The Binomial Theorem: Examples

Not only is the binomial expression raised to a power, the variable inside the binomial expression is also raised to a power. If I try to do this expansion completely in my head, I know that I will be more likely (than usual) to mess up the exponents.

So this isn't the time for me to worry about that square on the x inside the binomial expression. Instead, I need to start my answer by plugging the binomial's two terms, along with the exterior power, into the Binomial Theorem.

The first term in the binomial is x², the second term in 3, and the power n for this expansion is 6. So, counting from 0 to 6, the Binomial Theorem gives me these seven terms:

The binomial coefficients (that is, the ₆C_k expressions) can be evaluated by my calculator. I can apply exponent rules to simplify the variable terms. And I can plug the numerical terms into my calculator, too. Doing so results in:

Now I'll multiply the various factors together (again, making heavy use of my calculator). My final result is:

I'll plug "2x", "−5y", and "7" into the Binomial Theorem, counting up from zero to seven to get each term. (And I must be careful not to forget the "minus" sign that goes with the second term in the binomial.)

Then simplifying gives me:

Doing the multiplication in my calculator and simplifying each term, I end up with:

I've already expanded this binomial on the previous page, so let's pull up that first step in the expansion process and count to find the fourth term:

So the fourth term is not the one where I've counted up to 4, but the one where I've counted up just to 3. (Again, this is because, just as with Javascript, the counting starts with 0, not with 1.)

second power, so it has three terms:
     (x + y)² = x² + 2xy + y²

third power, so it has four terms:
     (x + y)³ = x³ + 3x²y + 3xy² + y³

fourth power, so it has five terms:
     (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

The expansion in this exercise, (3x − 2)¹⁰, has power of n = 10, so the expansion will have eleven terms, and the terms will count up, not from 1 to 10 or from 1 to 11, but from 0 to 10.

This is why the fourth term will not the one where I'm using "4" as my counter, but will be the one where I'm using "3".

To find the tenth term, I plug x, 3, and 12 into the Binomial Theorem, using the number 10 − 1 = 9 as my counter:

Since this binomial is to the power 8, there will be nine terms in the expansion, which makes the fifth term the middle one. So I'll plug 4x, −y, and 8 into the Binomial Theorem, using the number 5 − 1 = 4 as my counter.

I know that the first term is of the form aⁿ, because, for whatever n is, the first term is _nC₀ (which always equals 1) times aⁿ times b⁰ (which also equals 1). So 1296x¹² = aⁿ. By the same reasoning, the last term is bⁿ, so 625y⁸ = bⁿ.

And since there are alternating "plus" and "minus" signs, I know from experience that the sign in the middle has to be a "minus". (If all the signs had been "plusses", then the middle sign would have been a "plus" also. But in this case, I'm looking for a binomial in the form (a − b)ⁿ.)

I know that, for any power n, the expansion has n + 1 terms. Since this has 5 terms, this tells me that n = 4. So to find a and b, I only have to take the 4th root of the first and last terms of the expanded polynomial:

Then a = 6x³, b = 5y², there is a "minus" sign in the middle, and: