Question Video: Expanding a Binomial Using the Binomial Theorem

Expand ((π‘₯/4) βˆ’ (1/π‘₯))⁡.

04:10

Video Transcript

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

This is a binomial. It’s the sum or difference of two algebraic terms. And we’re looking to raise it to the fifth power. Because we’re raising it to a nonnegative integer power, that means we can use the binomial theorem. This says that π‘Ž plus 𝑏 to the 𝑛th power, where 𝑛 is a nonnegative integer, is π‘Ž to the 𝑛th power plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 and so on. Comparing our binomial to the general form, and we see we’re going to let π‘Ž be equal to π‘₯ over four, 𝑏 be equal to negative one over π‘₯, and 𝑛 is the power, so it’s five.

The first term in our expansion is π‘Ž to the 𝑛th power. So here, that’s 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‘Ž, and we reduce the power by one each time. But the power of 𝑏 increases by one each time. So our next term is five choose two times π‘₯ over four cubed times negative one over π‘₯ squared. We’re halfway there. We know that there will be 𝑛 plus one, so six terms in our expansion. Let’s 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, that’s 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 that’s 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 we’re 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 that’s equal to 10. This time, we’re 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.

Let’s 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‘₯.

There’s two more terms to go. Five choose four is five. This time, we’re raising this negative term to an even power. So we’re going to get a positive result. And it’s 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 we’re raising our negative value to an odd power. It’s negative one over π‘₯ to the fifth power.

And so we’re 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.

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