## Factoring Polynomials

Factoring a polynomial is the opposite process of multiplying polynomials. Recall that when we factor a number, we are looking for prime factors that multiply together to give the number; for example

6 = 2 × 3 , or 12 = 2 × 2 × 3.

When we factor a polynomial, we are looking for simpler polynomials that can be multiplied together to give us the polynomial that we started with. You might want to review multiplying polynomials if you are not completely clear on

how that works.

- When we factor a polynomial,

we are usually only interested in breaking it down into polynomials that have**integer**

coefficients and constants.

### Simplest Case: Removing Common Factors

The simplest type of factoring is when there is a factor common to every term. In that case, you can factor out that common factor. What you are doing is using the distributive law in reverse—you are sort of un-distributing the factor.

Recall that the distributive law says *a*(*b* + *c*) = *ab* + *ac*.

Thinking about it in reverse means that if you see *ab* + *ac*, you can write it as *a*(*b* + *c*).

**Example:** 2*x*^{2} + 4*x*

Notice that each term has a factor of 2*x*, so we can rewrite it as: 2*x*^{2} + 4*x* = 2*x*(*x* + 2)

### Difference of Two Squares

If you see something of the form *a*^{2} – *b*^{2}, you should remember the formula

_{}

**Example:** *x*^{2} – 4 = (*x* – 2)(*x* + 2)

- This only holds for a
**difference**of two squares. There is no way to factor a**sum**of two squares such as*a*^{2}+*b*^{2}into factors with real numbers.

### Trinomials (Quadratic)

A quadratic trinomial has the form ** ax^{2} + bx + c**,

where the coefficients *a*, *b*, and *c*, are real numbers (for simplicity we will only use integers, but in real life they could be any real number). We are interested here in factoring quadratic trinomials with integer coefficients into factors that have integer coefficients. Not all such quadratic polynomials can be factored over the real numbers, and even fewer into integers (they all can be factored of we allow for imaginary numbers and rational coefficients, but we don’t). Therefore, when we say a quadratic can be factored, we mean that we can write the factors with only integer coefficients.

If a quadratic can be factored, it will be the product of two first-degree binomials, except for very simple cases that just involve monomials. For example *x*^{2} by itself is a quadratic expression where the coefficient *a* is equal to 1, and *b* and *c* are zero. Obviously, *x*^{2} factors into (*x*)(*x*), but this is not a very interesting case.

A slightly more complicated case occurs when only the coefficient *c *is zero. Then you get something that looks like 2*x*^{2} + 3*x*

This can be factored very simply by factoring out (‘undistributing’) the common factor of *x*: 2*x*^{2} + 3*x* = *x*(2*x *+ 3)

The most general case is when all three terms are present, as in *x*^{2} + 5*x + *6

We look at two cases of this type. The easiest to factor are the ones where the coefficient of *x*^{2} (which we are calling ‘*a*’) is equal to 1, as in the above example. If *a* is not 1 then things get a little bit more complicated, so we will begin by looking at *a* = 1 examples.

#### Coefficient of *x*^{2} is 1

Since the trinomial comes from multiplying two first-degree binomials, lets review what happens when we multiply binomials using the FOIL method. Remember that to do factoring we will have to think about this process in reverse (you could say we want to ‘de-FOIL’ the trinomial).

Suppose we are given (*x* + 2)(*x* + 3)

Using the FOIL method, we get (*x* + 2)(*x* + 3) = *x*^{2} + 3*x* + 2*x* + 6

Then, collecting like terms gives (*x* + 2)(*x* + 3) = *x*^{2} + 5*x* + 6

Now look at this and think about where the terms in the trinomial came from. Obviously the *x*^{2} came from *x* times *x*. The interesting part is what happens with the other parts, the ‘+ 2’ and the ‘+ 3’. The last term in the trinomial, the 6 in this case, came from **multiplying **the 2 and the 3. Where did the 5*x* in the middle come from? We got the 5*x *by **adding** the 2*x* and the 3*x* when we collected like terms. We can state this as a rule:

- If the coefficient of
*x*^{2}

is one, then to factor the quadratic you need to find two numbers that:

1. Multiply to give the constant term (which we call *c*)

2. Add to give the coefficient of *x* (which we call *b*)

This rule works even if there are minus signs in the quadratic expression (assuming that you remember how to add and multiply positive and negative numbers).

### Special Case: Perfect Square Trinomial

Recall from special products of binomials that

_{}

and

_{}

The trinomials on the right are called *perfect squares* because they are the squares of a single binomial, rather than the product of two different binomials. A quadratic trinomial can also have this form:

(*x* + 3)^{2} = (*x* + 3)(*x* + 3) = *x*^{2} + 6*x* + 9

Notice that just as before the coefficient of *x* is the **sum **3 + 3, and the constant term is the **product** 3 ´ 3. One can also say that

- The coefficient of
*x*is twice the number 3 - The constant term is the number three squared

In general, if a quadratic trinomial is a perfect square, then

o The coefficient of *x *is twice the square root of the constant term

Or to put it another way,

o **The constant term is the square of half the coefficient of x**

In symbolic form we can express this as

_{}

It is helpful to be able to recognize perfect square trinomials. We will see them again when we talk about solving quadratic equations.

#### Coefficient of *x*^{2} is not 1

A quadratic is more difficult to factor when the coefficient of the squared term is not 1, because that coefficient is mixed in with the other products from FOILing the two binomials. There are two methods for attacking these: either you can use a systematic guess-and-check method, or a method called factoring by grouping. We will first look at the guess-and-check method (which we could call factoring by *grouping*).

If you need to factor a trinomial such as

2*x*^{2} + *x* – 3,

you have to think about what combinations could give the 2*x*^{2}as well as the other two terms. In this example the 2*x*^{2} must come from (*x*)(2*x*), and the constant term might come from either (–1)(3) or (1)(–3).The hard part is figuring out which combination will give the correct middle term. This gets messy because all those coefficients will be mixed in with the middle term when you FOIL the binomials. To see what is going on, let’s see what happens when we FOIL the following binomials:

_{}

What happened? There are several significant things to notice:

- The leading term in the trinomial (the 2
*x*^{2}) is just the product of the leading terms in the binomials. - The constant term in the trinomial (the –3) is theproduct of the constant terms in the binomials (so far this is the same as in the case where the coefficient of
*x*^{2}is 1) - The middle term in the trinomial (the
*x*) is the sum of the outer and inner products, which involves*all*the constants and coefficients in the binomials, in a messy way that is not always obvious by inspection.

Because 1 and 2 are relatively simple and 3 is complicated, it makes sense to think of the possible candidates that would satisfy conditions 1 and 2, and then test them in every possible combination by multiplying the resulting binomials to see if you get the correct middle term. This seems tedious, and indeed it can be if the numbers you are working with have a lot of factors, but in practice you usually only have to try a few combinations before you see what will work. As a demonstration, let’s see how we would attack the example by this method.

Given 2*x*^{2} + *x* – 3

We make a list of the possible factors of 2*x*^{2}: The only choice is (2*x*)(*x*).

Then we make a list of the possible factors of the constant term –3: it is either (1)(–3) or (–1)(3). (Notice that since we need a negative number, one factor must be negative and the other positive, but it doesn’t matter which one so we have to try it both ways).

The possible factors of the trinomial are the binomials that we can make out of these possible factors, taken in every possible order. From these possibilities, we see that the candidate binomials are:

(2*x* + 1)(*x* – 3)

(*x* + 1)(2*x* – 3)

(2*x* + 3)(*x* – 1)

(*x* + 3)(2*x* – 1)

If we start multiplying these out, we will find that the third one works, and then we are finished. All you really need to check is to see if the sum of the outer and inner multiplications will give you the correct middle term, since we already know that we will get the correct first and last terms.

In short, the method is:

1. List all the possible ways to get the coefficient of *x*^{2 }(which we call *a*) by multiplying two numbers

2. List all the possible ways to get the constant term (which we call *c*) by multiplying two numbers