Factoring by Grouping

Why Grouping Can Reveal Factors

A polynomial may contain four terms that do not have an obvious common factor across the entire expression. However, the terms may form two pairs, with each pair sharing a factor. If factoring each pair produces the same binomial, that shared binomial can be factored out of the whole expression.

This method is called factoring by grouping. It relies on the distributive property: first use the property within each pair to expose a repeated factor, then use it again to factor that repeated expression out.

The method is not a mechanical guarantee that every four-term polynomial will factor this way. It succeeds when an appropriate grouping produces matching binomial factors. If the brackets do not match, the steps so far do not establish a factorization by grouping.

The Basic Pattern

Suppose four terms can be written in two pairs as \(ax+ay+bx+by\). Factoring each pair gives \(a(x+y)+b(x+y)\). Both terms now contain the same factor, \(x+y\), so the entire expression can be factored.

The key is not merely to make two pairs. The key is to make pairs whose factored forms contain the same bracket. A four-term expression that has been split into pairs but does not produce matching brackets has not yet been factored by grouping.

In practice, begin with the terms in their given order and pair the first two terms and the last two terms. Factor out the greatest common factor from each pair. Include the signs as part of the coefficients: in particular, a negative term may mean that extracting a negative common factor from the second pair makes the brackets match.

Worked Examples

Why the Method Works

The justification is an application of the distributive property. Factoring a common factor out of a pair reverses distribution. If both pairs leave the same binomial, the distributive property can be applied once more to factor that binomial from the whole expression.

More explicitly, distribute the product \((a+b)(x+y)\):

So when grouping produces \(a(x+y)+b(x+y)\), rewriting it as \((a+b)(x+y)\) preserves the value of the expression. The two expressions are equal because the same distributive property works in reverse.

When the Brackets Do Not Match

After factoring each pair, compare the brackets as algebraic expressions, not just by their general appearance. For instance, \(x+3\) and \(x-3\) are different factors. They cannot be treated as one shared binomial without changing the expression.

A sign can sometimes be adjusted by extracting a negative common factor. For example, \(-x-3\) can be written as \(-(x+3)\). This changes the outside factor and makes the bracket \(x+3\). The negative sign must remain outside; it cannot simply be discarded.

If a first grouping does not produce matching brackets, that grouping has not completed the method. It may be possible to rearrange the terms and form different pairs, but a failed attempt is not itself proof that the polynomial cannot be factored. For this method, the required test remains the same: after factoring the pairs, the brackets must match.

A Common Sign Mistake

Consider the grouped form from the second worked example. A common error is to factor \(4x-6\) incorrectly as \(2(2x+3)\). That would change the sign of the constant term, because

not \(4x-6\). The correct factorization is \(2(2x-3)\), since

With the incorrect sign, the attempted grouping would be \(3x^2(2x-3)+2(2x+3)\), whose brackets do not match. With the corrected sign, it is \(3x^2(2x-3)+2(2x-3)\), and the common binomial is visible. Checking each pair by multiplying its outside factor back in is a useful way to catch this error before combining brackets.

What to Check Before Finishing

A correct-looking product is not enough: it must expand to the original polynomial. Use the check to confirm both the factors and the signs. In particular, verify the second pair carefully when it contains negative terms.

• Were all four terms included in exactly one of the two pairs?
• Does each pair expand back to the terms it came from?
• Are the remaining binomial brackets exactly the same?
• When you factor out a negative number, have you retained its sign?
• Does multiplying the final factors reproduce the original polynomial?

Factoring by grouping is useful because it turns a four-term expression into a product when its terms have the right structure. The decisive step is identifying the repeated bracket. If that bracket is not present, do not force the final step; check the pair factoring and signs first.