### Video Transcript

Multiplying binomials. In this video, we will learn how to
multiply two binomials using the FOIL method and the area method. Letβs start by reminding ourselves
of what a binomial is. A binomial is an expression that
contains the sum or difference of two terms. So, for example, three π₯ squared
plus five π₯ is a binomial. And so is π¦ minus seven.

So if we have two binomials and we
want to multiply them, what would that look like. For example, if we have the
binomial π₯ plus two and we multiply it with binomial π₯ plus three, we can write
this with a multiplication symbol in the middle. Notice that we have parentheses
around each binomial so that we know that we multiply all of each one by all of the
other one. We most commonly see this type of
problem without the multiplication symbol between the two binomials. We may also see that multiplication
of two binomials refer to as expanding, or expanding the product, or using the
distributive property.

Now, letβs talk about how to
actually multiply our two binomials. There are a number of different
methods for this. But here, weβre going to use two
different ones. The first method is the FOIL
method. Here, the acronym FOIL represents
First, Outer, Inner, and Last and refers to the position of the terms in the
binomial. The second method is the area or
grid method. And this method of multiplication
is commonly seen for multiplying integers.

So letβs take a look at an example
of multiplying two binomials and see how we would apply both of these methods.

Expand the product two π₯ plus one,
3π₯ minus two.

In this question, the word product
indicates that weβll be multiplying our two binomials. Weβre going to use two different
methods to demonstrate the expansion of these binomials. The first is the FOIL method. And the second is the area or grid
method. So starting with the FOIL method,
this means that we take our first, then the outer, then the inner, and then the last
terms and multiply. Looking at our binomials, the first
terms are the two π₯ and the three π₯. So we multiply those, writing two
π₯ times three π₯.

The outer terms will be our two π₯
and our negative two. So we add two π₯ times negative
two. The inner terms will be our plus
one and our three π₯. Therefore, we add our one times our
three π₯. And finally, the last two terms
will be the plus one and the negative two. So we add plus one times negative
two. We may not always need to write
this line of workings. But it can be helpful to illustrate
our method. And now, we work out the
products. Two π₯ times three π₯ is six π₯
squared. Two π₯ times negative two is
negative or minus four π₯ plus one times three π₯ is equivalent to plus three
π₯. And finally, plus one times
negative two is negative two.

The next thing to do at this stage
of the question is to simplify by collecting any like terms. Here, we have two terms in π₯. So we can write this as six π₯
squared minus π₯ minus two, because our negative four π₯ plus three π₯ is equivalent
to a negative π₯. Letβs see how we would get the same
answer using the grid method. To set up our grid, we put one
binomial horizontally and one binomial vertically. The first binomial will give us two
π₯ and one horizontally. And the second binomial of three π₯
minus two can be split into the terms three π₯ and negative two.

We then calculate the product of
each row and column, starting with three π₯ times two π₯. This will give us six π₯
squared. Next, three π₯ times one is three
π₯. In our second row, negative two
times two π₯ is negative four π₯. And finally, negative two times one
is negative two. We then simplify by adding all four
terms and seeing if we can collect any like terms. We start with six π₯ squared. We then have a plus three π₯ and a
negative four π₯, which simplifies to negative π₯ and then our final term of
negative two. So using either of our methods
would give us the answer of six π₯ squared minus π₯ minus two.

In the next example, weβll see a
problem where the binomials include a number of different variable terms. However, we can use the same
process and the same methods to get an answer for the expansion of these two
binomials.

Use the distributive property to
fully expand two π₯ plus π¦, π₯π¦ minus two π§.

In this question, the phrase βto
use the distributive propertyβ means that we can multiply the sum by multiplying
each addend separately and then adding the products. We could use a number of methods to
expand the brackets. Here, weβre going to look at two of
those. The first method is the FOIL
method, which is an acronym. We can recall that it stands for
the First, the Outer, the Inner, and the Last terms of our binomials.

So with our two binomials set, we
start by multiplying the first two terms, which is two π₯ times π₯π¦. Next, our outer terms will be two
π₯ times negative two π§. Itβs always worth taking extra care
when we have the variable π§, because we donβt want to confuse it with the digit
two. Next, our inner terms will be π¦
multiplied by π₯π¦. And finally, we add on the product
of our last terms, which is π¦ times negative two π§.

We now need to simplify our
products. The coefficient of our first term
is two. π₯ multiplied by π₯ will be π₯
squared. And we have our π¦. So the first term simplifies to two
π₯ squared π¦. The coefficient of our second term
is found by multiplying plus two by negative two, giving us negative four. And we then have π₯ times π§. Our third term simplified will be
π₯π¦ squared since we have π¦ times π¦. And our final term will be negative
two π¦π§. At this stage, we always check if
we can simplify our answer by collecting any like terms. And as there are none here, this
would be our final answer.

As an alternative method, we could
use the area or grid method to expand our binomials. To set up our grid, we split up our
binomials into their component terms with one on the row and one on the column. And it doesnβt matter which way
round we put them. To fill in each cell in our grid,
we multiply the row value by the column value.

So taking our first cell, we have
π₯π¦ times two π₯, which would be two π₯ squared π¦. For our next cell, we multiply π₯π¦
by π¦ to give us π₯π¦ squared. On our next row then, we have
negative two π§ times two π₯, which would give us negative four π₯π§. And our final grid value would be
negative two π¦π§. To find our answer from this grid,
we add our four products, giving us two π₯ squared π¦ minus four π₯π§ plus π₯π¦
squared minus two π¦π§, which is the same as we achieved using the FOIL method.

Letβs now look at a question where
we subtract the expansion of two binomials from another term.

Expand and simplify seven minus
three minus π¦, π¦ plus two.

In this question, we can see we
have two binomials which are multiplied together. Weβre going to expand these
binomials first and then subtract the answer from seven. Weβre going to use the grid method
to multiply these binomials. And notice the negative in front of
them isnβt included.

Setting up our grid, we have three
minus π¦ as our first binomial. Our second binomial can be split
into the terms π¦ and two. And we can write that with or
without the plus sign. It doesnβt matter which way round
do we put our binomials. Filling in our grid, we start with
π¦ times three, which is three π¦. Next, we have π¦ times negative π¦,
which will give us negative π¦ squared. On the next row, we have two times
three, which is six. And the final term is calculated by
two times negative π¦, which is negative two π¦.

To take our answer from the grid
then, we add together the four products weβve just calculated. Starting with our largest exponent
of π¦, we have negative π¦ squared. Next, we notice that we have two
terms in π¦, which we can collect together. So three π¦ plus negative two π¦
will give us plus π¦. And then, we add on our final term
from the grid which is plus six. So to answer the question then of
seven minus three minus π¦, π¦ plus two, we replace what weβve worked out in our
expansion, giving us seven minus minus π¦ squared plus π¦ plus six.

Itβs important to include all of
these in parentheses since we want to subtract all of these terms. It can be helpful to write the next
line where we distribute the negative across all the terms. So we have seven minus minus π¦
squared, which is plus π¦ squared. A negative times plus π¦ will give
us negative π¦. And finally, a negative times plus
six will give us negative six.

To simplify then, we check if there
are any like terms. We can see that we have a seven and
a negative six, which is equivalent to one, giving us a final answer of π¦ squared
minus π¦ plus one. We could write these terms in any
order. But by convention, we usually write
them in decreasing exponent values. Here, we have our π¦ squared first,
then our π¦ term, and then our constant.

In our next example, weβll look at
binomials that have higher exponent values. And weβre going to use the rules of
exponents to help us solve the problem.

Find π΄π΅ given π΄ equals five π₯
cubed minus three π₯ and π΅ equals negative six π₯ squared plus three π₯.

To find π΄π΅ in this question, that
means we need to multiply π΄ and π΅. So π΄π΅ can be found by multiplying
all of five π₯ cubed minus three π₯ by all of negative six π₯ squared plus three
π₯. Since π΄ and π΅ are both binomials,
we can use a method for multiplying or expanding two binomials. So setting up a multiplication or
an area grid, we can split up our binomials into their separate terms, remembering
to include the negatives wherever they appear.

To find the value in each cell, we
take the product of the row and the column value. The coefficient of the term in the
first cell can be found by multiplying negative six and five, which is negative
30. For our π₯ term, we multiply π₯
squared by π₯ cubed. And here, we may need to pause and
remind ourselves of one of the rules of exponents. And that is that π₯ to the power of
π times π₯ to the power of π is equal to π₯ to the power of π plus π. So our calculation π₯ squared times
π₯ cubed is equal to π₯ to the power of five. Therefore, the first term in our
grid is negative 30π₯ to the power of five.

In the next cell of our grid, we
have the coefficient, which will be equal to negative six times negative three,
which will be 18 since two negative values multiplied will always give a positive
value. For the π₯-value then, we have π₯
squared times π₯, which is equivalent to π₯ squared times π₯ to the power of one,
giving us π₯ cubed. And so the whole term will be 18π₯
cubed. In the next row, we have three π₯
times five π₯ cubed, which will give us 15π₯ to the fourth power. The final term in our grid is
calculated by three π₯ times negative three π₯. And thatβs equal to negative nine
π₯ squared.

To find the answer using the grid
then, we add our four products together, giving us negative 30π₯ to the fifth power
plus 15π₯ to the fourth power plus 18π₯ cubed minus nine π₯ squared. As there are no like terms, we
canβt simplify this any further. So this is our final answer for
π΄π΅.

Before we summarize what weβve
learned in this video, it may be useful to note that multiplying two binomials is
part of the method we use for multiplying three or more binomials. Itβs outside the scope of this
video. But as a rough guide, if we were
multiplying three binomials, for example, π₯ plus three times π₯ plus seven times
two π₯ minus five, we would expand two of the binomials, for example, to give us π₯
squared plus 10π₯ plus 21 and then multiply that by two π₯ minus five. Here, we could use the grid method
to find our final answer for this.

So in summary then, in this video,
we learned how to multiply two binomials using two methods, the FOIL method and the
area or grid method. We need to be careful when
multiplying negative values. And finally, we may need to use an
important exponent rule when multiplying binomials. That is that π₯ to the power of π
times π₯ to the power of π is equal to π₯ to the power of π plus π.