When a polynomial has four or more terms, the easiest way to factor it is to use *grouping. *In this method, you look at only two terms at a time to see if any techniques become apparent. For example, you may see a Greatest Common Factor (GCF) in two terms, or you may recognize a trinomial as a perfect square.

Sometimes you can group a polynomial into sets with two terms each to find a GCF in each set. You should try this method first when faced with a polynomial with four or more terms. This type of grouping is the most common method in pre-calculus.

For example, you can factor *x*^{3} + *x*^{2} – *x* – 1 by using grouping. Just follow these steps:

Break up the polynomial into sets of two.

You can go with (

*x*^{3}+*x*^{2}) + (–*x*– 1). Put the plus sign between the sets, just like when you factor trinomials.Find the GCF of each set and factor it out.

The square

*x*^{2}is the GCF of the first set, and –1 is the GCF of the second set. Factoring out both of them, you get*x*^{2}(*x*+ 1) – 1(*x*+ 1).Factor again as many times as you can.

The two terms you’ve created have a GCF of (

*x*+ 1). When factored out, you get (*x*+ 1)(*x*^{2}– 1).However,

*x*^{2}– 1 is a difference of squares and factors again as (x+1)(x-1). This gives you a final factorization of: (*x*+ 1)(*x*+ 1)(*x*– 1), or (*x*+ 1)^{2}(*x*– 1).

If this method doesn’t work, you may have to group the polynomial some other way. Of course, after all your effort, the polynomial may end up being prime, which is okay.

For example, look at the polynomial *x*^{2} – 4*xy *+ 4*y*^{2} – 16. You can group it into sets of two, and it becomes *x*(*x* – 4*y*) + 4(*y*^{2} – 4). This expression, however, doesn’t factor again. Bells and whistles should go off inside your head at this point, telling you to look again at the original. You must try grouping it in some other way. In this case, if you look at the first three terms, you’ll discover a perfect-square trinomial, which factors to (*x* – 2*y*)^{2} – 16. Now you have a difference of squares, which factors again to [(*x* – 2*y*) – 4][(*x* – 2*y*) + 4].