### Video Transcript

In this video, we will learn how to
factor an expression by completing the square. We will begin by recalling the form
of a perfect square trinomial. A perfect square trinomial is a
polynomial with three terms which can be represented in the form π squared plus or
minus two ππ plus π squared. They can be factored as shown. π squared plus two ππ plus π
squared is equal to π plus π all squared. And π squared minus two ππ plus
π squared is equal to π minus π all squared. In these trinomials, π and π may
be variables, constants, or products of variables and constants.

Letβs consider the trinomial π₯ to
the fourth power plus four π₯ squared plus four. In order to factor this, we begin
by letting the first and last terms be π squared and π squared, respectively. If π squared is equal to π₯ to the
fourth power, then π is equal to the square root of this, which is equal to π₯
squared. Likewise, since π squared is equal
to four, π is equal to two. And itβs worth noting at this point
that π equals negative π₯ squared and π equals negative two would also have been
solutions to π squared equals π₯ to the fourth power and π squared equals
four. However, since we want π plus π
all squared to be equal to π squared plus two ππ plus π squared, we canβt mix
the positive and negative root solutions. And in fact, we only use the
positive root solutions.

The middle term of our perfect
square trinomial is equal to two ππ. Substituting our expressions for π
and π, we have two multiplied by π₯ squared multiplied by two. This is equal to the middle term of
the initial expression four π₯ squared. The perfect square trinomial π₯ to
the fourth power plus four π₯ squared plus four in its factored form is π₯ squared
plus two all squared. It is worth noting from our
original definition that the expression π₯ to the fourth power minus four π₯ squared
plus four can be factored in the same way. This is equal to π₯ squared minus
two all squared.

In this video, we will use this to
help factor binomials that cannot be factored using other methods. For example, letβs consider the
expression π₯ to the fourth power plus four. In this case, our usual methods of
factoring do not work. However, we can factor this
expression using a method known as completing the square. From the previous trinomial we
factored, we know that when the middle term was four π₯ squared, we were able to
factor this as a perfect square trinomial. As such, if we artificially
introduce a term of four π₯ squared into this polynomial, we would create a perfect
square trinomial. It is this method that helps us
factor many polynomials of the form π squared plus π squared.

If we have two terms that are both
perfect squares, that is, in the form π squared plus π squared, we can create a
perfect square trinomial by completing the square as follows. Firstly, we determine the values of
π and π. In our example, we already know
that π is equal to π₯ squared and π is equal to two. Secondly, we calculate two
ππ. Once again, we already know that in
our example, this is equal to four π₯ squared. Our next step is to add two ππ
and negative two ππ to the expression. In our example, we have π₯ to the
fourth power plus four plus four π₯ squared minus four π₯ squared. Since weβre adding the positive and
negative of the same term, weβre effectively adding zero, which does not change the
polynomial. We can then rewrite the expression
as π₯ to the fourth power plus four π₯ squared plus four minus four π₯ squared.

The fourth and final step of
completing the square is to factor the perfect square trinomial. π₯ to the fourth power plus four π₯
squared plus four factors to π₯ squared plus two all squared. And the entire expression is as
shown. In this example, we can actually go
one stage further by noting that four π₯ squared is also a perfect square. It is equal to two π₯ all
squared. Rewriting our expression as π₯
squared plus two all squared minus two π₯ all squared, we note it is in the form π
squared minus π squared. This is known as the difference of
squares. And we know that any expression of
this type can be factored as π plus π multiplied by π minus π, where the values
of π and π in this example are π₯ squared plus two and two π₯, respectively. Our expression can therefore be
factored to π₯ squared plus two plus two π₯ multiplied by π₯ squared plus two minus
two π₯. Finally, rewriting the order within
the parentheses, we have π₯ to the fourth power plus four is equal to π₯ squared
plus two π₯ plus two multiplied by π₯ squared minus two π₯ plus two. It is worth noting that we can
check this answer by expanding or distributing the parentheses.

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

Factor 625π₯ to the fourth power
plus 64π¦ to the fourth power fully by completing the square.

In this question, we are asked to
factor the expression by completing the square. As such, we need to manipulate the
given binomial to include a perfect square trinomial in the form π squared plus two
ππ plus π squared, as this factors to π plus π all squared. We begin by letting the two terms
in our expression be π squared and π squared, respectively. This means that π is equal to the
square root of 625π₯ to the fourth power. And since the square root of 625 is
25, π is equal to 25π₯ squared. In the same way, π is equal to the
square root of 64π¦ to the fourth power. And this is equal to eight π¦
squared.

Our next step is to find an
expression for two ππ. This is equal to two multiplied by
25π₯ squared multiplied by eight π¦ squared. This is equal to 400π₯ squared π¦
squared. It is this term we need to add to
our expression to create a perfect square trinomial. Since we are adding 400π₯ squared
π¦ squared, we also need to subtract this from the initial expression in order for
the expression to remain the same. We are now in a position where the
first three terms form a perfect square trinomial. And this factors to 25π₯ squared
plus eight π¦ squared all squared. And the entire expression can be
written as shown.

Next, we notice that 400π₯ squared
π¦ squared is a perfect square. It can be written as 20π₯π¦ all
squared. Our expression simplifies to 25π₯
squared plus eight π¦ squared all squared minus 20π₯π¦ all squared. This is written in the form π
squared minus π squared, which is known as the difference of squares. And we know that this can be
factored to π plus π multiplied by π minus π. Our expression can therefore be
written as 25π₯ squared plus eight π¦ squared plus 20π₯π¦ multiplied by 25π₯ squared
plus eight π¦ squared minus 20π₯π¦. Rearranging the order of the terms,
we have that 625π₯ to the fourth power plus 64π¦ to the fourth power is equal to
25π₯ squared plus 20π₯π¦ plus eight π¦ squared multiplied by 25π₯ squared minus
20π₯π¦ plus eight π¦ squared. We could check this answer by
distributing the parentheses, where all terms apart from the first and last will
cancel.

We will now look at one more
example where other factoring methods are required first. But in the resulting expression,
one of the factors can be factored using completing the square.

Factor π₯ to the eighth power minus
16π¦ to the eighth power by completing the square.

In this question, it may not
immediately be obvious how we can solve the problem by completing the square. However, we do notice that our
expression is of the form π squared minus π squared. This is known as the difference of
squares and can be factored as π plus π multiplied by π minus π. Letting π squared equal π₯ to the
eighth power, we know that π is equal to π₯ to the fourth power. And letting π squared equals 16π¦
to the eighth power, we know that π is the square root of this, which is equal to
four π¦ to the fourth power. The original expression can
therefore be rewritten as π₯ to the fourth power all squared minus four π¦ to the
fourth power all squared. And factoring this using the
difference of squares, we have π₯ to the fourth power plus four π¦ to the fourth
power multiplied by π₯ to the fourth power minus four π¦ to the fourth power.

The second part of our expression
is once again written in the form π squared minus π squared. π₯ to the fourth power minus four
π¦ to the fourth power is equal to π₯ squared plus two π¦ squared multiplied by π₯
squared minus two π¦ squared. These two parentheses cannot be
factored any further. So we now need to consider the
expression π₯ to the fourth power plus four π¦ to the fourth power. It is this expression that weβll be
able to factor by completing the square. This is in the form π squared plus
π squared. And we need to manipulate this so
it is in the form of a perfect square trinomial.

We know that any perfect square
trinomial of the form π squared plus two ππ plus π squared can be factored into
the form π plus π all squared. Since π squared is equal to π₯ to
the fourth power, π is equal to π₯ squared. Likewise, since π squared is equal
to four π¦ to the fourth power, π is equal to two π¦ squared. The term two ππ is therefore
equal to two multiplied by π₯ squared multiplied by two π¦ squared. And this is equal to four π₯
squared π¦ squared.

We need to add the positive and
negative of this to the expression in our first set of parentheses to create our
perfect square trinomial. The full expression can be written
as shown. And we can now factor the perfect
square trinomial contained in the first set of parentheses. This is equal to π₯ squared plus
two π¦ squared all squared. And rewriting four π₯ squared π¦
squared as two π₯π¦ all squared, we have π₯ squared plus two π¦ squared all squared
minus two π₯π¦ all squared multiplied by π₯ squared plus two π¦ squared multiplied
by π₯ squared minus two π¦ squared.

We note that the expression within
the brackets is the difference of squares. And using the rule we saw earlier,
this factors to π₯ squared plus two π¦ squared plus two π₯π¦ multiplied by π₯
squared plus two π¦ squared minus two π₯π¦. We can then reorder the terms in
these parentheses as shown, giving us a final answer of π₯ squared plus two π₯π¦
plus two π¦ squared multiplied by π₯ squared minus two π₯π¦ plus two π¦ squared
multiplied by π₯ squared plus two π¦ squared multiplied by π₯ squared minus two π¦
squared. This is the fully factored form of
π₯ to the eighth power minus 16π¦ to the eighth power.

We will now finish this video by
recapping the key points. We saw in this video that a perfect
square trinomial is of the form π squared plus or minus two ππ plus π squared
and can be factored as π plus or minus π all squared. If weβre asked to factor a
polynomial that we cannot factor by other means and the polynomial contains two
terms of the form π squared plus π squared, we can complete the square to form a
perfect square trinomial. The term that must be added to
create this is positive or negative two ππ. It is important to note that both
two ππ and negative two ππ must be included so that we are effectively adding
zero. Whilst we did not see an example of
this type in this video, if possible, we need to factor out the greatest common
factor first, as this reduces the complexity of the polynomial that needs to be
factored.

Often after completing the square,
factoring the resulting expression or parts of the expression using the difference
of squares will be possible. As with our second example, it may
also be the case that we need to do this prior to completing the square. In all cases, once the polynomial
has been factored in one way, it is important to check to see if the resulting
expressions can be factored further.