − 4x2 − 3x + 12, the second pair is 3x + 12, so take out 3.
✓ InsteadThe sign belongs to the term: the pair is −3x + 12. Take out −3 to get −3(x − 4), which matches the first pair’s (x − 4).
✗ Mistakex3 + 5x2 − x − 5 = x2(x + 5) − (x + 5) = (x + 5)(x2) (the −1 disappears).
✓ Instead−(x + 5) means −1(x + 5), so the −1 stays: (x + 5)(x2 − 1). That factors further into (x + 5)(x + 1)(x − 1).
✗ MistakeIf the binomials do not match, just write them as the same binomial anyway.
✓ InsteadDifferent binomials mean that arrangement does not work. Check your signs, try rearranging the terms, or accept that grouping fails. Never change a binomial to force a match.
✗ MistakeTo split the middle term, find two numbers with sum a × c and product b.
✓ InsteadThe product is a × c and the sum is b. For 2x2 + 7x + 6, look for product 12 and sum 7, which gives 3 and 4.

Teach it back

Explain to a friend how to factor x3 − 9x2 − 2x + 18 by grouping, why the second pair needs a negative GCF, and how to check your answer.

Explaining it out loud shows what you really know. Did you…

  • I paired the terms as (x3 − 9x2) + (−2x + 18) and took out x2 from the first pair: x2(x − 9).
  • I took out −2 (not 2) from the second pair, so it became −2(x − 9) and the binomials matched.
  • I pulled out (x − 9) to get (x − 9)(x2 − 2).
  • I checked by expanding: (x − 9)(x2 − 2) = x3 − 2x − 9x2 + 18, which is the original polynomial.

Your 3-week memory plan

  1. TodayRead the lesson, then try the flashcards.
  2. TomorrowRecall it from memory first, then check.
  3. In 3 daysTake the true/false and practice questions.
  4. In 1 weekMix it with older topics and teach it back.
  5. In 3 weeksA last quick review before it fades.

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Factoring Out the Greatest Common FactorFactoring TrinomialsDifference of SquaresMultiplying PolynomialsThe Distributive Property