## Factor by Grouping

Factoring quadratic expressions is a key skill that can prevent students from being efficient in future mathematics. There are many ways, tricks, and short-cuts that can help a student arrive at an answer, but we will focus on just one method here.

The method we will use called Guess and Check. It has been chosen because it most directly ties to what happens when binomials are multiplied together. This method will reinforce what you’ve previously learned and tie it into a new consequence of that understanding.

Read through the notes below, taking notes of your own. Try the problems there, on your own. Then, watch the videos and try the practice problems in the tab labeled Practice Problems. If you find this helpful, please share it on social media!

Here we will learn how factor a quadratic whose leading coefficient is NOT 1, even after a monomial has been factored out. Here are two examples, each representing a slight variation of our new process.

5*x*^{2} + 8*x* + 3 and 5*x*^{2} + 2*x* – 3

What makes these substantially different is the sign of *c*, the constant term (term without a variable). If you remember, the last term of a quadratic expression like *ax*^{2} + *bx + c*, is the product of the last terms of two binomials. If *c* is positive, those two numbers that multiplied to make *c*, are the same sign! If *c* is negative, the two numbers are different signs.

Here are some examples of that.

Remember, the way we get each term can be described with the acronym FOIL (First Outer Inner Last). The first term “F” is the product of the first term in each binomial. The last term “L” is the product of the last term in each binomial. The middle term is often combined from the “Outside + Inside.”

**In ax^{2} + bx + c, we get ax^{2} from F, we get c from L, and bx from O + I.**

Let’s talk about how to approach factoring these quadratic expressions when *c* is positive, and *a* is not 1, first. Remember, always see if you can factor out a constant first because that is far simpler than what follows.

Example 1: 5*x*^{2} + 8*x* + 3

We are going to use a method called Guess and Check to factor this quadratic expression. We know that the product of two binomials will produce this quadratic, and they’ll look like this:

(*kx + m*)(*jx + n*)

The product of the First will be 5*x*^{2}. That only has one set of factors, 5*x* and 1*x*. The last term,* c*, is 3. The only way to get 3 is either 3 × 1 or 1 × 3. Since *c* is positive, the signs are the same. The coefficient *b* is also positive, so we know both signs (of the factors) are positive. This gives us two possible factors of our original quadratic expression.

(5*x* + 3)(*x* + 1) OR (5*x* + 1)(*x* + 3)

We know this because the First term is 5*x*^{2}, which is 5*x* × *x*, and the Last term is 3, which is either 3 × 1 or 1 × 3. To find out which is right we just check the “O + I.”

Our original expression is 5*x*^{2} + 8*x* + 3. The second option yields that.

5*x*^{2} + 8*x* + 3 = (5*x* + 3)(*x* + 1)

## Guess and Check

- Guess what the First term of each binomial could be.
- Guess what the Last term of each binomial could be.
- Check of the O + I produces the
*b*of the original problem.- If not, switch the placement of your guess in step 2. Check again.
- If still not, try other factors.

Let’s see another example, with just two possible factors.

Example 2: 3*x*^{2} – 22*x* + 7

Here the *c* term is positive, which means that the signs of the factors the same. Since the signs of the factors are the same, and they add to a negative number, they must both be negative.

Step 1: Guess the First. The First is 3*x*^{2}, so the factors must be 3*x* and *x*.

Step 2: Guess the Last: The last is 7, so it will be either 7 and 1, or 1 and 7.

Step 3: Set up your guess, and check the O + I.

(3*x* – 7)(*x* – 1)

Here the Outside is – 3*x*, and the Inside is – 7*x*, which makes – 10*x*. We need – 22*x*, so let’s switch our guess of the Last.

(3*x* – 1)(*x* – 7)

Here the Outside is – 21*x*, and the Inside is – *x*, which combine to make – 22*x*. This is the correct combination.

3*x*^{2} – 22*x* + 7 = (3*x* – 1)(*x* – 7)

Let’s look at a trickier example, where the First or Last is NOT a prime number. This creates more possible factors, which makes things trickier.

Example 3: 4*x*^{2} – 12*x* + 5

Note that *c *is positive and *b* is negative. Both factors will be negative again. The guess for the last is easy, once again, because 5 is prime. It is either 1 and 5, or 5 and 1. The First is 4, which has two pairs of factors. It could be 4*x* and *x*, or 2*x* and 2*x*. We do NOT know which it is. We’ll set up our guesses and check, systematically. Here’s how.

First guess will use 4*x* and *x* for F, and 1 and 5.

(4*x* – 1)(*x* – 5)

This is an appropriate guess because the F and L work. When checking the O + I we get – 21*x*, and we need – 12*x* (because that’s the original expression).

If your guess doesn’t work, switch the last numbers before switching to new sets of factors. In this case it will yield the correct factors, but it keeps your work organized and saves time. Often you can check mentally, as you did if you verified that switching here doesn’t give us the factors.

Let’s switch to 2*x* and 2*x* for the first time, *ax*^{2}.

(2*x* – 5)(2*x* – 1)

This works because the Outside is – 2*x* and the inside is – 10*x*, which combine to make – 12*x*.

We will keep the examples at this level of complexity (with numbers of possible factors). In the future we will tackle things with more factors, like 12*x*^{2} – 7*x* – 12.

Let’s practice some of these quickly to see how they work before moving on.

Factor completely.

## When *c* is Negative

If the last term of the quadratic expression is negative you’ll still use guess and check, but you’ll figure out the signs LAST. Here’s how.

Example 4: 2*x*^{2} + 11*x* – 6

Remember, the – 6 is the last term, and it comes from the product of the last terms of our two binomials. They multiply to make a negative, meaning they are different signs. We still guess the factors of the First and Last, and set up a guess. We just don’t write any signs yet. Let’s see how it works.

(2*x* 6)(*x* 1)

This is a valid guess because the first two make 2*x*^{2} and the last two make 6. The O + I will give us the following.

2*x* & 6*x*

These are different signs, so their sum is 4*x*, either positive or negative 4*x*. We need a 11*x*, so this isn’t the combination we’re looking for. When this happens, switch those last terms and try again.

(2*x* 1)(*x* 6)

Here the Outside gives us 12*x* and the inside gives us *x*. They’re different signs, so their sum is either 11*x* or – 11*x*, which is what we need. Let’s see how to assign the signs.

+ 12*x* & – *x* → + 11*x*

You’ll have to think carefully here to assign the signs correctly. We definitely need a positive 12 and a negative 1 (*x* of course). The 12 came from the product of 2 and 6, so the 6 is positive. The 1*x* needs to be negative and it came from the product of 1 and *x*, so the 1 is negative.

2*x*^{2} + 11*x* – 6 = (2*x* – 1)(*x* + 6)

Check it again to make sure! Negatives are tricky, so take your time when *c* is negative.