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Factoring by grouping: pair the terms, pull out the shared binomial

Four terms, two pairs, one shared binomial. The points get lost when the sign pulled from the second pair is wrong, when the two binomials almost match and get forced together anyway, and when the answer is left as a sum instead of a product.

Short answer

Split the four terms into two pairs. Factor the greatest common factor out of each pair so that both pairs leave the same binomial in parentheses. Then factor that binomial out of the whole thing: x2(x + 3) + 2(x + 3) = (x + 3)(x2 + 2). If the two binomials don't match, check the sign you pulled out of the second pair, or try a different pairing. The answer is a product; if there is still a plus sign between two groups, you aren't done.

Why it works

Grouping is the distributive property run backwards, twice. After the first round you have something like x2(x + 3) + 2(x + 3). Call the repeated binomial u for a moment: that's x2u + 2u, and u is a common factor, so it comes out as u(x2 + 2). Put (x + 3) back in for u and you have the answer. The binomial has to be exactly the same in both groups, including its sign, or it isn't a common factor at all.

The same move finishes the AC method: once the middle term of a trinomial is split into two, you have four terms and you group them.

Example 1: the basic case

Factor x3 + 3x2 + 2x + 6.

  1. (x3 + 3x2) + (2x + 6)Pair the first two terms and the last two.
  2. x2(x + 3) + 2(x + 3)Factor x² out of the first pair and 2 out of the second. Both leave (x + 3).
  3. (x + 3)(x2 + 2)Factor out the shared (x + 3). Check by multiplying: x³ + 2x + 3x² + 6. Same four terms.

Example 2: a minus sign in the second pair

Factor 2x3 − 5x2 − 6x + 15.

  1. (2x3 − 5x2) + (−6x + 15)Pair up. The second pair starts with a negative term, which is where the sign has to be watched.
  2. x2(2x − 5) − 3(2x − 5)The first pair gives (2x − 5). To make the second pair give the same thing, pull out −3: −3 · 2x = −6x and −3 · (−5) = +15.
  3. (2x − 5)(x2 − 3)Factor out (2x − 5). The leftover factor is x² and −3, in that order. Check: 2x³ − 6x − 5x² + 15.

Rule of thumb: if the third term is negative, the factor you pull out of the second pair is negative too. Then the sign of the fourth term flips inside the parentheses, and that flip is what makes the binomials match.

Example 3: the −1, and factoring all the way

Factor x3 + 4x2 − x − 4.

  1. x2(x + 4) − 1(x + 4)The second pair, −x − 4, has no common factor except −1. Write the 1 so it doesn't vanish: −1 · x = −x and −1 · 4 = −4.
  2. (x + 4)(x2 − 1)Factor out (x + 4). The −1 becomes the second term of the other factor.
  3. (x + 4)(x − 1)(x + 1)x² − 1 is a difference of squares, so it factors again. "Factor completely" means keep going until nothing does.

Example 4: take out the GCF first

Factor 4x3 + 8x2 − 12x − 24.

  1. 4(x3 + 2x2 − 3x − 6)Every term is a multiple of 4. Pulling it out first keeps the numbers inside small.
  2. 4[x2(x + 2) − 3(x + 2)]Group inside the brackets. The second pair needs −3 so it leaves (x + 2).
  3. 4(x + 2)(x2 − 3)Factor out (x + 2), and keep the 4 in front. x² − 3 isn't a difference of integer squares, so this is complete.

Common mistakes, and the exact line they happen on

1. Pulling a positive out of the second pair when it needed a negative

Factor 2x3 − 5x2 − 6x + 15 (Example 2).

  1. x2(2x − 5) + 3(−2x + 5)This is the mistake. The arithmetic is right: 3(−2x + 5) really is −6x + 15. But (−2x + 5) isn't (2x − 5); it's the opposite of it. There is no common binomial to factor out, so the next line can't happen.

Pull out −3 instead and the second group becomes −3(2x − 5). If you've already written the line above, you can rescue it: (−2x + 5) = −(2x − 5), so +3(−2x + 5) = −3(2x − 5).

2. Losing the −1, or flipping the wrong sign with it

Factor x3 + 4x2 − x − 4 (Example 3).

  1. x2(x + 4) − (x − 4)This is the mistake. −(x − 4) = −x + 4, not −x − 4. Factoring out a negative flips every sign inside, including the 4's. The correct group is −1(x + 4).
  2. (x + 4)x2The other version of the mistake. From the correct line x²(x + 4) − (x + 4), the −1 that was the whole second group got dropped. Multiply back: (x + 4)x² = x³ + 4x², only two of the four terms.

Writing − 1(x + 4) with the 1 visible costs nothing and keeps it from disappearing on the next line.

3. Stopping at a sum

In Example 1 the line x2(x + 3) + 2(x + 3) is halfway, not an answer. It has two terms added together, so it isn't factored. Factored means a product, and the only way this line is a product is after (x + 3) comes out. The same goes for stopping at (x + 4)(x2 − 1) in Example 3 when the question says "completely."

4. Declaring it doesn't factor because the first pairing didn't work

Factor xy + 6 + 2x + 3y.

  1. (xy + 6) + (2x + 3y)This is the mistake. Neither pair has a common factor, and the student writes "prime." But the terms are just in an awkward order.
  2. xy + 2x + 3y + 6Rearrange so each pair shares something: the first two share x, the last two share 3.
  3. x(y + 2) + 3(y + 2)Now both pairs leave (y + 2).
  4. (x + 3)(y + 2)Factor out (y + 2). Check: xy + 2x + 3y + 6.

With four terms there are three ways to pair them. Try a second pairing before giving up.

Practice

  1. Factor x3 + 5x2 + 3x + 15.

    Show answer

    x2(x + 5) + 3(x + 5) = (x + 5)(x2 + 3).

  2. Factor completely: 3x3 − 2x2 − 12x + 8.

    Show answer

    x2(3x − 2) − 4(3x − 2) = (3x − 2)(x2 − 4) = (3x − 2)(x − 2)(x + 2).

  3. Factor ab − 3a + 2b − 6.

    Show answer

    a(b − 3) + 2(b − 3) = (a + 2)(b − 3).

  4. Factor completely: 2x3 + 10x2 − 8x − 40.

    Show answer

    GCF first: 2(x3 + 5x2 − 4x − 20) = 2[x2(x + 5) − 4(x + 5)] = 2(x + 5)(x2 − 4) = 2(x + 5)(x − 2)(x + 2).

  5. Factor x2 + 7x + 12 by splitting the middle term and grouping.

    Show answer

    3 and 4 multiply to 12 and add to 7: x2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4).

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