### Video Transcript

In this video, we will learn how to
expand the square of the difference or sum of two monomials. A monomial is just a single algebraic
term such as π₯, two, or π¦. The difference or sum of two monomials
gives an expression such as π₯ plus two or π minus π, which is referred to as a
binomial. Usually, when we see words with bi- at
the start, there is some link with the number two, for example, bicycles which have two
wheels. So, a binomial is simply an algebraic
expression with two terms.

Squaring a binomial means to multiply a
binomial by itself. So, for example, weβre looking for the
result of π₯ plus two multiplied by π₯ plus two, which we can write as π₯ plus two all
squared. This is an important algebraic skill
because it allows us to manipulate more complex algebraic expressions and will often be
required in problems in other areas of mathematics, such as problems involving area of
two-dimensional shapes. In this video, weβll look at a variety of
different methods for squaring binomials, so you can pick the one that makes the most sense
to you.

So, letβs begin by looking at our first
example.

Expand π plus four squared.

So, we have a binomial, π plus four. And the first really important thing to
notice is that we are squaring this entire binomial. This means that weβre taking the
expression π plus four and multiplying it by itself. So, another way of writing this would be
as π plus four multiplied by π plus four. The word expand is another way of saying
distribute. So, we want to expand the brackets or
distribute the parentheses. We need to multiply both terms in the
first binomial by both terms in the second.

Letβs look at a physical interpretation
of this first using area. Suppose we have a length of π plus four
units. We can break this down into a length of
π units and a length of four units. When we multiply π plus four by itself,
this is equivalent to finding an expression for the area of a square with side lengths of π
plus four units. We can divide this area up into four
smaller regions and find each of their individual areas. Theyβre each either rectangles or
squares, so we find their areas by multiplying their dimensions together.

The first region has an area of π
multiplied by π which is π squared. The second region has an area of π
multiplied by four, which is four π. The third region has an area of four
multiplied by π, which is another lot of four π. And the final region has an area of four
multiplied by four, which is 16. The total area can be found by summing
the four individual areas, giving π squared plus four π plus four π plus 16. Now, we must remember to simplify this
expression by grouping like terms. And the only like terms are those in the
center of our expansion, plus four π plus four π, which makes positive eight π. Our expression, therefore, simplifies to
π squared plus eight π plus 16. And this is the expanded form of π plus
four all squared.

Now, notice that, at this stage here, we
have four terms in our expansion, and the middle two are exactly the same. Once weβve grouped the like terms, we
have three terms in our expansion. This will always be the case when
squaring a binomial. So, if you find you have a different
number of terms or you donβt have two terms which are exactly the same, then something is
gone wrong. So, you need to check your work. So, thatβs one approach to consider this
expansion as the area of a square with side lengths of π plus four units. This is also sometimes referred to as a
grid method because it follows the same principles as the grid method for
multiplication.

A second method we could consider would
be to use the distributive property of multiplication. If weβre multiplying π plus four by π
plus four, then we can write this as π multiplied by π plus four plus four multiplied by
π plus four. So, weβve distributed the terms in our
first binomial over the second. We then have to multiply out or expand or
distribute each of these single sets of parentheses. π multiplied by π gives π squared. π multiplied by four gives four π. We then have four multiplied by π, which
gives another lot of four π, and finally four multiplied by four, which gives 16.

Our expression is now identical to the
one that we had at this stage here in our first method. And the only remaining step is to group
the like terms in the center of our expansion. As before, we find that π plus four all
squared is equal to π squared plus eight π plus 16. So, thatβs two possible methods, but
there is a third. And this one is perhaps the most
popular. Itβs called the FOIL method. And the word FOIL is an acronym, meaning
that each of its letters stands for something. Itβs just a way of ensuring that we
multiply all of the correct pairs of terms together, and we donβt miss any out.

The letter F in the word FOIL stands for
first. So, this means we multiply the first term
in the first binomial by the first term in the second. Thatβs π multiplied by π, which is, of
course, π squared. The letter O stands for outers or
outside. So, we multiply the terms that are on the
outside of our expansion. Thatβs the π in the first binomial and
the four in the second, π multiplied by four, which is four π. Youβve probably guessed that the I stands
for inners or inside, so we multiply together the terms in the inside of the expansion. Thatβs the four in the first binomial and
the π in the second, four multiplied by π, which gives four π. And finally, L, which stands for
lasts. We multiply together the last term in
each binomial. Thatβs the four in the first binomial and
the four in the second, four multiplied by four, which is 16.

As before, we should always have four
terms in our expansion at this point. And we have two like terms in the center
of the expansion, which can be combined. Using all three methods then, weβve
arrived at the same result. When we expand π plus four all squared,
which means to multiply the binomial π plus four by itself, we get the answer π squared
plus eight π plus 16. It is usual, although not essential, to
write the terms in our expansion in this order, that is, decreasing powers of π. So, we have our π squared term first,
then our π term, and then finally the constant term.

Before we move on to looking at other
examples, letβs address a really common mistake. We saw in our first example that the
correct expansion of π plus four all squared is π squared plus eight π plus 16. Unfortunately, the mistake that gets made
so very often is to think that when we square a binomial such as π plus four, we simply
square each term, giving π squared plus four squared. This would, of course, simplify to π
squared plus 16. But as we can see, this is not the same
as the answer weβve already found. Weβre missing that term of positive eight
π.

This is equivalent to forgetting about
two regions of the area in the first method we used. We have the π squared and the 16. But we donβt have the two areas that
correspond to multiplying π by four each time. But this is such a common mistake, even
made by students of advanced mathematics when theyβre not thinking properly. So, make sure that you donβt make it. Another common mistake worth mentioning
would be to think that π multiplied by π gives two π. Remember, though, that we are multiplying
the terms not adding, so the correct answer is π squared.

Letβs now go on and consider some more
examples.

Expand negative π₯ plus two π¦ all
squared.

In this question then, we have a binomial
expression, negative π₯ plus two π¦. And we are squaring it. That means weβre multiplying this
binomial by itself. So, weβre looking for the result of
multiplying negative π₯ plus two π¦ by negative π₯ plus two π¦. There are numerous different methods that
we can use. In this question, Iβm going to choose to
use the FOIL method. Now, we just need to be a little bit
careful because one of the terms in our binomial is negative. And we donβt want to let this trip us
up. We need to be really careful with the
signs when weβre multiplying each pair of terms together.

So, F, remember, stands for firsts. We multiply the first term in each
binomial together. Thatβs negative π₯ multiplied by negative
π₯, which gives π₯ squared. Remember, a negative multiplied by a
negative gives a positive. Then, the letter O stands for outers or
outsides. We multiply the terms on the outside of
our expansion. Thatβs the negative π₯ in the first
binomial by the positive two π¦ in the second, giving negative two π₯π¦. I stands for inners or inside. So, we multiply the terms in the center
of our expansion. Thatβs the two π¦ in the first binomial
and the negative π₯ in the second, giving another lot of negative two π₯π¦. Finally, the letter L stands for lasts,
so we multiply the last term in each binomial together. Thatβs positive two π¦ multiplied by
positive two π¦, which is four π¦ squared.

So, after completing all four of our
multiplications, we now have four terms in our expansion, π₯ squared minus two π₯π¦ minus
two π₯π¦ plus four π¦ squared. Remember, there should always be two
identical terms in the center of our expansion. And indeed, there are. We have negative two π₯π¦ minus another
lot of two π₯π¦. We can therefore simplify our expansion
by grouping like terms, and we have our final answer to the problem. The simplified expansion of negative π₯
plus two π¦ all squared is π₯ squared minus four π₯π¦ plus four π¦ squared.

Great, weβre gaining some confidence in
applying these methods now. Letβs go on and look at another example
which is slightly more complex, but only because the terms in the binomial weβre squaring
are not quite so straightforward.

Expand four π₯ squared plus three over
two π¦ squared all squared.

Now, at first glance, this question looks
relatively complicated. But, in fact, what we have is a
binomial. Thatβs the sum of two algebraic
terms. And then we are squaring it. That means weβre multiplying this
binomial by itself. So, weβre looking for an algebraic
expression for the result of multiplying four π₯ squared plus three over two π¦ squared by
four π₯ squared plus three over two π¦ squared. Although this may look complicated, the
standard processes we follow are exactly the same. Letβs use the distributive method. So, we take the two terms in our first
binomial and we distribute them over the second, giving four π₯ squared multiplied by four
π₯ squared plus three over two π¦ squared plus three over two π¦ squared multiplied by four
π₯ squared plus three over two π¦ squared.

We now have two single parentheses to
distribute. Four π₯ squared multiplied by four π₯
squared gives 16π₯ to the fourth power because four multiplied by four is 16. And π₯ squared multiplied by π₯ squared
is π₯ to the fourth power. Remember, we add the exponents. We then have four π₯ squared multiplied
by three over two π¦ squared. And we can simplify the coefficient here
in the moment. For now, we have four multiplied by three
over two π₯ squared π¦ squared.

Thatβs the first set of parentheses
expanded. Now, letβs consider the second. We have positive three over two π¦
squared multiplied by four π₯ squared. And again, weβll simplify the coefficient
here in a moment. For now, weβll just write it as three
over two multiplied by four π₯ squared π¦ squared. Finally, we have three over two π¦
squared multiplied by three over two π¦ squared, which weβll write as three over two
multiplied by three over two π¦ to the fourth power. Again, remember, we add the
exponents.

Notice that we have four terms in our
expansion at this stage and the two central terms are identical to one another, although the
four and the three over two have been written in the opposite order. Now, we can simplify. 16π₯ to the fourth power requires no
simplification. In our second term, we have four
multiplied by three over two. So, we can cancel a factor of two in the
numerator and denominator to give two multiplied by three over one, which is simply six.

Our second term is therefore six π₯
squared π¦ squared. By exactly the same reasoning, our third
term is also six π₯ squared π¦ squared. And to simplify the coefficient in our
final term, we multiply the numerators together, giving nine, and multiply the denominators,
giving four. So, our final term is nine over four π¦
to the fourth power.

All that remains is to simplify our
expansion by grouping the like terms in the center. Six π₯ squared π¦ squared plus six π₯
squared π¦ squared is 12π₯ squared π¦ squared. So, we have our simplified expansion;
four π₯ squared plus three over two π¦ squared all squared is equal to 16π₯ to the fourth
power plus 12π₯ squared π¦ squared plus nine over four π¦ to the fourth power.

So, whilst that example may have looked a
little more complicated, the processes we use are exactly the same. In our final example, weβll see how to
apply these methods to a question involving problem solving.

If π₯ plus π¦ all squared is equal to 100
and π₯π¦ equals 20, what is the value of π₯ squared plus π¦ squared?

So, weβve been given two pieces of
information about these numbers π₯ and π¦ and asked to use them to determine the value of π₯
squared plus π¦ squared. Now, your first thought may be that π₯
plus π¦ all squared is just equal to π₯ squared plus π¦ squared. In which case, the value weβre looking
for is the value given in the question; itβs 100. But if this is the case, why have we also
been given the value of π₯π¦?

In fact, if we were to answer the
question this way, weβd have made one of the most common mistakes in mathematics because
weβve incorrectly expanded the binomial. Remember that π₯ plus π¦ all squared
means π₯ plus π¦ multiplied by π₯ plus π¦. So, we are, in fact, multiplying a
binomial by itself, not just squaring each of the individual terms.

Letβs see what happens if we correctly
expand π₯ plus π¦ all squared. Using the FOIL method which, remember,
stands for firsts, outers, inners, lasts, this gives π₯ squared plus π₯π¦ plus π₯π¦ plus π¦
squared, which simplifies to π₯ squared plus two π₯π¦ plus π¦ squared. What we now have is an equation
connecting π₯ plus π¦ all squared, whose value we know, π₯π¦, whose value we know, and π₯
squared plus π¦ squared, whose value we wish to calculate.

Substituting 100 for π₯ plus π¦ all
squared and 20 for π₯π¦, we have 100 equals π₯ squared plus π¦ squared plus two multiplied
by 20. That simplifies to 100 equals π₯ squared
plus π¦ squared plus 40. And subtracting 40 from each side, we
find that π₯ squared plus π¦ squared is equal to 60. So, weβve solved the problem. By correctly expanding the binomial π₯
plus π¦ all squared and then substituting the values given in the question, we found that π₯
squared plus π¦ squared is equal to 60.

Letβs now review the key points weβve
seen in this video. One really useful method for squaring
binomials is the FOIL method, where each letter stands for a different pair of terms we need
to multiply: firsts, outers, inners, lasts. The grid or area method can be helpful
for visualizing why we have to multiply each term in the first binomial by each term in the
second. We picture this as finding an expression
for the area of the square whose side lengths are each equal to the binomial weβre
squaring. We will always have four terms in our
initial expansion. The two central terms will be
identical. And we then simplify by collecting like
terms to give an algebraic expression with three terms.

The general form when expanding a
binomial of the form π₯ plus π¦ all squared is π₯ squared plus two π₯π¦ plus π¦ squared. And finally, an incredibly common
misconception which we must do everything in our power to avoid: when we square a binomial,
we donβt simply square the individual terms. π₯ plus π¦ all squared is not equal to π₯
squared plus π¦ squared. Instead, we need to use one of the formal
methods for squaring a binomial such as the FOIL method, the grid or area method, or the
distributive property.