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Algebra I: High School20 chapters | 168 lessons | 1 flashcard set

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Lesson Transcript

Instructor:
*Jeff Calareso*

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

How do you factor an expression if all the terms don't share a common factor? Grouping! In this lesson, we'll learn how to use grouping to help factor expressions.

In algebra, we love to build and we love to take apart. It's like we're using building blocks to make a spaceship, only instead of blocks, we have terms. And once we build our spaceship, we love to smash it. Is that just me? I built and smashed a lot of spaceships as a kid. The smashing part is **factoring**. Factoring is the process of finding the factors. If our space ship looks like 5*x* + 10, we factor out a 5 and get 5(*x* + 2). But what if we encounter something like this?

**3 x^2 + 2x + 12x + 8**

That's a pretty awesome spaceship. We have four terms here. Is there a common factor? Well, three of the terms have an *x*, but not the last one. The last three terms have a factor of 2, but not 3*x*^2. There's no common factor. It's like someone glued our spaceship together! That's not cool. What if we want to build something else? We need a way to break it apart. How do you factor an expression if you can't factor anything out of each term?

The answer: **grouping**. With grouping, we break up the expression into smaller groups that can be factored. Then we do what we already know how to do. Let's take our expression and learn about the steps involved with grouping.

The first step is to group the first two terms and the last two terms. Think of these as separate sets of expressions. Here, let's do (3*x*^2 + 2*x*) and have that plus (12*x* + 8). We basically have two two-term expressions, or binomials. Think of this like delicately taking the spaceship apart and splitting up the pieces by shape or color. This is a less fun way of deconstructing your blocks but, hey, to each his own.

The second step is to factor the greatest common factor from each binomial. We got stuck trying to factor something from all four terms, but we can factor something from these binomials. With (3*x*^2 + 2*x*), we can factor out an *x* to get *x*(3*x* + 2). With (12*x* + 8), we can factor out a 4 to get 4(3*x* + 2). Okay, now we have *x*(3*x* + 2) + 4(3*x* + 2). That leads to our final step.

The third and final step is to factor out the common binomial. Do you see how we have two (3*x* + 2)'s? We can factor that out. That gets us (*x* + 4)(3*x* + 2). And that's it. It's a spaceship no more! But wait, let's check our work. (*x* + 4)(3*x* + 2) doesn't look anything like our original expression. How do we know if it's correct? Well, we rebuild it with **FOIL**. *x*(3*x*) is 3*x*^2, *x*(2) is 2*x*, 4(3*x*) is 12*x*, and 4(2) is 8. That's 3*x*^2 + 2*x* + 12*x* + 8. And that is our original expression. We did it!

Let's practice another one.

**5 y^3 - 3y^2 + 10y - 6**

Wow, look at all those *y*'s. This is a cool spaceship. Well, let's break it apart by following our steps.

**Step 1**: Group it as (5*y*^3 - 3*y*^2) + (10*y* - 6).

Okay, **Step 2**: What can we factor from (5*y*^3 - 3*y*^2)? Well, nothing from 5 and 3, but we can factor out a *y*. Not just that, but *y*^2. We get *y*^2(5*y* - 3). What about (10*y* - 6)? We can't factor out a *y*, but 10 and 6 have a common factor in 2. We get 2(5*y* - 3). Hey, look at that: 5*y* - 3 again! We found some common parts that go together.

That leads us to **Step 3**: factor out that 5*y* - 3. So we have (*y*^2 + 2)(5*y* - 3). Let's check that. *y*^2(5*y*) is 5*y*^3, that's good. And *y*^2(-3) is -3*y*^2; good again. 2(5*y*) is 10*y*; I like where this is going. Finally, 2(-3) is -6. That's what we wanted!

So that's all well and good with a four-term expression. But what if we have this?

**2 x^2 + 5x + 3**

Hmm, there are three terms and no common factor. It's like someone built a spaceship with castle parts and, well, what do you do with that? You just need to know what you're working with. This equation is the same as *ax*^2 + *bx* + *c*. We want to take our *a* and *c*, which is 2 and 3, and multiply them together: *a* x *c* = 6. Now, look for the factors of 6 that add up to *b*, which is 5.

What are the factors of 6: 1 and 6, 2 and 3. Well, 1 + 6 is 7, so that's no good. 2 + 3 is 5. That works! So we split 5 into 2*x* + 3*x*. That makes our new expression 2*x*^2 + 2*x* + 3*x* + 3. That's more familiar isn't it?

Now we can group and factor. Note that if we wrote it as 2*x*^2 + 3*x* + 2*x* + 3, we'd see that grouping wouldn't help us factor the 2*x* + 3. You can arrange the terms however you need to to best factor. Just don't lose sight of the signs. Here it's all addition, so no worries. Okay, with 2*x*^2 + 2*x* + 3*x* + 3, (2*x*^2 + 2*x*) becomes 2*x*(*x* + 1). (3*x* + 3) becomes 3(*x* + 1).

Now we can factor out *x* + 1 to give us (*x* + 1)(2*x* + 3). We did it! We're masters at taking apart these expressions. Let's check out our work. *x*(2*x*) is 2*x*^2, *x*(3) is 3*x*, 1(2*x*) is 2*x*, and 1(3) is 3. If we combine 3*x* and 2*x*, we get 5*x*, and that's our original expression.

To summarize, we followed three steps to factor by grouping. First, group the first two terms and the last two terms. We're basically making two separate binomials. Next, factor the greatest common factor from each binomial. Finally, factor out the common binomial. If we only have three terms, as in *ax*^2 + *bx* + *c*, we can add a step. We find *a* times *c*, then find the factors of that product that add up to *b*. We split up *bx* based on these numbers and proceed as normal.

After absorbing this lesson's facts, you might recall how to do the following:

- Outline the processes of factoring and grouping
- Interpret the three steps associated with grouping to factor expressions
- Arrange and split terms

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Algebra I: High School20 chapters | 168 lessons | 1 flashcard set

- What is Factoring in Algebra? - Definition & Example 5:32
- How to Find the Prime Factorization of a Number 5:36
- Using Prime Factorizations to Find the Least Common Multiples 7:28
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- Factoring Out Variables: Instructions & Examples 6:46
- Combining Numbers and Variables When Factoring 6:35
- Transforming Factoring Into A Division Problem 5:11
- Factoring By Grouping: Steps, Verification & Examples 7:46
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