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When you factor an algebraic expression, you rewrite the sums and differences of the terms as a product. For instance, you write the three terms x² – x – 42 in factored form as (x – 7)(x + 6). The expression changes from three terms to one big, multiplied-together term. You can factor two terms, three terms, four terms, and so on for many different purposes. The factorization comes in handy when you set the factored forms equal to zero to solve an equation. Factored numerators and denominators in fractions also make it possible to reduce the fractions.

You can think of factoring as the opposite of distributing. You have good reasons to distribute or multiply through by a value – the process allows you to combine like terms and simplify expressions. Factoring out a common factor also has its purposes for solving equations and combining fractions. The different formats are equivalent – they just have different uses.

Factoring two terms

When an algebraic expression has two terms, you have four different choices for its factorization – if you can factor the expression at all. If you try the following four methods and none of them work, you can stop your attempt; you just can’t factor the expression:

In general, you check for a greatest common factor before attempting any of the other methods. By taking out the common factor, you often make the numbers smaller and more manageable, which helps you see clearly whether any other factoring is necessary.

To factor the expression 6x⁴ – 6x, for example, you first factor out the common factor, 6x, and then you use the pattern for the difference of two perfect cubes:

A quadratic trinomial is a three-term polynomial with a term raised to the second power. When you see something like x² + x + 1 (as in this case), you immediately run through the possibilities of factoring it into the product of two binomials. In this case, you can just stop. These trinomials that crop up when factoring cubes just don’t cooperate.

Keeping in mind my tip to start a problem off by looking for the greatest common factor, look at the example expression 48x³y² – 300x³. When you factor the expression, you first divide out the common factor, 12x³, to get 12x³(4y² – 25). You then factor the difference of perfect squares in the parentheses: 12x³(4y² – 25) = 12x³(2y – 5)(2y + 5).

Here’s one more: The expression z⁴ – 81 is the difference of two perfect squares. When you factor it, you get z⁴ – 81 = (z² – 9)(z² + 9). Notice that the first factor is also the difference of two squares – you can factor again. The second term, however, is the sum of squares – you can’t factor it. With perfect cubes, you can factor both differences and sums, but not with the squares. So, the factorization of z⁴ – 81 is (z – 3)(z + 3)(z² + 9).

Taking on three terms

When a quadratic expression has three terms, making it a trinomial, you have two different ways to factor it. One method is factoring out a greatest common factor, and the other is finding two binomials whose product is identical to those three terms:

You can often spot the greatest common factor with ease; you see a multiple of some number or variable in each term. With the product of two binomials, you either find the product or become satisfied that it doesn’t exist.

For example, you can perform the factorization of 6x³ – 15x²y + 24xy² by dividing each term by the common factor, 3x: 6x³ – 15x²y + 24xy² = 3x(2x² – 5xy + 8y²).

You want to look for the common factor first; it’s usually easier to factor expressions when the numbers are smaller. In the previous example, all you can do is pull out that common factor – the trinomial is prime (you can’t factor it any more).

Trinomials that factor into the product of two binomials have related powers on the variables in two of the terms. The relationship between the powers is that one is twice the other. When factoring a trinomial into the product of two binomials, you first look to see if you have a special product: a perfect square trinomial. If you don’t, you can proceed to unFOIL. The acronym FOIL helps you multiply two binomials (First, Outer, Inner, Last); unFOIL helps you factor the product of those binomials.

Finding perfect square trinomials

A perfect square trinomial is an expression of three terms that results from the squaring of a binomial – multiplying it times itself. Perfect square trinomials are fairly easy to spot – their first and last terms are perfect squares, and the middle term is twice the product of the roots of the first and last terms:

To factor x² – 20x + 100, for example, you should first recognize that 20x is twice the product of the root of x² and the root of 100; therefore, the factorization is (x – 10)². An expression that isn’t quite as obvious is 25y² + 30y + 9. You can see that the first and last terms are perfect squares. The root of 25y² is 5y, and the root of 9 is 3. The middle term, 30y, is twice the product of 5y and 3, so you have a perfect square trinomial that factors into (5y + 3)².

Resorting to unFOIL

When you factor a trinomial that results from multiplying two binomials, you have to play detective and piece together the parts of the puzzle. Look at the following generalized product of binomials and the pattern that appears:

So, where does FOIL come in? You need to FOIL before you can unFOIL, don’t ya think?

The F in FOIL stands for “First.” In the previous problem, the First terms are the ax and cx. You multiply these terms together to get acx². The Outer terms are ax and d. Yes, you already used the ax, but each of the terms will have two different names. The Inner terms are b and cx; the Outer and Inner products are, respectively, adx and bcx. You add these two values. (Don’t worry; when you’re working with numbers, they combine nicely.) The Last terms, b and d, have a product of bd. Here’s an actual example that uses FOIL to multiply – working with numbers for the coefficients rather than letters:

Now, think of every quadratic trinomial as being of the form acx² + (ad + bc)x + bd. The coefficient of the x² term, ac, is the product of the coefficients of the two x terms in the parentheses; the last term, bd, is the product of the two second terms in the parentheses; and the coefficient of the middle term is the sum of the outer and inner products. To factor these trinomials into the product of two binomials, you can use the opposite of the FOIL, which I call unFOIL.

Here are the basic steps you take to unFOIL the trinomial acx² + (ad + bc)x + bd:

1. Determine all the ways you can multiply two numbers to get ac, the coefficient of the squared term.

2. Determine all the ways you can multiply two numbers to get bd, the constant term.

3. If the last term is positive, find the combination of factors from Steps 1 and 2 whose sum is that middle term; if the last term is negative, you want the combination of factors to be a difference.

4. Arrange your choices as binomials so that the factors line up correctly.

5. Insert the + and – signs to finish off the factoring and make the sign of the middle term come out right.

To factor x² + 9x + 20, for example, you need to find two terms whose product is 20 and whose sum is 9. The coefficient of the squared term is 1, so you don’t have to take any other factors into consideration. You can produce the number 20 with , , or . The last pair is your choice, because 4 + 5 = 9. Arranging the factors and x’s into two binomials, you get x² + 9x + 20 = (x + 4)(x + 5).

Factoring four or more terms by grouping

When four or more terms come together to form an expression, you have bigger challenges in the factoring. You see factoring by grouping in the previous section as a method for factoring trinomials; the grouping is pretty obvious in this case. But what about when you’re starting from scratch? What happens with exponents greater than two? As with an expression with fewer terms, you always look for a greatest common factor first. If you can’t find a factor common to all the terms at the same time, your other option is grouping. To group, you take the terms two at a time and look for common factors for each of the pairs on an individual basis. After factoring, you see if the new groupings have a common factor. The best way to explain this is to demonstrate the factoring by grouping on x³ – 4x² + 3x – 12 and then on xy² – 2y² – 5xy + 10y – 6x + 12.

The four terms x³ – 4x² + 3x – 12 don’t have any common factor. However, the first two terms have a common factor of x², and the last two terms have a common factor of 3:

Notice that you now have two terms, not four, and they both have the factor (x – 4). Now, factoring (x – 4) out of each term, you have (x – 4)(x² + 3).

Factoring by grouping only works if a new common factor appears – the exact same one in each term.

The six terms xy² – 2y² – 5xy + 10y – 6x + 12 don’t have a common factor, but, taking them two at a time, you can pull out the factors y², –5y, and –6. Factoring by grouping, you get the following:

The three new terms have a common factor of (x – 2), so the factorization becomes (x – 2)(y² – 5y – 6). The trinomial that you create also factors (see the previous section):

Factored, and ready to go!