### Video Transcript

In this video, we will learn how to
factor expressions by grouping.

Letβs begin by revisiting the idea
of factoring an expression by identifying its highest common factor. Looking at the expression three π₯
plus nine π¦ minus six, we see that each term is divisible by three. This means that three is a common
factor of the expression, and we can take this outside of our parentheses. Three π₯ plus nine π¦ minus six can
be rewritten as three multiplied by π₯ plus three π¦ minus two. Since the terms inside the
parentheses have no common factors apart from one, we can conclude that three is the
highest common factor and the expression is now in its most simplified form.

Letβs now consider a second example
where the entire expression does not have a common factor apart from one. We can therefore begin by grouping
pairs of terms. The terms two and two π₯ have a
common factor of two. So we can rewrite this as two
multiplied by one plus π₯. The terms π¦ and π₯π¦ have a common
factor of π¦. Therefore, this can be rewritten as
π¦ multiplied by one plus π₯. We can therefore rewrite the
original expression as shown.

At this stage, we notice that our
two terms have a common factor of one plus π₯. Factoring this out, we can rewrite
the expression as one plus π₯ multiplied by two plus π¦. The expression two plus two π₯ plus
π¦ plus π₯π¦ is therefore equal to one plus π₯ multiplied by two plus π¦. We could check this answer by
distributing our parentheses using the FOIL method. We will now generalize the method
used in this example to help us factor by grouping.

If we have an expression with an
even number of terms that did not all share a common factor, then in certain
situations we can apply the following approach to fully factor the expression. Firstly, we group the terms into
pairs that each share a common factor. Secondly, we factor out the common
factor from each pair. Finally, we completely factor the
expression by identifying a common factor in the factored terms.

Letβs now consider another example
where we can apply this method.

Factorize fully π§π minus π§π₯
plus π minus π₯.

The expression in this question
does not have a common factor greater than one, so we cannot directly factor all of
it. However, we can factor it by
grouping pairs of terms with common factors. For example, the first and third
terms have a common factor of π, and the second and fourth terms have a common
factor of negative π₯. Rewriting the expression as shown,
we begin by taking out these common factors. π§π plus π can be rewritten as π
multiplied by π§ plus one, and negative π§π₯ minus π₯ can be rewritten as negative
π₯ multiplied by π§ plus one.

We now have an expression where the
term in the parentheses, π§ plus one, is common. Factoring this out gives us π§ plus
one multiplied by π minus π₯. And as this expression cannot be
factored further, it is in its fully simplified form. The expression π§π minus π§π₯ plus
π minus π₯ factorized fully is π§ plus one multiplied by π minus π₯.

It is worth noting that we could
have grouped different pairs of terms in the original expression. For example, the first two terms in
the initial expression have a common factor of π§. So we can rewrite these as π§
multiplied by π minus π₯. The third and fourth terms have a
common factor of one. So we can rewrite π minus π₯ as
one multiplied by π minus π₯. At this stage, since π minus π₯ is
common, we can factor this out, giving us π minus π₯ multiplied by π§ plus one. And since multiplication is
commutative, this is the same expression as we found using the first method.

We will now consider a more
complicated example of this type.

Factorize fully nine π₯π minus
four πΏπ§ plus four πΏπ₯ minus nine ππ§.

In the expression given, there is
no common factor among all four terms apart from one. And we will therefore factorize the
expression by grouping pairs of terms. If we consider the first and last
terms, we note that they have a common factor of nine π. So we can factor this out. Nine π₯π minus nine ππ§ can be
rewritten as nine π multiplied by π₯ minus π§. We can use the same method when
grouping the second and third terms. These have a common factor of four
πΏ. Negative four πΏπ§ plus four πΏπ₯
can be rewritten as four πΏ multiplied by negative π§ plus π₯. Altering the order of the terms in
the parentheses, we have four πΏ multiplied by π₯ minus π§.

We now have an equivalent form to
the original expression. At this stage, our two terms have a
common factor of π₯ minus π§. Factoring this out, we have π₯
minus π§ multiplied by nine π plus four πΏ. This is the fully factored form of
nine π₯π minus four πΏπ§ plus four πΏπ₯ minus nine ππ§.

Whilst it is not required in this
question, we could check our answer by distributing the parentheses using the FOIL
method. Multiplying the first terms gives
us nine π₯π. Multiplying the outer terms gives
us four πΏπ₯. The inner terms have a product of
negative nine ππ§. And the last terms multiply to give
us negative four πΏπ§. This is the same expression as in
the question.

So far, in this video, we have
considered expressions where no variable has been raised to a power. Sometimes, we may also be able to
factor quadratic or cubic expressions by grouping terms that we can factor. Letβs now consider some specific
situations where this is the case.

Factorize fully π₯ cubed minus two
π₯ squared plus five π₯ minus 10.

The first thing we notice is that
the expression is cubic since the largest exponent of a variable is three. Normally, there would be no easy
way to factor this expression fully. But we notice there is a common
ratio between the coefficients of the first two terms and last two terms. In the first two terms, this ratio
is one to negative two. And in the last two terms, the
ratio is five to negative 10, which simplifies to one to negative two. This suggests that we might be able
to factor the expression by grouping the first two and last two terms together.

The highest common factor of π₯
cubed and negative two π₯ squared is π₯ squared. And as such, factoring π₯ cubed
minus two π₯ squared gives us π₯ squared multiplied by π₯ minus two. The highest common factor of five
π₯ and negative 10 is five. And we can rewrite five π₯ minus 10
as five multiplied by π₯ minus two. This means that we can rewrite the
original expression as shown.

We now have two terms with a common
factor of π₯ minus two. Factoring this out gives us π₯
minus two multiplied by π₯ squared plus five. As multiplication is commutative,
we can rewrite this as π₯ squared plus five multiplied by π₯ minus two, which is the
fully factorized form of π₯ cubed minus two π₯ squared plus five π₯ minus 10. It is worth noting that we can
check this answer by distributing our parentheses using the FOIL method, and doing
so would give us the expression in the question.

Before looking at one final
example, letβs recall two special types of cubic expressions. Firstly, we recall the factored
form of the sum of two cubes π cubed plus π cubed. This is equal to π plus π
multiplied by π squared minus ππ plus π squared. We also recall the factored form of
the difference of two cubes. π cubed minus π cubed is equal to
π minus π multiplied by π squared plus ππ plus π squared. In our final example, part of our
expression can be factored using one of these formulae.

Factorize fully π cubed plus π
cubed plus π plus π.

In order to factorize this
expression, we begin by noticing that the first two terms are the sum of two
cubes. We recall that π cubed plus π
cubed is equal to π plus π multiplied by π squared minus ππ plus π
squared. This means that we can rewrite the
expression π cubed plus π cubed plus π plus π as shown. Noting that π plus π can be
rewritten as one multiplied by π plus π, it is clear that our two terms have a
common factor of π plus π. Factoring this out gives us π plus
π multiplied by π squared minus ππ plus π squared plus one.

And as the expression in the second
parentheses cannot be factored further, we have the fully factorized form of π
cubed plus π cubed plus π plus π. This is equal to π plus π
multiplied by π squared minus ππ plus π squared plus one. And we could check this answer by
redistributing the parentheses.

We will now finish this video by
summarizing the key points. We saw in this video that we can
factor expressions that lack a common factor by grouping them into pairs that share
common factors. We also saw that we can factor
cubic expressions that have a shared ratio between the coefficients of terms using
factoring by grouping. Finally, we saw that if we can
apply a factoring formula to part of an expression β such as the sum or difference
of cubes, the difference of squares, or perfect square trinomials β then we can
group that expression and factor it separately.