### Video Transcript

In this video, weβll learn how to
apply the binomial theorem to distribute expressions of the form π plus π to the
πth power for positive integer values of π. We know that we can distribute
small powers of binomials such as second or third powers by applying a strategic
method like the FOIL or grid method. This process, however, becomes
pretty unpleasant when working with higher powers, for instance, π plus π to the
ninth power. And so our job in this video is to
find a quicker way to do this.

Letβs consider the expression π
plus π cubed to see how this might work. We know that π plus π cubed can
be written as π plus π times π plus π times π plus π. And when we distribute these using
our usual techniques, we get the expression π cubed plus three π squared π plus
three ππ squared plus π cubed. But where does each of these terms
come from? Letβs begin by considering the
first term, π cubed. This term contains no πβs at
all. And so we look back to our earlier
expression, and we say, βWell, how many ways are there for us to achieve this? How many ways are there to choose
zero πβs from our three binomials?β

Well, in fact, we see that thereβs
only one. We have to choose an π from every
single one of our binomials. Formally, though, we say that this
is three choose zero. Thatβs the number of ways of
choosing zero distinct items from a set of three where order doesnβt matter. And we do indeed know that that is
equal to one. So thatβs the coefficient of our
first term. We might even choose to write this
as three choose zero times π cubed times π to the zeroth power since π to the
zeroth power is simply equal to one.

Then what about our second
term? This time, we ask ourselves, βHow
many ways can we choose one π from our three binomials?β And we might do this by hand, but
we can actually use the πCπ notation. And we can say that this is three
choose one. Thatβs the number of ways of
choosing one item from a set of three where order doesnβt matter. Now if weβre choosing one π, then
we must have two πβs. So our coefficient of our second
term is three choose one. And the entire term is three choose
one times π squared times π to the power of one.

And what about our next term? This time, weβre choosing two πβs
from a set of three binomials. And so the coefficient of our term
is now going to be three choose two. Thatβs the number of ways of
achieving this. Then, of course, if weβre choosing
two πβs, we want to choose one π. So the term is three choose two
times π to the power of one times π squared. And at this point, you might want
to think about where that final term has come from. Thatβs the π cubed term.

Did you work that one out? This time, weβre choosing three
πβs and zero πβs. So our coefficient is three choose
three. And then our term is three choose
three times π to the zeroth power times π cubed. And so thatβs where all our terms
have actually come from. So letβs put this back into the
original expression and see if we can find a way to generalize this.

Letβs take the general form of a
binomial raised to the πth power. Thatβs π plus π to the πth power
where π is a positive integer. The first term will contain no πβs
from a total of π groups. And so there will be π πβs. The coefficient then is π choose
zero. Thatβs the number of ways of
choosing zero πβs from π groups. And the term is π choose zero
times π to the πth power times π to the power of zero.

Our next term contains one π, so
the coefficient will be π choose one. Thatβs the number of ways of
choosing one π from π groups. We have one less π. So thatβs π to the power of π
minus one. In our third term, weβre choosing
two πβs from a total of π groups. And weβre going to have one less π
again. So this third term is π choose two
times π to the power of π minus two times π squared. We also know that the final term
will contain π πβs and zero πβs. Now the number of ways of choosing
π πβs from a group of π is π choose π. So this final term is π choose π
times π to the power of zero times π to the πth power.

But what do we do about any term
between the third and the final? Letβs say we want to find the term
that contains π πβs. We know then that there will be π
minus π πβs. The number of ways of choosing π
πβs from a group of π is π choose π. And so we can say that the general
term of the binomial expansion is π choose π times π to the power of π minus π
times π to the power of π. And that is our binomial
expansion.

Notice that π choose zero and π
choose π are equal to one as is π to the zeroth power and π to the zeroth
power. We can actually rewrite this a
little. And we get π to the πth power
plus π choose one times π to the power of π minus one times π all the way up to
π to the πth power. We might notice that the power of
π decreases by one each time and the power of π increases by one each time. The sum of the powers of π and π
also will always add up to π. And this value here will always
match the power of π.

Finally, we note that we can also
write this using Ξ£ notation. Itβs the sum from π equals zero to
π of π choose π times π to the power of π minus π times π to the πth power,
where π choose π or πCπ, written as shown depending on where you are in the
world, is π factorial divided by π factorial times π minus π factorial. Remember, of course, this formula
does only hold for nonnegative integer values of π. So now that we have that
all-important formula, letβs look at how to apply it on a very simple example.

Use the binomial theorem to find
the expansion of one plus π₯ to the fourth power.

Remember, the binomial theorem
allows us to distribute parentheses of the form π plus π to the πth power, where
π is a nonnegative integer. And when we do, we get π to the
πth power plus π choose one π to the power of π minus one π plus π choose two
π to the power of π minus two π squared and so on all the way up to π to the
πth power.

So weβre going to use this formula
to distribute one plus π₯ to the fourth power. And before we do, weβre going to
define π, π, and π in this example. Comparing the expression one plus
π₯ to the fourth power with the general form of a binomial, and we can see weβre
going to let π be equal to one, π be equal to π₯, and π be equal to four. And whilst itβs not instantly
obvious, we should note that when weβre performing this expansion, there will always
be π plus one terms. So here, π is equal to four. So weβre expecting to write out
five terms.

Those terms might eventually cancel
or simplify, but in the first instance, we should be writing five terms. The first term in our expansion is
π to the πth power. So we see that simply one to the
fourth power. Then our next term is four choose
one times one to the power of four minus one times π₯, or one cubed times π₯. In the same way, our third term is
four choose two times one squared times π₯ squared. We notice that weβre decreasing the
power of one each time and weβre increasing the power of π₯. So our next term is four choose
three times one times π₯ cubed. And our very final term is π₯ to
the fourth power.

Weβll evaluate our coefficients by
recalling the formula for π choose π. Itβs π factorial over π factorial
times π minus π factorial. So four choose one is four
factorial over one factorial times four minus one factorial. And thatβs four factorial over one
factorial times three factorial. Now what weβre going to do is
recall that four factorial is four times three times two times one. So we can actually write it as four
times three factorial. One factorial is also one, so we
can simplify this fraction by dividing the numerator and denominator by three
factorial. And weβre left with four divided by
one, which is of course simply four.

In a similar way, the coefficient
of our third term four choose two is four factorial over two factorial times four
minus two factorial. Weβll rewrite four factorial as
four times three times two factorial, but two factorial is just two. And so we see that we simplify our
fraction by dividing the numerator and the denominator by two times two, which is of
course four. And so weβre left with three times
two over one. And thatβs six. We perform a similar process for
four choose three. And we find the fourth term in our
expansion has a coefficient of four.

You might wish to pause the video
now and convince yourself thatβs true by applying the π choose π formula. Finally, we notice that one to the
fourth power, one cubed, and so on, those are all equal to one. And so we can simplify each of our
terms as shown. And therefore, the expansion of one
plus π₯ to the fourth power is one plus four π₯ plus six π₯ squared plus four π₯
cubed plus π₯ to the fourth power.

Now that weβve seen a very simple
application, weβll see how our formula holds for binomials that include negatives
and radical terms.

Expand π₯ minus the square root of
two all cubed.

We have π two-toned expression
here. In other words, we have a binomial
which weβre raising to the third power. Since that power is a nonnegative
integer, we know we can apply the binomial theorem. This says that π plus π to the
πth power where π is a nonnegative integer is equal to π to the πth power plus
π choose one π to the power of π minus one π and so on. And so we begin by comparing our
expression to the general binomial form.

Weβre going to let π be equal to
π₯. Then weβre going to be really
careful with π. A common mistake is to think that
the sign of this term doesnβt matter. In fact, it does. And weβre going to say that since
our general form is π plus π, our value of π must be negative root two. And then π is equal to three. The first term in our expansion is
π to the πth power. So that must be equal to π₯
cubed. Then our second term is π choose
one, so thatβs three choose one, times π to the power of π minus one, so thatβs π₯
to the power of three minus one, itβs π₯ squared, times negative root two times π,
which is negative root two.

Our third term is three choose two
times π₯ times negative root two squared. And since we know we always have π
plus one terms in the first line of our expansion, we know that our next term is
going to be the final term. Itβs the fourth term. That final term is π to the πth
power, so itβs negative root two cubed. Now three choose one is three
factorial over one factorial times three minus one factorial. And that gives us three. Now three choose two is also three,
so weβre going to replace each of our coefficients three choose one and three choose
two with three.

Once weβve done that, all that we
need to do is to evaluate our increasing powers of negative root two. Now thatβs quite
straightforward. With our second term, itβs just
negative root two. So we can rewrite this as negative
three root two π₯ squared. Negative root two squared, though,
is positive two. So our third term becomes three
times two times π₯, which is simply six π₯. Then our final term can be written
as negative root two times negative root two squared. So thatβs negative root two times
two or negative two root two. The expansion then of π₯ minus root
two cubed is π₯ cubed minus three root two π₯ squared plus six π₯ minus two root
two.

Now the great thing about this
formula is that it holds for fractional terms too. Letβs see what that might look
like.

Expand π₯ over four minus one over
π₯ to the fifth power.

This is a binomial. Itβs the sum or difference of two
algebraic terms. And weβre looking to raise it to
the fifth power. Because weβre raising it to a
nonnegative integer power, that means we can use the binomial theorem. This says that π plus π to the
πth power, where π is a nonnegative integer, is π to the πth power plus π
choose one π to the power of π minus one π and so on. Comparing our binomial to the
general form, and we see weβre going to let π be equal to π₯ over four, π be equal
to negative one over π₯, and π is the power, so itβs five.

The first term in our expansion is
π to the πth power. So here, thatβs π₯ over four to the
fifth power. Our next term is then five choose
one times π₯ over four to the fourth power times negative one over π₯. Remember, we take π, and we reduce
the power by one each time. But the power of π increases by
one each time. So our next term is five choose two
times π₯ over four cubed times negative one over π₯ squared. Weβre halfway there. We know that there will be π plus
one, so six terms in our expansion. Letβs write out the remaining
three.

They are five choose three times π₯
over four squared times negative one over π₯ cubed plus five choose four times π₯
over four times negative one over π₯ to the fourth power plus negative one over π₯
to the fifth power. Our job now is to simplify each of
these terms. For our first term, thatβs fairly
straightforward. We know that we can simply apply
the fifth power to both parts of our fraction. So we get π₯ to the fifth power
over four to the fifth power. And thatβs π₯ to the fifth power
over 1024. Five choose one is simply equal to
five. We know that this term is going to
be negative since weβre multiplying by negative one over π₯. And then π₯ over four to the fourth
power is π₯ to the fourth power over 256.

We then see that we can divide
through by one power of π₯. So we get five times π₯ cubed over
256 times one, meaning our second term is negative five π₯ cubed over 256. The coefficient of our third term
is five choose two. Now thatβs equal to 10. This time, weβre going to have a
positive term since negative one over π₯ squared is simply positive one over π₯
squared. Similarly, π₯ over four cubed is π₯
cubed over 64. And now we might notice that we can
divide through by π₯ squared and by two. 64 divided by two is 32. So we get five times π₯ over 32
times one, giving us five π₯ over 32.

Letβs keep going. Five choose three is once again
10. The coefficient of this term is
going to be negative since negative one over π₯ cubed is negative one over π₯
cubed. And we get negative 10 times π₯
squared over 16 times one over π₯ cubed. And once again, we can divide
through by π₯ squared. We can also divide through by two,
giving us five and eight. So we get five times an eighth
times one over π₯, which is five over eight π₯.

Thereβs two more terms to go. Five choose four is five. This time, weβre raising this
negative term to an even power. So weβre going to get a positive
result. And itβs five times π₯ over four
times one over π₯ to the fourth power. And this time, we can also divide
through by π₯. So we get five times a quarter
times one over π₯ cubed. So this term is five over four π₯
cubed. The coefficient of our last term
will be negative since weβre raising our negative value to an odd power. Itβs negative one over π₯ to the
fifth power.

And so weβre finished with our
binomial expansion. π₯ over four minus one over π₯ to
the fifth power is equal to π₯ to the fifth power over 1024 minus five π₯ cubed over
256 plus five π₯ over 32 minus five over eight π₯ plus five over four π₯ cubed minus
one over π₯ to the fifth power.

Weβll now recap the key points from
this lesson. In this video, we learned that we
can expand binomials raised to nonnegative integer powers by using the formula π
plus π to the πth power equals π to the πth power plus π choose one π to the
power of π minus one π plus π choose two π to the power of π minus two π
squared all the way up to π to the πth power. This can be written using Ξ£
notation as the sum from π equals zero to π of π choose π times π to the power
of π minus π times π to the πth power, where π choose π is π factorial over
π factorial times π minus π factorial. We saw that the first line of our
expansion will always contain π plus one terms, though those terms might cancel or
simplify, and that this process works for radicals, negatives, and fractional
terms.