### Video Transcript

Find the third term in the expansion of 10π₯ squared plus two over three π₯ to the fourth power.

Here we have a binomial, a two-term expression, raised to the power of four. And so weβre going to quote the binomial theorem to help us. This says that π plus π to the πth power for positive integer values of π is 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 nasty to work with, so we can alternatively quote the expanded form. We get π to the πth power plus π choose one π to the power of π minus one times 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. Notice how we reduce by one the exponent of π each time and we increase the exponent of π.

Letβs compare our binomial to the general form. We can see we can let π be equal to 10π₯ squared. π is the second part of our binomial, two over three π₯. And π is the exponent; itβs four. Now weβre not going to perform the full expansion of this expression. Weβre simply finding the third term. Thatβs π choose two π to the power of π minus two times π squared. π choose two will be four choose two. π to the power of π minus two is 10π₯ squared to the power of four minus two. And π to the power of two is two over three π₯ all squared.

Next, weβll evaluate four choose two. We recall the formula π choose π is π factorial over π factorial times π minus π factorial. And this means four choose two is four factorial over two factorial times four minus two factorial or four factorial over two factorial times two factorial. We can write four factorial as four times three times two times one. And we can write two factorial as two times one. And then we should be able to see that we can divide both our numerator and our denominator by four. So weβre left with three times two times one over one times one. Thatβs simply six over one, which is six.

We can also simplify the second part of our expression since four minus two is two. So we have six times 10π₯ squared squared times two over three π₯ squared. Distributing the two over by the 10 and the π₯ squared gives us 100π₯ to the fourth power. We know 10 squared is 100 and here we multiply the exponents. Similarly, with two over three π₯, we square the numerator and the denominator. So we get four over nine π₯ squared. Then, we notice we can cancel by a factor of three and by π₯ squared. Then two times 100 times four is 800. So we have 800π₯ squared over three.

And we found the third term in the expansion of 10π₯ squared plus two over three π₯ to the fourth power. Itβs 800 over three π₯ squared.