Lesson · Polynomials and Factoring · Factoring

Factoring by Grouping

With four terms and no common factor, split into pairs, take the GCF out of each, and hope the same binomial is left behind. What makes the method work is that the shared binomial is itself a common factor.

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x³ + 3x² − 4x − 12 = x²(x + 3) − 4(x + 3) both pairs leave (x + 3) = (x + 3)(x² − 4) the binomial is the common factor = (x + 3)(x − 2)(x + 2) keep going until nothing factors

The method in four lines

Core ideas

What this lesson establishes

Grouping is the first technique with a condition attached. It applies when the pairs cooperate, and knowing when they do not is as important as the method itself.

The shared binomial is an ordinary common factor

After the first step you have x²·P − 4·P with P = (x + 3). That is ab + ac again, with a = P. Nothing new is being assumed — the common factor just happens to be a bracket.

It only works if BOTH pairs leave the same binomial

If one pair leaves (x + 3) and the other (x − 3), there is no common factor and the step has failed. You may not adjust one of them to match; you regroup, or you conclude this technique does not apply.

Failure of the method is not failure to factor

If no pairing works, all you have learned is that grouping did not crack it. Whether the polynomial factors at all is settled by the root theorems in “Synthetic Division and the Remainder Theorem” and “The Factor Theorem and Rational Roots”, not here.

The material

Pairing terms so a common factor appears

The method, the sign trap in the second pair, and what to do when the brackets disagree.

Grouping is the standard move for four terms. Take the GCF out of the first two and out of the last two, and if the same binomial is left in both places, take that out.

x³ + 3x² − 4x − 12 (x³ + 3x²) + (−4x − 12) pair them x²(x + 3) + (−4)(x + 3) GCF of each pair (x + 3)(x² − 4) (x + 3) is common

Theorem · Why the last step is legal

For any polynomials A, B and P, A·P + B·P = (A + B)·P.

This is the distributive law with the common factor P a binomial rather than a monomial. The law never cared what P was.

The sign in the second pair

When the third term is negative, take the negative out with the GCF. In the example above the second pair is −4x − 12; factoring out −4 rather than 4 gives −4(x + 3), which matches the first bracket. Factoring out +4 gives 4(−x − 3), which does not — and the method appears to fail when it has not.

The rule of thumb: make the second bracket match the first. If it cannot be made to match by choosing the sign, the pairing is genuinely wrong.

Regrouping

The order of the terms is yours to choose, and sometimes the given order is unhelpful. Consider ax + by + ay + bx. Pairing as given leaves a-something and b-something with nothing shared. Reorder:

ax + ay + bx + by = a(x + y) + b(x + y) = (x + y)(a + b)

Rearranging terms of a sum is always legal, so trying another pairing costs nothing but a line. With four terms there are only three genuinely different pairings, so the search is short and finite.

Finish the job

A factorisation is complete only when no factor factors further. In the worked example (x + 3)(x² − 4) is not the end: x² − 4 is a difference of squares, so the complete answer is (x + 3)(x − 2)(x + 2).

Stopping early is the most common way to lose marks on a factoring question that was otherwise done correctly. After every step, look at each factor again.

Example · When grouping does not apply

x³ + x² + 2x + 6 gives x²(x + 1) + 2(x + 3). The brackets differ, and no reordering fixes it. The honest conclusion is "grouping fails here" — not "this does not factor". That second claim needs “The Factor Theorem and Rational Roots”.

Interactive

Factoring by grouping

All three pairings tried, not just the obvious one

Method

Factoring four terms by grouping

Step 1 is the same as it was in “Factoring Out the Greatest Common Factor”, and it is still not optional.

Take out the overall GCF first

Grouping is stated for a polynomial with no common factor across all four terms. If there is one, remove it and group what is left.

Pair the terms and take the GCF of each pair

Usually first-two and last-two. Where the second pair begins with a minus, factor out the negative so the bracket has a chance of matching.

Compare the two brackets

Identical brackets mean the method has worked. Different brackets mean try another pairing — there are only three — before concluding anything.

Take out the shared binomial, then factor each factor

A·P + B·P = (A + B)·P. Then look at both results: a difference of squares or a factorable trinomial often remains, and the answer is not complete until nothing factors.

Grouping

Score 0 / 0

Choose the answer you can justify.

Two groupings, one with a sign trap

The first needs the negative taken out of the second pair; the second is clean but finishes early.

A. x³ + 3x² − 4x − 12 (x³ + 3x²) + (−4x − 12) x²(x + 3) + (−4)(x + 3) take out −4, not +4 (x + 3)(x² − 4) (x + 3)(x − 2)(x + 2) x² − 4 is a difference of squares check at x = 1: 1 + 3 − 4 − 12 = −12 (4)(−1)(3) = −12 B. 6x³ − 9x² + 4x − 6 3x²(2x − 3) + 2(2x − 3) (2x − 3)(3x² + 2) 3x² + 2 is positive for every real x, so it has no real linear factor.

In A, factoring +4 out of the second pair would have given 4(−x − 3), and the two brackets would not have matched. The method would have looked broken; only the sign choice was.

In B the second factor 3x² + 2 cannot be factored over the reals — it is positive for every x, so it has no root and therefore no linear factor. Recognising when to stop is part of the technique.

For a faded grouping, begin 2x³ + 4x² − 8x − 16 with the supplied first move 2(x³ + 2x² − 4x − 8). Pair the four terms, choose the sign of the second GCF, and factor every piece. The complete result is 2(x + 2)²(x − 2); stopping at 2(x + 2)(x² − 4) leaves a difference of squares unfactored.

Common mistakes

Factoring + out of a negative second pair−4x − 12 should give −4(x + 3). Taking out +4 produces a bracket that cannot match, and the method is abandoned for no reason.

Forcing two different brackets to be the same(x + 3) and (x − 3) are different polynomials. If the pairs do not agree, regroup; never adjust one bracket to fit.

Stopping at the first product(x + 3)(x² − 4) is a factorisation but not a complete one. Check every factor for a further factorisation before you call it finished.

Check your understanding

✓

Finish when four terms make you reach for a pairing automatically.

You should try the pairings without hesitation, get the sign right on the second pair first time, and always look again at both factors afterwards. You should also be comfortable saying "grouping does not work here" without claiming more than that.

Note: Grouping earns its place twice. Here it handles four terms, and in “The AC Method” the AC method deliberately splits a three-term polynomial into four so that this technique can be used on it. Getting the signs reliable now pays for itself two lessons from now.