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[SOUND] Let's look at factoring by grouping.

Â [SOUND] For example, let's factor the following expression by grouping.

Â Now, to factor means we want to write this expression as a product of other

Â numbers or algebraic expressions. Now looking here, there is nothing in

Â common to all four of these terms, and when that's the case, we try to group

Â terms together and look for common factors.

Â And we can begin by trying to group the first two terms together and the last two

Â terms together. In other words, this is equal to

Â (4v^5+v^4) and then plus (20v+5). And now the greatest common factor in the

Â first two terms is v^4, factoring that out, we're left with

Â (4v+1). And the greatest common factor in the

Â second two terms, is a 5, factoring that out, we're left with

Â (4v+1), which is the same binomial expression in

Â the first grouping. Which is why factoring by grouping works

Â here, because now, we can factor that out of both of these.

Â In other words, this is equal to (4v+1*v^4+5),

Â which would be our answer. Let's look at another example.

Â [SOUND] Again, let's factor this expression by grouping and we'll start in

Â the same way, we'll group the first two terms together

Â as well as the last two terms. In other words, this is equal to

Â (5v^3-4w^2)+(-25w+20). And now, the greatest common factor in

Â the first two terms is a w^2, and when we factor that out, we're left

Â with (5w-4). What is the greatest common factor in this second grouping? Well, we

Â get factor either a five out or a negative five out.

Â But remember, our hope is that we're going to get the same binomial leftover.

Â So what happens when we factor out a five here? We'd be left with (-5w+4), which is

Â not that same binomial. However, if we factor out a negative 5,

Â we're left with (5w-4), which is that same binomial.

Â So that's what we want to factor out is the negative five,

Â so this is minus 5(5w-4). And now, we can factor this binomial out

Â of each of these groupings, which gives us (5w-4)(w^2-5), which would be our

Â answer. Alright. Let's see one more example.

Â [SOUND] Again, let's factor this by grouping and we'll begin in the same way,

Â we'll group together the first two terms and then the last two terms, which gives

Â us. (3pr-qr) and then plus (6ps-2qs).

Â 3:57

Now, the greatest common factor in these first two terms is an r, when we factor

Â that out, we're left (3p-q). Now remember, our hope is is to get the

Â same binomial leftover in the second grouping.

Â And if we pull out a 2s, we'll be left with that same binomial, won't we?

Â (3p-q). So let's do that.

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So we have plus 2s(3p-q), and then factoring that out of each of these

Â groupings, gives us (3p-q)*(r+2s),

Â which would be our answer. Now, in all these examples, we've been

Â starting with grouping the first two terms and then the last two terms.

Â But what if we started a little bit different here? So we still have the six

Â expression, 3pr-qr+6ps-2qs, but instead of grouping the first two terms and the

Â last two terms, what if we group the first term and third term together and

Â then the second term and the fourth term? In other words, let's write this as

Â 5:53

(3pr+6ps)+(-qr-2qs). Now, the greatest common factor in the

Â first two terms is a 3p, so let's pull that out,

Â and we're left with (r+2s). Now remember, in the second grouping over

Â here, our hope is that after factoring we have the same binomial leftover.

Â So if we factor out a -q here, we'll be left with (r+2s),

Â which is the same binomial. So let's do that.

Â So we have -q(r+2s). And now we can factor out that binomial,

Â which leaves us with (r+2s)(3p-q). And by commutativity, that's the same as

Â this, isn't it? Okay. And this is how we factor by grouping.

Â Thank you and we'll see you next time. [SOUND]

Â