Lesson Video: The Binomial Theorem Mathematics

In this video, we will learn how to expand any binomial expression of the form (π‘Ž + 𝑏)^𝑛.

17:52

Video Transcript

In this video, we’ll learn how to apply the binomial theorem to distribute expressions of the form π‘Ž plus 𝑏 to the 𝑛th power for positive integer values of 𝑛. 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‘Ž plus 𝑏 to the ninth power. And so our job in this video is to find a quicker way to do this.

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

Well, in fact, we see that there’s only one. We have to choose an π‘Ž from every single one of our binomials. Formally, though, we say that this is three choose zero. That’s the number of ways of choosing zero distinct items from a set of three where order doesn’t matter. And we do indeed know that that is equal to one. So that’s the coefficient of our first term. We might even choose to write this as three choose zero times π‘Ž cubed times 𝑏 to the zeroth power since 𝑏 to the zeroth power is simply equal to one.

Then what about our second term? This time, we ask ourselves, β€œHow many ways can we choose one 𝑏 from our three binomials?” And we might do this by hand, but we can actually use the 𝑛Cπ‘Ÿ notation. And we can say that this is three choose one. That’s the number of ways of choosing one item from a set of three where order doesn’t matter. Now if we’re choosing one 𝑏, then we must have two π‘Žβ€™s. So our coefficient of our second term is three choose one. And the entire term is three choose one times π‘Ž squared times 𝑏 to the power of one.

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

Did you work that one out? This time, we’re choosing three 𝑏’s and zero π‘Žβ€™s. So our coefficient is three choose three. And then our term is three choose three times π‘Ž to the zeroth power times 𝑏 cubed. And so that’s where all our terms have actually come from. So let’s put this back into the original expression and see if we can find a way to generalize this.

Let’s take the general form of a binomial raised to the 𝑛th power. That’s π‘Ž plus 𝑏 to the 𝑛th power where 𝑛 is a positive integer. The first term will contain no 𝑏’s from a total of 𝑛 groups. And so there will be 𝑛 π‘Žβ€™s. The coefficient then is 𝑛 choose zero. That’s the number of ways of choosing zero 𝑏’s from 𝑛 groups. And the term is 𝑛 choose zero times π‘Ž to the 𝑛th power times 𝑏 to the power of zero.

Our next term contains one 𝑏, so the coefficient will be 𝑛 choose one. That’s the number of ways of choosing one 𝑏 from 𝑛 groups. We have one less π‘Ž. So that’s π‘Ž to the power of 𝑛 minus one. In our third term, we’re choosing two 𝑏’s from a total of 𝑛 groups. And we’re going to have one less π‘Ž again. So this third term is 𝑛 choose two times π‘Ž to the power of 𝑛 minus two times 𝑏 squared. We also know that the final term will contain 𝑛 𝑏’s and zero π‘Žβ€™s. Now the number of ways of choosing 𝑛 𝑏’s from a group of 𝑛 is 𝑛 choose 𝑛. So this final term is 𝑛 choose 𝑛 times π‘Ž to the power of zero times 𝑏 to the 𝑛th power.

But what do we do about any term between the third and the final? Let’s say we want to find the term that contains π‘Ÿ 𝑏’s. We know then that there will be 𝑛 minus π‘Ÿ π‘Žβ€™s. The number of ways of choosing π‘Ÿ 𝑏’s from a group of 𝑛 is 𝑛 choose π‘Ÿ. And so we can say that the general term of the binomial expansion is 𝑛 choose π‘Ÿ times π‘Ž to the power of 𝑛 minus π‘Ÿ times 𝑏 to the power of π‘Ÿ. And that is our binomial expansion.

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

Finally, we note that we can also write this using Ξ£ notation. It’s the sum from π‘Ÿ equals zero to 𝑛 of 𝑛 choose π‘Ÿ times π‘Ž to the power of 𝑛 minus π‘Ÿ times 𝑏 to the π‘Ÿth power, where 𝑛 choose π‘Ÿ or 𝑛Cπ‘Ÿ, written as shown depending on where you are in the world, is 𝑛 factorial divided by π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. Remember, of course, this formula does only hold for nonnegative integer values of 𝑛. So now that we have that all-important formula, let’s 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‘Ž plus 𝑏 to the 𝑛th power, where 𝑛 is a nonnegative integer. And when we do, we get π‘Ž to the 𝑛th power plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 plus 𝑛 choose two π‘Ž to the power of 𝑛 minus two 𝑏 squared and so on all the way up to 𝑏 to the 𝑛th power.

So we’re going to use this formula to distribute one plus π‘₯ to the fourth power. And before we do, we’re going to define π‘Ž, 𝑏, and 𝑛 in this example. Comparing the expression one plus π‘₯ to the fourth power with the general form of a binomial, and we can see we’re going to let π‘Ž be equal to one, 𝑏 be equal to π‘₯, and 𝑛 be equal to four. And whilst it’s not instantly obvious, we should note that when we’re performing this expansion, there will always be 𝑛 plus one terms. So here, 𝑛 is equal to four. So we’re 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‘Ž to the 𝑛th 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 we’re decreasing the power of one each time and we’re 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.

We’ll evaluate our coefficients by recalling the formula for 𝑛 choose π‘Ÿ. It’s 𝑛 factorial over π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. So four choose one is four factorial over one factorial times four minus one factorial. And that’s four factorial over one factorial times three factorial. Now what we’re 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 we’re 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. We’ll 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 we’re left with three times two over one. And that’s 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 that’s true by applying the 𝑛 choose π‘Ÿ 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 we’ve seen a very simple application, we’ll 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‘Ž two-toned expression here. In other words, we have a binomial which we’re raising to the third power. Since that power is a nonnegative integer, we know we can apply the binomial theorem. This says that π‘Ž plus 𝑏 to the 𝑛th power where 𝑛 is a nonnegative integer is equal to π‘Ž to the 𝑛th power plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 and so on. And so we begin by comparing our expression to the general binomial form.

We’re going to let π‘Ž be equal to π‘₯. Then we’re going to be really careful with 𝑏. A common mistake is to think that the sign of this term doesn’t matter. In fact, it does. And we’re going to say that since our general form is π‘Ž plus 𝑏, our value of 𝑏 must be negative root two. And then 𝑛 is equal to three. The first term in our expansion is π‘Ž to the 𝑛th power. So that must be equal to π‘₯ cubed. Then our second term is 𝑛 choose one, so that’s three choose one, times π‘Ž to the power of 𝑛 minus one, so that’s π‘₯ to the power of three minus one, it’s π‘₯ squared, times negative root two times 𝑏, 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 𝑛 plus one terms in the first line of our expansion, we know that our next term is going to be the final term. It’s the fourth term. That final term is 𝑏 to the 𝑛th power, so it’s 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 we’re going to replace each of our coefficients three choose one and three choose two with three.

Once we’ve done that, all that we need to do is to evaluate our increasing powers of negative root two. Now that’s quite straightforward. With our second term, it’s 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 that’s 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. Let’s see what that might look like.

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.

We’ll 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‘Ž plus 𝑏 to the 𝑛th power equals π‘Ž to the 𝑛th power plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 plus 𝑛 choose two π‘Ž to the power of 𝑛 minus two 𝑏 squared all the way up to 𝑏 to the 𝑛th power. This can be written using Ξ£ notation as the sum from π‘Ÿ equals zero to 𝑛 of 𝑛 choose π‘Ÿ times π‘Ž to the power of 𝑛 minus π‘Ÿ times 𝑏 to the π‘Ÿth power, where 𝑛 choose π‘Ÿ is 𝑛 factorial over π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. We saw that the first line of our expansion will always contain 𝑛 plus one terms, though those terms might cancel or simplify, and that this process works for radicals, negatives, and fractional terms.

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