What is factoring by grouping?
Factoring by grouping is a method for polynomials with four terms. You treat the first two terms and the last two terms as separate groups, factor each one, and hope the same binomial shows up in both.
It relies on the distributive property running backward. If two pieces both contain (x + 3), you can pull (x + 3) out the same way you would pull out a common number. Grouping is also the second half of the ac method for trinomials, so it comes up more often than you might expect.
How to factor by grouping, step by step, like the calculator does
Write the polynomial in descending powers, group the terms in pairs, factor each pair, then factor out the common binomial. Here is x³ + 3x² + 2x + 6, the same example the calculator shows.
Check by multiplying: (x + 3)(x² + 2) = x³ + 2x + 3x² + 6, which matches once you reorder the terms. The factor x² + 2 has no real roots, so the calculator labels it irreducible over the rationals and stops.
If the terms come in a jumbled order, like x³ + 2x + 3x² + 6, put them in descending powers first. The calculator does this for you and gives the same answer, (x + 3)(x² + 2).
How can you tell if grouping will work?
For a cubic ax³ + bx² + cx + d, grouping in the usual order works when a·d = b·c. That quick test tells you whether the two pairs will share a binomial before you do any factoring.
In x³ + 3x² + 2x + 6, a·d = 1 × 6 = 6 and b·c = 3 × 2 = 6. They match, so grouping works. In x³ − 4x² + 4x − 16, a·d = −16 and b·c = −16, so it works too, giving (x − 4)(x² + 4).
Now try x³ + 2x² + 3x + 4. Here a·d = 4 but b·c = 6, so the pairs x²(x + 2) and (3x + 4) share nothing. This one has no rational roots at all, and the calculator reports it as irreducible over the rationals.
| Polynomial | Test a·d = b·c | Result |
|---|---|---|
| x³ + 3x² + 2x + 6 | 1 × 6 = 3 × 2 | (x + 3)(x² + 2) |
| 2x³ + 6x² + 5x + 15 | 2 × 15 = 6 × 5 | (x + 3)(2x² + 5) |
| x³ − 5x² − 2x + 10 | 1 × 10 = (−5)(−2) | (x − 5)(x² − 2) |
| x³ + 2x² + 3x + 4 | 4 ≠ 6 | Does not group |
What if the third term is negative?
Factor a negative out of the second pair so the binomials match. This is where most sign errors happen when factoring by grouping.
Look at 2x³ + 4x² − x − 2. The first pair gives 2x²(x + 2). For the second pair, take out −1, not 1: −x − 2 = −1(x + 2). Now both pieces share (x + 2), and the answer is (x + 2)(2x² − 1).
If you had written the second pair as 1(−x − 2), the binomials would not match and it would look like grouping failed. When the third term is negative, pull out the negative.
How is grouping used to factor trinomials?
In the ac method, you split the middle term of a trinomial into two terms, which turns it into a four-term polynomial you can group. So learning to factor by grouping also teaches you the last step of trinomial factoring.
Take 6x² + 7x + 2. Multiply a·c = 6 × 2 = 12, then find two numbers that multiply to 12 and add to 7. Those are 3 and 4.
The split always passes the a·d = b·c test, because you chose the two middle numbers to multiply to a·c. Here 6 × 2 = 3 × 4 = 12. The order of 3x and 4x does not matter: 6x² + 4x + 3x + 2 groups to the same answer.
What are common mistakes when factoring by grouping?
The most common mistake is stopping at x²(x + 3) + 2(x + 3). That is still a sum of two terms, so it is not factored yet. You have to pull out (x + 3) to get a product.
Another mistake is skipping the GCF of the whole polynomial. In 4x³ + 12x² + 8x + 24, take out 4 first to get 4(x³ + 3x² + 2x + 6), then group. Last, check the leftover factor. If it is a difference of squares like x² − 4, keep factoring.