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Binomial Theorem - Problem 2

Teacher/Instructor Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

So behind me I have the binomial theorem which is a pretty daunting formula on the surface but we’re going to talk about it and hopefully make a lot more sense of all the chaos behind me. So what this is telling us is how we can multiply x plus y to the n, okay?

So we’re always going to start with x to the n, whatever this first term to the n power, just multiply those all through. We then go down to the next term which is this little group right here. Our power on our x has decreased by 1, so instead of dealing with x to the n we’re now dealing with x to the n minus 1, and we introduce one y to balance out for that missing x. And this coefficient right here is n choose one, if you remember this n factorial over n minus 1 factorial times 1 factorial, so it’s just the choose formula. Going down the row, our degree on x is once again dropped by 1, our degree on y has increased by 1 and now instead of choosing 1 we’re choosing 2, okay?

That power, the thing we’re choosing is always going to match up with our power on our y term. Continuing on, we would do n choose 3x n minus 3y³, just continue on and on and on until we end up with a single x and then no x at all, just the y's, okay?

So it’s a pretty involved formula but when pushed going to serve things all sort of, it’s just a repetitive pattern, you can figure it out. So let’s take a look at one example, let’s get rid of this and let’s look at 2x minus 3 to the fourth. So actually to make our life easier, I’m going to actually change this x to an a so we don’t have 2x going on, two different xs going on, we just have one x we’re referring to.

So what we really have here is 2a we can call x and this 3 we can call y. So using our binomial theorem, what we would end up with is x to the fourth and I’m going to sort of make this substitution so we’ll talk about it with xs and then plus the a and the 2a and the 3 back in to sort of see how this all works out.

So our first term will just be x to the fourth, our second term, our x term degree drops by 1 and our y degree increases by 1 and now we are going to have 4 choose one as our coefficient. Continuing on, our x degree drops by 1, our y degree increases by 1. We now have 4 choose 2. Continue this pattern one more time, 4 choose 3, x, y to the third, and our last term is just going to be plus y to the fourth. 4 choose 4, I could have a 4 choose 4 right here but this is just 4 factorial over 4 factorial, it just cancels to 1.

So this is our binomial; theorem for if we were dealing with just x and y. But we have to go back in now is to go and plug in 2a for x and -3 for y. So doing that replacement, this ends up being 2a to the fourth plus 4 choose 1, 2a to the third times -3, 4 choose 2, 2a -3². And just continuing on plugging in all these values, 4 choose 3, 2a, -3 to the third and lastly plus -3 to the fourth.

Okay, not an easy process but believe it or not, it is actually easier than having to multiply four of these together, okay? I’m not going to go any further from this problem, we know what, how to do these 4 choose 1, 4 choose 2, 4 choose 3, we know how to expand 2a to the fourth, so on so fourth. So this is our general set up with a little bit more expansion, we could actually figure out exactly what it is but we have a good idea of how this is all going to develop.

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