### Video Transcript

Write the coefficients of the terms that result from the expansion of π₯ plus π¦ to the fourth power.

Here we have a binomial expression. Itβs the sum of two binomial terms. And this expression is being raised to the fourth power. And so, weβre going to use the binomial expansion to find the coefficients of the terms that result from evaluating π₯ plus π¦ to the fourth power. The binomial theorem says that, for positive integer values of π, π plus π to the πth power can be written 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.

Now, this can be quite unpleasant to work with. So, we could alternatively consider its expanded form. Itβs 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 through to π to the πth power.

Notice how the powers of π decrease by one in each term, whereas the powers of π increase by one. Letβs compare π plus π to the πth power to the binomial expansion weβre looking for. Thatβs π₯ plus π¦ to the fourth power. We can see that weβll let π be equal to π₯, π, we can say, is equal to π¦, and the exponent π is equal to four.

Then, using the second form of our binomial expansion, we see that the first term in the expansion of π₯ plus π¦ to the fourth power is π₯ to the fourth power. Then, our second term is four choose one π₯ to the power of four minus one π¦, or four choose one π₯ cubed π¦. Remember, weβre reducing the power of π₯ each time and increasing the power of π¦. Our third term is four choose two π₯ squared π¦ squared. Then, we have four choose three π₯π¦ cubed. And our fourth and final term is π¦ to the fourth power.

We should be able to see quite clearly that the coefficients, the numbers of π₯ to the fourth power and π¦ to the fourth power that we have, are one. The coefficients of our other terms are a little bit trickier, though. Theyβre given by four choose one, four choose two, and four choose three. But, what does this mean? Well, π choose π is π factorial over π factorial times π minus π factorial. And so, four choose one is four factorial over one factorial times four minus one factorial. But of course, we can rewrite our denominator as one factorial times three factorial.

Next, we recall that four factorial is four times three times two times one, and three factorial is three times two times one. We divide our numerator and denominator by three, two, and one, in other words, by three factorial. And we see that four choose one is four divided by one, which is four. And we have our second coefficient, the coefficient the number of π₯ cubed π¦ that we have in our expression, is four. Now, in fact, we donβt need to perform the same process for four choose three. If we substitute π is equal to four and π is equal to three into our formula for π choose π, we can see itβs four factorial over three factorial times one factorial. This is the same as four choose one. And so, four choose three is also four. Thatβs the coefficient of π₯π¦ cubed.

We do, however, need to evaluate four choose two. Itβs four factorial over two factorial times two factorial. Letβs write four factorial as four times three times two factorial. Then, weβll write one of our two factorials as two times one. And we see that we can divide through by two factorial. We can also, however, divide through by two. And we find that four choose two is just two times three divided by one, which is equal to six. And so, we found the expansion of π₯ plus π¦ to the fourth power. Itβs one π₯ to the fourth power plus four π₯ cubed π¦ plus six π₯ squared π¦ squared plus four π₯π¦ cubed plus one π¦ to the fourth power.

And in fact, had we spotted that the coefficients of π₯ and π¦ in our original binomial were equal to one, we might have realized that the coefficients of the terms that result from its expansion are given by π choose π for values of π from zero to π, so zero to four. Either way, we end up with the coefficients being one, four, six, four, and one.