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[SOUND] Let's look at factoring by grouping.
[SOUND] For example, let's factor the following expression by grouping.
Now, to factor means we want to write this expression as a product of other
numbers or algebraic expressions. Now looking here, there is nothing in
common to all four of these terms, and when that's the case, we try to group
terms together and look for common factors.
And we can begin by trying to group the first two terms together and the last two
terms together. In other words, this is equal to
(4v^5+v^4) and then plus (20v+5). And now the greatest common factor in the
first two terms is v^4, factoring that out, we're left with
(4v+1). And the greatest common factor in the
second two terms, is a 5, factoring that out, we're left with
(4v+1), which is the same binomial expression in
the first grouping. Which is why factoring by grouping works
here, because now, we can factor that out of both of these.
In other words, this is equal to (4v+1*v^4+5),
which would be our answer. Let's look at another example.
[SOUND] Again, let's factor this expression by grouping and we'll start in
the same way, we'll group the first two terms together
as well as the last two terms. In other words, this is equal to
(5v^3-4w^2)+(-25w+20). And now, the greatest common factor in
the first two terms is a w^2, and when we factor that out, we're left
with (5w-4). What is the greatest common factor in this second grouping? Well, we
get factor either a five out or a negative five out.
But remember, our hope is that we're going to get the same binomial leftover.
So what happens when we factor out a five here? We'd be left with (-5w+4), which is
not that same binomial. However, if we factor out a negative 5,
we're left with (5w-4), which is that same binomial.
So that's what we want to factor out is the negative five,
so this is minus 5(5w-4). And now, we can factor this binomial out
of each of these groupings, which gives us (5w-4)(w^2-5), which would be our
answer. Alright. Let's see one more example.
[SOUND] Again, let's factor this by grouping and we'll begin in the same way,
we'll group together the first two terms and then the last two terms, which gives
us. (3pr-qr) and then plus (6ps-2qs).
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Now, the greatest common factor in these first two terms is an r, when we factor
that out, we're left (3p-q). Now remember, our hope is is to get the
same binomial leftover in the second grouping.
And if we pull out a 2s, we'll be left with that same binomial, won't we?
(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
groupings, gives us (3p-q)*(r+2s),
which would be our answer. Now, in all these examples, we've been
starting with grouping the first two terms and then the last two terms.
But what if we started a little bit different here? So we still have the six
expression, 3pr-qr+6ps-2qs, but instead of grouping the first two terms and the
last two terms, what if we group the first term and third term together and
then the second term and the fourth term? In other words, let's write this as
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(3pr+6ps)+(-qr-2qs). Now, the greatest common factor in the
first two terms is a 3p, so let's pull that out,
and we're left with (r+2s). Now remember, in the second grouping over
here, our hope is that after factoring we have the same binomial leftover.
So if we factor out a -q here, we'll be left with (r+2s),
which is the same binomial. So let's do that.
So we have -q(r+2s). And now we can factor out that binomial,
which leaves us with (r+2s)(3p-q). And by commutativity, that's the same as
this, isn't it? Okay. And this is how we factor by grouping.
Thank you and we'll see you next time. [SOUND]
Dr. Sarah EichhornAssociate Dean, Distance Learning & Tenured Lecturer
Dr. Rachel Cohen Lehman
