Hi and welcome. My name is Anthony Varela, and today we're going to divide polynomials. Now we're going to go over two different methods for doing so. The first is polynomial long division and the second is synthetic division. So we're going to start with polynomial long division, which can look very scary at first, but it follows a cycle of steps that repeat, and it's very similar to the process for regular long division. So let's start this out.
The first thing that we're going to do is divide two terms. So we're going to look at the first term in our dividend here, and we're going to divide that by the first term in our divisor. So we're ignoring the minus 2 for now. So when we divide 2x cubed by x, we're just eliminating a factor of x, so now we have 2x squared. Well, 2x squared, we're going to write on top. That's part of our quotient, and we're going to continue to add to our quotient.
So now what we do is we take that 2x squared and we multiply it by our entire divisor, so now we're including the minus 2 there. So when we multiply 2x squared by x minus 2, you can see that we get 2x cubed minus 4x squared. So what do we do with that now? Well, we write that below and we set up a subtraction. Very similar to regular long division. So I'm going to group this in parentheses so I don't mix up my signs. And notice that 2x cubed minus 2x cubed, our difference is 0, so we really just have 3 minus negative 4, that would be 3 plus 4, so 7x squared.
After we subtract, we bring down our next term. So now we take the first term here, 7x squared, and divide that by our first term of our divisor, x. So dividing 7x squared by x, we get 7x. Following our cycle of steps, we write that as part of our quotient, and we multiply that by our entire divisor. So 7x multiplied by x minus 2 gives us 7x squared minus 14x. Following our cycle of steps, we write that down below to set up subtraction.
So when we subtract these two polynomials, once again, our difference here is 0, and negative 11 minus negative 14 is 3, so we have 3x there. Bring down our next term. So let's repeat this cycle again. We take this first term 3x and divide it by x. 3x over x equals 3. That's part of our quotient. We multiply it by x minus 2, so 3 times x minus 2 gives us 3x minus 6. We write that down below and subtract. And notice here, these two polynomials are identical, so our difference is 0.
So we have a remainder of 0 because we don't have any more terms to bring down. We're all done, and whatever we're left up with then is our remainder. So we can go ahead and check that we've done this right if we take our quotient and multiply it by the divisor, we should get this polynomial right here. And in the interest of time, I will let you on your own time go ahead and verify that the distribution all works out. And we could add these two polynomials, and what do you know, we get our original dividend here from our polynomial long division problem.
All right, so long division of polynomials can be a very scary thing. There's lots of variables and hard processes to remember, so we're going to go over synthetic division. It looks different, but we're going to get the same result. So there is what we just worked through, we're going to use the same polynomials here. And setting up synthetic division looks a bit odd at first. It works primarily with the coefficients of our polynomials, which makes it a little bit easier on the eyes. There aren't so many x cubes and x squares. We're dealing just with the coefficients.
So here I am writing down 2, 3, negative 11, and negative 6. Now you should note, for example, let's say we didn't have an x squared term. It jumped right from x cubed to x. We would want to make sure that we write a 0 because we want these to be place holders for our descending degrees. So that's just a little note to keep in mind. Now, we also need to write a number in this small box, and we call that an a value, and it comes from this general x minus a expression. So we can see here because our divisor is x minus 2, our a value is 2.
Something to also watch out for is if we divided by x plus 2, our a value would be negative 2. So we're going to write down a positive 2 because we're dividing by x minus 2, and now here are the steps for synthetic division. First, we just bring down that first coefficient right outside of the box here. Now if that number 2, we multiply it by our a value. So 2 times 2 is 4, and 4 goes inside the box underneath this next coefficient, and then we add vertically so 3 plus 4 is 7. Now these steps of multiplying by a and adding vertically repeat until we have no more numbers to multiply and add.
So we take 7, we multiply that by 2, and write 14 underneath our negative 11. And we add vertically, so we get a positive 3, and we continue to multiply that number 3 by our a value 2, and write it underneath our negative 6. And then we add vertically again, and we get 0. So now it's time to interpret our coefficients. Now these numbers here are the coefficients to our quotient. So we have 2x cubed plus 7x plus 3, and our 0 here is our remainder.
Well, we're going to go over synthetic division one more time, and this time, if you've noticed, I've changed this to a plus 2. So we're going to see how this affects our process and our quotient. So we're going to have the same setup here with our coefficients 2, 3, negative 11, and negative 6. Our a value, however, is a negative 2, so we're going to write that down there. Following the steps of synthetic division, bring down that first coefficient. Then we're going to continue to multiply that bottom number by a and add vertically.
So 2 times negative 2 is negative 4. Adding vertically, we get negative 1. Multiplying that by negative 2, we get positive 2. Add vertically, we have a negative 9. Multiply that by negative 2, and we have a positive 18. And add vertically, so we have a 12. That is a non-zero number, so we have a remainder. So our quotient so far looks like this. 2x squared minus x minus 9, and then we have to tack on our remainder, which I know is positive, so plus-- and what does the remainder look like?
Well, it's a fraction, and this number right here is the numerator and the divisor is the denominator. So my fraction is 12 over x plus 2, and that's my quotient for this polynomial division. So let's review synthetic division and long division of polynomials. Well, first, we looked at long division, which follows a cycle of steps that is very similar to regular long division of numbers, but we also looked at another method called synthetic division, which deals primarily with those coefficients. The important thing then at the end is to interpret those numbers as coefficients in your quotient.
So thanks for watching this tutorial on synthetic division and long division of polynomials. Hope to see you next time.
