### Video Transcript

In this video, we will learn how to
multiply the sum of two terms by their difference to get the polynomial known as the
difference of two squares. We will begin by looking at a quick
definition.

The product of two binomials is a
difference of two squares if it is in the form π₯ plus π¦ multiplied by π₯ minus
π¦. If we expand these two brackets,
otherwise known as distributing the parentheses, we get π₯ squared minus π₯π¦ plus
π₯π¦ minus π¦ squared. This can be shown using the FOIL
method. We begin by multiplying the first
terms. π₯ multiplied by π₯ is π₯
squared. We then multiply the outside
terms. π₯ multiplied by negative π¦ is
negative π₯π¦. The inside terms give us positive
π₯π¦. Finally, the last terms, π¦
multiplied by negative π¦, gives us negative π¦ squared. This confirms the expansion
stated.

This simplifies to π₯ squared minus
π¦ squared as the middle two terms cancel. Negative π₯π¦ plus π₯π¦ is equal to
zero. The result π₯ plus π¦ multiplied by
π₯ minus π¦ is equal to π₯ squared minus π¦ squared is called the difference of two
squares. This is a useful result that allows
us to quickly expand expressions that are written in this form. We will now look at some questions
that can be answered using this result.

Expand the product π₯ plus one
multiplied by π₯ minus one.

We might immediately notice that
this expression is the difference of two squares. However, if we didnβt notice this,
we could expand the brackets or distribute the parentheses using the FOIL
method. We begin by multiplying the first
terms. π₯ multiplied by π₯ is π₯
squared. Multiplying the outside terms gives
us negative π₯. Multiplying the inside terms gives
us positive π₯. Finally, multiplying the last terms
gives us negative one. π₯ plus one multiplied by π₯ minus
one is equal to π₯ squared minus π₯ plus π₯ minus one. Our middle two terms cancel as
negative π₯ plus π₯ is zero. So we are left with π₯ squared
minus one. This is the expansion of the
product π₯ plus one multiplied by π₯ minus one.

Had we recognized that the
expression was in the form of the difference of two squares, we can recall the
result that π₯ plus π¦ multiplied by π₯ minus π¦ is equal to π₯ squared minus π¦
squared. The value of π₯ in our expression
is π₯, and the value of π¦ is one. π₯ plus one multiplied by π₯ minus
one is therefore equal to π₯ squared minus one squared. This once again simplifies to π₯
squared minus one.

The next question involves a more
complicated expression.

Use the difference of two squares
identity to expand three π plus seven multiplied by three π minus seven.

We recall that the difference of
two squares identity states that any expression of the form π₯ plus π¦ multiplied by
π₯ minus π¦ is equal to π₯ squared minus π¦ squared. In this question, π₯ is equal to
three π and π¦ is equal to seven. Three π plus seven multiplied by
three π minus seven is therefore equal to three π squared minus seven squared. Three π multiplied by three π is
equal to nine π squared. Seven squared is equal to 49. The expansion of three π plus
seven multiplied by three π minus seven in its simplest form is nine π squared
minus 49.

If this question hadnβt asked us to
use the difference of two squares method, we could have expanded the two parentheses
or brackets in a different way. One way would be to use the grid or
box method. Three π multiplied by three π is
nine π squared. Three π multiplied by seven is
21π. Three π multiplied by negative
seven is negative 21π. Finally, seven multiplied by
negative seven is negative 49. Distributing the parentheses gives
us nine π squared plus 21π minus 21π minus 49. As the 21πβs cancel, weβre left
with 9π squared minus 49.

Our next question has an expression
involving radicals or surds.

Simplify root three plus two
multiplied by root three minus two.

We notice that this expression is
written in the form π₯ plus π¦ multiplied by π₯ minus π¦. Using our difference of two squares
result, this is equal to π₯ squared minus π¦ squared. Our value for π₯ is root three, and
our value for π¦ is two. π₯ squared would therefore be equal
to root three squared. We know that square routing and
squaring are inverse operations. Therefore, π₯ squared is equal to
three. π¦ squared is equal to four as two
multiplied by two is four. The expression root three plus two
multiplied by root three minus two is equal to three minus four. This is equal to negative one.

We will now look at some more
complicated problems involving the difference of two squares.

If π₯ plus π¦ is equal to two and
π₯ minus π¦ equals six, what is the value of π₯ squared minus π¦ squared?

The expression π₯ squared minus π¦
squared is known as the difference of two squares. This factors or factorizes into two
parentheses or brackets: π₯ plus π¦ multiplied by π₯ minus π¦. In this question, weβre told that
π₯ plus π¦ is equal to two and π₯ minus π¦ is equal to six. Multiplying these numbers gives us
an answer of 12. If π₯ plus π¦ is equal to two and
π₯ minus π¦ is equal to six, then π₯ squared π¦ squared is equal to 12. It is the product of these two
numbers.

Use the identity π₯ minus π¦
multiplied by π₯ plus π¦ is equal to π₯ squared minus π¦ squared to evaluate 33
squared minus 31 squared without a calculator.

If we consider the right-hand side
of our identity, our value for π₯ in this question is 33 and π¦ is equal to 31. Substituting these values into the
left-hand side of the identity gives us 33 minus 31 multiplied by 33 plus 31. 33 minus 31 is equal to two. 33 plus 31 is equal to 64. We need to multiply two by 64. This is equal to 128. Therefore, 33 squared minus 31
squared is 128.

The final question that we look at
is an extension of this type.

Using the difference of two
squares, evaluate 21 multiplied by 19 without a calculator.

The difference of two squares
result or identity states that π₯ plus π¦ multiplied by π₯ minus π¦ is equal to π₯
squared minus π¦ squared. In this question, weβre trying to
multiply two numbers, 21 and 19. We can begin by letting π₯ plus π¦
equal 21 and π₯ minus π¦ equal 19. We have a pair of simultaneous
equations. And we can eliminate π¦ by
adding. This gives us two π₯ is equal to
40. Dividing both sides of this
equation by two gives us π₯ equals 20. Substituting this back into the
first equation gives us 20 plus π¦ is equal to 21. Subtracting 20 from both sides of
this equation gives us π¦ equals one.

The calculation 21 multiplied by 19
can be rewritten as 20 plus one multiplied by 20 minus one. Using the difference of two squares
result, this is equal to 20 squared minus one squared. 20 squared is equal to 400, and one
squared equals one. As 400 minus one is equal to 399,
21 multiplied by 19 must also be equal to 399.

We will now summarize the key
points from this video. The difference of two squares
result states that π₯ plus π¦ multiplied by π₯ minus π¦ is equal to π₯ squared minus
π¦ squared. This enables us to quickly expand
the product of two binomials of this type. For example, to work out the
product of four π plus seven π and four π minus seven π, we simply square four
π, square seven π, and find the difference between them. The square of four π is 16π
squared, and the square of seven π is 49π squared. Therefore, four π plus seven π
multiplied by four π minus seven π is 16π squared minus 49π squared.

We can also use this rule to help
us factor or factorize an expression in the form π₯ squared minus π¦ squared into
two parentheses or brackets. The square root of 25π squared is
five π. So this will be the first term in
both our parentheses. The square root of 81π squared is
nine π, so we have five π plus nine π multiplied by five π minus nine π. We can also use this rule for
expressions involving radicals or surds. And as shown in the last couple of
questions, we can use it to evaluate calculations.