# Dividing Polynomials using Synthetic Division - Problem 2

Using synthetic division to divide polynomials. So whenever we have two polynomials we could always look to see if we can use synthetic division if two things occur.

First and foremost our divisor has to be a single power of x, there can’t be any x² or anything like that. So looking at that we only have a single x so we’re good to go on that account. The other restriction is that we can’t have any coefficients on our x term. So looking at this problem, we have a 2, what that means is we can’t actually do synthetic division on this problem. If we wanted to divide it out, we would have to use long division. For the course of this example though, I’m going to pretend this 2 doesn’t exist. So as is, we can’t do synthetic division on this problem. If we change it up a little bit we’ll be able to. So let’s delete this 2, okay?

Now we have a coefficient of one on this x term, we can use our synthetic division. How synthetic division works, take your denominator and look and see what gives you a denominator of 0. So we would need to plug in one. Write that number down and make a bracket. So we have coefficients of the numerator, of your dividend in the same row as the number we just wrote down. So our coefficient of x³ is 1, we don’t have an x² term so we’re going to have to have a place holder, so the coefficient is actually 0, we need to throw it in there just to make sure each degree is accounted for. Continuing down the row we then have 2 and -4.

So how synthetic division works, first we want to drop down our first term, that becomes a 1. We then multiply 1 times 1 and that result goes here and then add, 0 plus 1, 1. Continuing the process, 1 times 1 is 1, add 3, 1 times 3 is 3, leaving us with -1. Okay, so we have a remainder of -1.

Decoding the rest of our quotient, the rest of our answer, this is going to be our constant term, this is 3, this is our x term, this is going to be x, and here is our x². These both have coefficients of 1 so there is no number in front of them.

Taking into account our remainder, we then will add plus -1 over our divisor, x minus 1. So by looking at it, remember we had this 2 here to begin with and we couldn’t do synthetic division, but by changing the problem up a little bit; which you’re not probably able to do on most of your problems, but just to use this is a demo we erased that 2 and went through the synthetic division process.

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