### Video Transcript

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.